Motion of System of Particles and Rotational Motion - SelfStudys

CENTRE OF MASS

In a system of extended bodies there is one special

point that has some interesting and simple properties

no matter how complicated the system is. This point is

called the center of mass.

For a system of n particles whose position vectors are

1 2 nr r r

, , .........., as shown in the figure, the position

vector for the center of mass cmr

is defined as

1 1 2 2 n ncm

1 2 n

m r m r ........ m rr

m m ........... m

or i icm

m rr

M

O

m1

m2 m3

x

y

1r

2r

3r

where M = mi is the total mass of the system.

The components of the above equation may be written

as

xcm

= i im x

M

; ycm

= i im y

M

; zcm

= i im z

M

The location of the centre of mass is independent of

the reference frame used to locate it. The centre of

mass of the system of particles depends only on the

masses of the particles and the positions of the

particles relative to one another.

A rigid body, such as a meter stick, can be thought of

as a system of closely packed particles. Hence it also

has the centre of mass. The number of particles in the

body is so large and their spacing so small, however,

that we can treat the body as though it has a

continuous distribution of mass. For continuous

distribution of mass, the centre of mass is defined as

cm

1r rdm

M

where r is the position vector of the centre of mass of

a small mass element dm.

The components of this equation are

xcm

= 1

xdmM , y

cm =

1ydm

M

and zcm

= 1

zdmM

Often we deal with homogeneous objects having a

point, a line, or a plane of symmetry. Then the centre

of mass will lie at the point, on the line, or in the plane

of symmetry. For example, the centre of mass of a

homogeneous rod will lie at the centre of the rod, the

centre of mass of a homogenous sphere will be at the

centre of the sphere, the centre of mass of a cone will

be at the axis of the cone, etc.

SOLVED EXAMPLE

Example-1Find the center of mass of the four point masses as

shown in figure.

2

x(m) 4 -2 -4

y(m)

-2

-4

2

4 m2 = 4kg

m3 = 5kg

m1 = 2kg

m4 = 1kg

Motion of System of Particles and

Rotational Motion

Sol. The total mass M = 12 kg, from the component

equations, we have

xcm

= 2kg 3m + 4kg 3m + 5kg -4m + 1kg -3m 5

= -12kg 12 m

ycm

= 2kg -1m + 4kg 3m + 5kg 4m + 1kg -2m

12kg

= 28

12m

The position vector of the centre of mass is

cmr

= –0.42 i + 2.3 j m

Example-2

Two particles of masses 1 kg and 2 kg are located at

x = 0 and x = 3 m. Find the position of their centre of

mass.

Sol.

m =1kg1 COM m =2kg2

x=0 x=x x=3

r =x1 r =(3–x)2

Since, both the particles lie on x-axis, the COM will

also lie on x-axis. Let the COM is located at x = x, then

r1 = distance of COM from the particle of mass 1 kg = x

and r2 = distance of COM from the particle of mass

2 kg = (3 – x)

Using 1 2

2 1

r m

r m or

x 2

3 – x 1 or x = 2 m

thus, the COM of the two particles is located at x = 2m

Example-3The position vector of three particles of masses m

1 = 1 kg,

m2 = 2 kg and m

3 = 3 kg are 1

ˆ ˆ ˆr (i 4j k) m

,

2ˆ ˆ ˆr (i j k) m

and 3

ˆ ˆ ˆr (2i j 2k) m

respectively..

Find the position vector of their center of mass.Sol. The position vector of COM of the three particles will

be given by

1 1 2 2 3 3COM

1 2 3

m r m r m rr

m m m

Substituting the values, we get

COM

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(1) (i 4j k) (2) (i j k) (3)(2i–j–2k)r

1 2 3

1 ˆ ˆ ˆ(3i j k) m2

Ans.

CENTRE OF MASS OF CONTINUOUS MASS SYSTEM(i) Center of Mass of a Uniform Rod

Suppose a rod of mass M and length L is lying alongthe x-axis with its one end at x = 0 and the other at

x = L. Mass per unit length of the rod = M

L

Hence, dm, (the mass of the element dx situated at

x = x is) = M

Ldx

The coordinates of the element dx are (x, 0, 0). Therefore,

x-coordinate of COM of the rod will be

x=Lx=xx=0

dx

xCOM

=

L

0x dm

dm

=

L

0

M(x) dx

LM

= L

0

1 Lx dx

L 2

The y-coordinate of COM is

yCOM

= y dm

dm

= 0

Similarly, zCOM

= 0

i.e., the coordinates of COM of the rod are L

,0, 02

,

i.e. it lies at the center of the rod.

SOLVED EXAMPLE

Example-4Mass is non uniformly distributed over the rod oflength ‘’. Its linear mass density varies linearly with

length as = kx. The position of centre of mass is

given by–

(A) 5

2(B)

3

(C) 4

3(D)

3

2

Sol. (D) dx

x

= kx

Let the cross section = A

mass of element dx at distance x

dm = (kx) Adx

Xcm

=

0

0

Adx)kx(

Adx)kx(x

=

0

0

2

xdx

dxx

=

0

2

0

3

2

x

3

x

= 3

2

Example-5A thin rod of length 3L is bent at right angles at a

distance L from one end. Locate the centre of mass

with respect to the corner. Take L = 1.2 m

Sol. The centre of mass of each arm is at its midpoint. The

centre of mass of the two arms can be found by treating

each arm as a point particle at its centre of mass. From

the diagram we see that x1 = L/2, y

1= 0 and x

2 = 0,

y2 = L. If we take m

1 = m, then m

2 = 2m. From equation,

we have

xc.m.

= 1 1 2 2m x m x L

M 6

yc.m.

= 1 1 2 2m y m y 2L

M 3

The position of the centre of mass is

c.m.r = 0.2 i + 0.8 jm. Note that the centre of mass does

not lie within the body itself.

(ii) Center of mass of a Semicircular Ring

Figure shows the object (semi circular ring). By

observation we can say that the x-coordinate of the

center of mass of the ring is zero as the half ring is

symmetrical about y-axis on both sides of the origin.

Only we are required to find the y-coordinate of the

center of mass.

Y

ycm

d y=Rsin

Rd

X

To find ycm

we use

ycm

=1

M dm y ...(i)

Here for dm we consider an elemental arc of the ring at

an angle from the x-direction of angular width d. If

radius of the ring is R then its y coordinate will be R

sin, here dm is given as

dm = M

R × R d

So from equation .....(i),

we have

ycm

= 1

M0

MRd

R

(R sin) =

R

0

sin d

ycm

= 2R

.....(ii)

(iii) Center of mass of Semicircular Disc

Figure shows the half disc of mass M and radius R.

Here, we are only required to find the y-coordinate of

the center of mass of this disc as center of mass will be

located on its half vertical diameter. Here to find ycm

,

we consider a small elemental ring of mass dm of radius

x on the disc (disc can be considered to be made up

such thin rings of increasing radii) which will be

integrated from 0 to R. Here dm is given as

dm = 2

2M

R( x)dx

Y

ycm

XR

dx

x

Now the y-coordinate of the element is taken as 2x

,

as in previous section, we have derived that the center

of mass of a semi circular ring is concentrated at 2R

Here ycm

is given as

ycm

= 1

M

R

0

dm 2x

=

1

M

R2

20

4 Mx dx

R

ycm

= 4R

3

(iv) Center of mass of a solid Hemisphere

The hemisphere is of mass M and radius R. To find its

center of mass (only y-coordinate), we consider an

element disc of width dy, mass dm at a distance y from

the center of the hemisphere. The radius of this

elemental disc will be given as, r = 2 2R y

The mass dm of this disc can be given as

dm = 3

3M

2 R × r2 dy = 3

3M

2R (R2 – y2) dy

ycm

of the hemisphere is given as

ycm

= 1

M

R

0

dm y = 1

M

R

30

3M

2R (R2 – y2) dy y

=

R2 2

30

3(R y ) y dy

2R

ycm

= 3R

8

(v) Center of mass of a Hollow Hemisphere

A hollow hemisphere of mass M and radius R. Now we

consider an elemental circular strip of angular width

d at an angular distance from the base of the

hemisphere. This strip will have an area.

dS = 2R cos Rd

Its mass dm is given as

dm = 2

M

2 R 2R cos Rd

Here y-coordinate of this strip of mass dm can be taken

as R sin. Now we can obtain the center of mass of the

system as.

ycm

=1

M

2

0

dm R sin

=1

M

22

20

M2 R cos d

2 R

R sin

= R2

0

sin cos d

ycm

= R

2

(vi) Center of mass of a Solid Cone

A solid cone has mass M, height H and base radius R.

Obviously the center of mass of this cone will lie

somewhere on its axis, at a height less than H/2. To

locate the center of mass we consider an elemental

disc of width dy and radius r, at a distance y from the

apex of the cone. Let the mass of this disc be dm,

which can be given as

dm = 2

3M

R H × r2 dy

here ycm

can be given as

ycm

= 1

M

H

0

y dm

= 1

M

2H

20

3M Rydy y

HR H

= 3

3

H

H3

0

y dy = 3H

4 (from verfex)

(vii) Center of mass of some common systems

Rectangular plate (By symmetry)

xc =

b

2; y

c =

L

2

A triangular plate (By qualitative argument)

at the centroid : yc =

h

3

A circular cone (hollow)

yc =

h

3

CENTER OF MASS OF COMBINATION OF MASSES

SOLVED EXAMPLE

Example-6Two circular disc having radius R and mass density and 2 respectively are placed as shown in figure. Thenfind out the position of COM of the system (from point‘O’).

O O'R R

A B

T

Sol. Mass of disc A mA = R2

Mass of disc B mB = 2R2

Due to symmetry the COM of disc A lie at point O andCOM of disc B lie at point O. So we realize the aboveproblem in a following way

CmA mB

x

O O'

2R

Centre of mass due to both the disc lie at point C(assume), having distance x from m

A

B

A B

m (2R)x

m m

;

2

2 2

2 R (2R)x

( R 2 R )

4Rx

3

So the centre of mass lie in the disc B having distance

4R

3from O.

Example-7Find out the position of centre of mass of the figureshown below.

R

BA

Plate

Crectangularplate

2R

2R

2R

Sol. We divide the above problem in two parts

(i) First find out position of centre of mass of bothsemicircular plate and rectangular plate separately.

(ii) Then find the position of centre of mass of givenstructure .

Centre of mass of semicircular disc lie at 4R

3

4R

AB3

Centre of mass of rectangular plate lie at the centre ofplate at point C

BC = R mSC

mR

2

SC

Rm

2

; 2

Rm 4 R

mR

Cr1msc

Let us assume COM is at r1 distance from m

R

2

1 22

R 4R. R

2 3r

R. 4R

2

r1 =

R (3 4)

3( 8)

Ans.

CAVITY PROBLEMS :

If some mass or area is removed from a rigid body thenthe position of centre of mass of the remaining portionis obtained by assuming that in a remaining part +m& – m mass is there. Further steps are explained byfollowing example.

Example-8Find the position of centre of mass of the uniform laminashown in figure. If the mass density of the laminais .

a x

y

Sol. We assume that in remaining portion a disc of radiusa/2 having mass density + is there then we alsoinclude one disc of a/2 radius having – mass density.So now the problem change in following form

+

O O'a a/2

–

A B

a/2

O O' –

So the centre of mass of both disc A & B lie in their

respective centre such as O & O'.

Now

C.O.M. of the lamina A

A B

m a / 2

m m

O C O'mA a/2 mB

mA

= ( a2)m

B= – () (a/2)2

= – 2a

4

Xcom

= 2

22

a .a / 2

aa –

4

= 3

2

a / 2

3a / 4=

3

2

a 4 2a

2 33a

i.e., C.O.M lie on leftward side from point O.

MOTION OF THE CENTRE OF MASS

Now we can discuss the physical importance of the

centre of mass concept. Consider the motion of a

group of particles m1, m

2, ……., m

n and whose total

mass is M which is a constant. From the definition of

the centre of mass, we have

M cmr

= m1 1r

+ m

2 2r

+………………+ m

n nr

Differentiating this equation w.r.t. time, we obtain

cm 1 1 2 2 n nMv m v m v ......... m v

where 1v 1dr

dt

is the velocity of the first particle,

etc., and cmv cmdr

dt

is the velocity of the centre of

mass.

Differentiating the above equation w.r.t. time, we obtain

cm 1 1 2 2 n nMa m a m a ......... m a

where 1a

is the acceleration of the first particle, etc.,

and cma

is the acceleration of the centre of mass. Now,,

from Newton’s second law the force 1F

acting on the

first particle is given by 1 1 1F m a

. Likewise,

2F

= 2 2m a

, etc. We can then write the above equation

as cm 1 2 nMa F F ......... F

Among all these forces the internal forces are exerted

by the particles on each other. However, from Newton’s

third law, these internal forces will occur in equal and

opposite pairs, so that they contribute nothing to the

sum. The right hand sum in the above equation then

represents the sum of only the external forces acting

on all the particles (system). We can then rewrite the

above equation as

cmMa

= externalF

This states that the centre of mass of a system of

particles moves as though all the mass of the system

is concentrated at the centre of mass and all the external

forces were applied at that point.

One important situation is that in which externalF

= 0

then cmv

= constant.

Example-9A dog of mass 10 kg stands on a stationary boat of

mass 40 kg so that he is 20 m from the shore. He then

walks 8 m on the boat towards the shore and halts.

How far is he from the shore now? Assume that there

is no friction between the boat and water.

Sol. In the horizontal direction, there is no force on the

system (dog + boat). Therefore, the centre of mass of

the system does not move in the horizontal direction.

Since the dog moves towards the shore, the boat

moves away from the shore to keep centre of mass

stationary. Let d be the distance by which the boat

moves backwards and let x be the initial distance of

the boat from the shore. The initial x-coordinate of the

centre of mass = 10 20 40.x

10 40

The final x-coordinate of the centre of mass

= 10 20 8 d 40.x

10 40

Equating the two, we get

d = 5

8m

The dog is 20 – 8 + 5

8 = 12.62 m from the shore.

Example-10Two balls with masses m

1 = 3 kg and m

2 = 5 kg have

initial velocities v1 = v

2 = 5 m/s in the directions as

shown in the figure. They collide at the origin.(a) Find the velocity of the center of mass 3 s beforethe collision(b) Find the position of the center of mass 2 s after thecollision

m1 v1

y vcm

m2

v2

x

37o

*

*

*

* *

*

Trajectory of center of mass

Sol. (a) The given time is of no consequence since vcm

isconstant for all times.From equation, in component form

vcmx

= 1 1x 2 2xm v m v

M

o3kg 5cos37 m / s 5kg 0m / s

1.58kg

vcmy

= 1 1y 2 2ym v m v

M

o3kg 5sin37 m / s 5kg 5m / s

28kg

m/s

Thus, cmv

= -1.5 i + 2 j m/s

(b) Since the collision occurs at the origin, the positionof the center of mass 2s later is

c.m.r = cmv

t = -3 i + 4 j m

Example-11A 75 kg man stands at the rear end of a platform ofmass 25 kg and length 4m, which moves initially at 4m/s over a frictionless surface. At t = 0, he walks at 2m/s relative to the platform and then stops at the frontend. During the period of walking, find thedisplacement of(a) the platform,(b) the man(c) the center of mass

Sol. Initially the man, the platform, and the cm have thesame velocity, 4m/s. When he begins to walk forward,his increase in momentum must be compensated by adecrease in the platform’s momentum. Let us say thatthe velocity of the platform relative to the ground whilehe is walking i.e.

PG pˆv = v i

The velocity of man relative to the ground is then

MG MP PGv v v

= (2 + vP) j .

From the conservation of momentum, we have(75 + 25) 4 = 75( 2 + v

p) + 25 v

P

Thus, the velocity of the platform is vP = 2.5 m/s, and

the velocity of the man is vm = 4.5 m/s.

Since the velocity of man relative to platform is 2m/s,

therefore, it takes t = 4

22

m

m/ss for him to walk from

the rear to the front.Displacement of platform is x

p = v

pt = (2.5)(2) = 5m

Displacement of the man is xm = v

mt = (4.5)(2) = 9m

Displacement of the center of mass xcm

= vcm

t = (4)(2)= 8 m

Example-12

A particle of mass m1 = 4 kg moves at 5 i m/s, while

m2 = 2 kg moves at 2 i m/s. Find

(a) the kinetic energy of center of mass(b) the kinetic energy with respect to center of mass.

Sol. The velocity of the centre of mass is obtained as

vcm

= 4kg 5m/s + 2kg 2m/s

= 46kg

m/s

The following figure shows the velocities relative tothe centre of mass

v1 = v

1- v

cm = +1 m/s

v2 = v

2 - v-

cm = -2 m/s

cm

4 m/s 5 m/s 2 m/s

m1

m1

1 m/s

cm -2 m/s m2

m2

(a) (b)

* *

Laboratory Frame Centre of mass Frame

Thus

(a) Kcm

= 1

2(m

1 + m

2)v

cm2 = 48 J

(b) K = 1

22

1 1m v ' + 1

22

2 2m v ' = 6 J

IMPULSE

In the previous chapter, we have learnt the concept of

work which was an integral of force with respect to

displacement. Now we are going to learn another

concept, called impulse.

Impulse is defined as the integral of force with respect

to time.

I

= f

i

t

t

dtF

Since force is a vector and time is a scalar, the result of

the integral in above equation is a vector. If the force

is constant (both in magnitude and direction), it may

be removed from the integral so that the integral is

reduced to

I =

f

i

t

t

F dt = F(tf - t

i) = Ft

Graphically, the impulse is the area between the force

curve and the F = 0 axis, as shown in figure.

The SI unit of impulse is Ns.

If more than one force are acting on a particle, then the

net impulse is given by the time integral of the net

force.

netI

=

f

i

t

t

dt

netF

O t

F

SOLVED EXAMPLE

Example-13

Find the impulse due to the force F

= ˆ â bti j , where

a = 2N and b = 4 N/s, if this force acts from ti = 0 to t

f =

0.3 s.

Sol. I

=

f

i

t

t

dtF =

0.3

0

ˆ âi + btj dt

or I

=

0.3 0.3

0 0

ˆ âi dt + b j tdt

or I

=

0.320.3

00

btˆ ât i + j2

or I

= (2)(0.3) i + 24 0.3

j2

thus I

= (0.6 i + 0.18 j ) Ns

Example-14Figure shows the variation of force acting on a bodywith time. Calculate the impulse of this force.

Sol. Impulse of a force is the area under the graph.

O t

F (N)

500

1 2 3

(ms)

Thus

I = 31 1500 1 500 2 1 500 3 2 10

2 2

or I = [250 + 500 + 250]10-3

or I = 1 Ns

IMPULSE – MOMENTUM THEOREM

In the previous chapter, we have learnt that work done

by a force brings about change in kinetic energy of a

particle. Let us see what physical quantity changes

due to impulse of a force.

According to Newton’s second law, the net force acting

on a particle is equal to the product of mass and

acceleration.

ma netF

Since a =

d

dt

v

, therefore

netF

= mdv

dt

Substituting the value of net force, we get

netI

=

f

i

t

t

dvm dt

dt

ornetI

=

f

i

v

v

mdv

Notice when we change the variable of integrationfrom t to v, we must also change the time limits of theintegral to the corresponding limits of v.For constant mass,

netI

=

f

i

v

v

m dv

ornetI

= f i f im v v mv mv

The quantity

p mv

is a vector and is called the momentum of a particle of

mass m moving with velocity v .

ThusnetI

= f ip p

ornetI

= p

The above equation shows that the net impulse of

forces acting on a particle is equal to the change in

momentum of the particle. This is called the Impulse-

Momentum Theorem.

SOLVED EXAMPLE

Example-15A 2 kg bock is moving at a speed of 6 m/s. How large aforce F is needed to stop the block in a time of 0.5 ms?

Sol. Impulse on block = Change in momentum of blockFt = mv

f - mv

i

F(5 10-4) = 2(0) - (2)(6)or F = -2.4 104 NThe negative sign indicates that the force opposesthe motion.

Example-16

A ball falling with velocity iˆ ˆv = -0.65i - 0.35j

m/s is

subjected to a net impulse I

= (0.6 i + 0.18 j ) Ns. If

the ball has a mass of 275 g, calculate its velocityimmediately following the impulse.

Sol. Using Impulse - Momentum Theorem

mvf - mv

i = I

or f i

Iv v

m

Thus, fv

= -0.65 i - 0.35 j + ˆ ˆ0.6 i + 0.18 j

0.275

or fv

= (-0.65 i - 0.35 j ) + ˆ ˆ2.18i + 0.655j

or fv

= (1.53 i + 0.305 j ) m/s

NEWTON’S SECOND LAW AND THE RATE OF CHANGE

OF MOMENTUM

From the impulse momentum theorem

netI

= p

Using definition of Impulse, we get

f

i

t

net

t

F dt

= p

For constant net force,

f

i

t

net

t

F dt

= p

ornetF

t = p

If the force is not constant, then we can find the average

force as

net avF

= p

t

For short intervals p

t

approximates the

instantaneous force. In the limit as t approaches zero

the fraction p

t

becomes the derivative.

Thus,netF

= dp

dt

The above equation is the momentum statement of

the Newton’s Second Law.

The net force acting on a particle is equal to the time

rate of change of its linear momentum.

For a particle of constant mass the above equation

reduces to the common statement of Newton’s Second

Law.

netF

= dmv

dt

ornetF

= mdv

madt

SOLVED EXAMPLE

Example-17An 150 g ball is thrown at 30 m/s. It is struck by a bat,which gives it a velocity of 40 m/s in the oppositedirection. If the time of contact is 10-2s, what is theaverage force on the ball?

Sol. If we choose the original direction as +x axis, then

p

= f imv mv

= m(–40 i – 30 i )

The average force is

Fav

= –2

ˆΔp –0.15kg×70i ˆ= = –1050i NΔt 10

Notice that this is much larger than the weight (1.5 N)

of the ball.

CONSERVATION OF LINEAR MOMENTUM

When the sum of the forces on an object is zero, the

equation

net

dpF

dt

tells us that the time derivative of momentum is zero.

That is, dp

0dt

This implies p

= constant

Consequently, one can state a conservation law for

momentum:

When the net force on a particle is zero, its momentum

is constant.

The real utility of the momentum conservation concept

comes about when it is applied to a collection of

particles. For a system of particles, the total momentum

is simply the vector sum of the momentum of each of

the particles in the system. That is

P

= N

ii 1

p

Now consider the net force on a system of particles.

There are two kinds of forces:

(i) Internal forces, resulting from the forces between

the particles within the system, and

(ii) External forces, arising from the forces between

the particles in the system and objects outside the

system.

For example, consider a system of two blocks joined

together with a spring. If the system is allowed to fall

freely under gravity, then the gravitational force acting

on each block is the external force and the spring force

acting on each block is internal.

When we calculate the net force on a system of

particles by performing the vector sum, then the

summation of all the internal forces is zero.

intF 0

Thus the momentum statement of the Newton ’s second

law may be written as

netF

= ext int

dF F

dt

P

or ext

dF

dt

P

Now, if the summation of external forces is zero, then

d0

dt

P

Thus, P

= constant

In the absence of a net external force, the momentum

of a system is conserved.

The conservation of momentum law can be used to

relate the initial motion of particles within a system to

the motion of those same particles some time later.

The law emphasizes the equality of momentum before

and after something happens within the system.

Thus, the conservation law is usually written as

initial finalP P

N N

i fi 1 f 1

p p

SOLVED EXAMPLE

Example-18Suppose that a bullet, of mass m = 10 g and speed u, is

fired into a block of mass M = 2 kg suspended as in

figure. The bullet embeds in the block and raises it by

a height H = 5 cm.

H M u m

(a) How can one determine u from H?

(b) What is the thermal energy generated?

Sol. If the collision occurs within a very short time interval,

the ropes will remain essentially vertical as the bullet

comes to rest. Thus, there will be no external horizontal

force. We may apply the conservation of linear

momentum along this direction.

mu = (m + M)V ...(i)

where V is the common velocity after the collision.

Since the collision is completely inelastic part of the

bullet’s initial kinetic energy, 1

2mu2, is converted into

thermal energy. Only the kinetic energy of the bullet-

block system that remains after the collision is available

to raise the system by height H. From the conservation

of mechanical energy we have

1

2(m + M)V2 = (m + M)gH ...(ii)

Hence, V = (2gH)1/2. Substituting this into equation

(i), we find

u = m M 2gH

m

Putting m = 10-2 kg; M = 2 kg; H = 0.05 m; g = 10m/s2

We get u = 2

21 2 10 0.05

10

or u = 201 m/s

(b) The kinetic energies before and after the collision

are

Ki =

1

2mu2 = 202 J and K

f =

1

2(m + M)V2 = 1 J

The change in the kinetic energy as a result of the

collision is –201J. Virtually all the bullet’s kinetic energy

is converted into thermal energy.

COLLISION

Collision is a brief event between objects that contact

each other. The interaction between two or more

objects is called a collision if there exists three

identifiable stages to this interaction: before, during

and after. In the before and after stage the interaction

forces are zero or approaches zero asymptotically.

Between these two states the interaction forces are

large and often the dominating forces governing the

object’s motion. The magnitude of the interacting force

is often unknown. Therefore, the conservation of

momentum statement is useful for relating the initial

velocities before the interaction to the final velocities

after the interaction without requiring a detailed

knowledge of the interaction forces.

Types of Collision

Collisions may be either elastic or inelastic. Linear

momentum is conserved in both cases. A perfectly

elastic collision is defined as one in which the total

kinetic energy of the particles is also conserved.

2 2 2 21 1 2 2 1 1 2 2

1 1 1 1m u m u m v m v

2 2 2 2

In an inelastic collision, the total kinetic energy of the

particle changes. Some of the kinetic energy is stored

as potential energy associated with a change in internal

structure or state, and is not immediately recovered.

Some of the energy may be used to raise the system

(e.g. an atom) to a state with higher energy. Or, it may

be converted into thermal energy of vibrating atoms

and molecules or into light, sound or some other form

of energy.

In a completely inelastic collision, the two bodies

couple or stick together.

Coefficient of Restitution (e)

The elasticity of collision may be measured in terms of

a dimensionless parameter called the coefficient of

restitution (e).

It is defined as the ratio of velocity of separation to

the velocity of approach of the two colliding bodies

e = appraochofvelocity

separationofvelocity

u1 u2

v2 v1

Before

After

Velocity of separation = v2 – v

1

Velocity of approach = u1 – u

2

e = 1 2

1 2

v v

u u

u1 u2

v2 v1

Before

After

Velocity of separation = v1 + v

2

Velocity of approach = u1 + u

2

e = 1 2

1 2

v v

u u

For an elastic collision: e = 1

For an inelastic collision: 0 < e < 1

For completely inelastic collision: e = 0

ELASTIC COLLISION IN ONE DIMENSION

Given figure shows a one - dimensional elastic collision

between two balls.

Applying momentum conservation,

m1

u1

m2

u2

Before

m1

v1 v2

m2

After

Pi = P

f

m1u

1 + m

2u

2 = m

-1v

1 + m

2v

2

or m1(u

1- v

1) = m

2(v

2 – u

2)

Taking e = 1, velocity of separation = velocity of

approach

or v2 – v

1 = u

1- u

2

Let us consider two special cases of elastic collisions:

First, when the particles have equal mass, and second

when one of them, say m2, is initially at rest.

(i) Equal masses: m1 = m

2 = m

The above equations become u1 + u

2 = v

1 + v

2 and

v2 – v

1 = u

1 – u

2

Solving above equations

v1 = u

2and v

2 = u

1

i.e. in case of one dimensional elastic collision of

particles of equal mass, the particles exchange their

velocities.

m

u1

Before After

u2 u2 u1

m m m

(ii) Unequal masses

m1 m

2. Target at Rest: u

2 = 0

In this case, we get

v1 =

1 2 1

1 2

m m u

m m

and v

2 =

1 1

1 2

2m u

m m

(a) When m1 >> m

2, we may ignore the mass of m

2 in

comparison with m1. This leads to

v1 = u

1 and v

2 = 2u

1, which means that m

1 maintains

(approximately) its initial velocity u1 but it imparts

double this value to m2.

u1

m1

u1 2u1

m2 m1 m2

Before After

u2 = 0

(b) When m1 << m

2 we may ignore m

1 in comparison

with m2. We then find that v

1 = u

1 and v

2 = 0. Thus, m

1

reverses its velocity, leaving m2 essentially unmoved

as in figure.

m1

u1 -u1

m2 m2 m1

Before After

INELASTIC COLLISION IN ONE DIMENSION

In case of inelastic collision, after collision the two

bodies move with same velocity (or stick together).

v u1

m1 m2

Before After

u2

If two particles of masses m1 and m

2 moving with

velocities u1 and u

2 (< u

1) respectively along the same

line collide “head-on” and after collision they have a

common velocity v, then by conservation of linear

momentum,

m1u

1 + m

2u

2 = (m

1 + m

2)v

or v = 1 1 2 2

1 2

m u m u

m m

The kinetic energy of the system before collision is

2 2i 1 1 2 2

1 1K m u m u

2 2

and kinetic energy after collision is

2f 1 2

1K m m v

2

Therefore, loss in kinetic energy during the collisionis

Ki – K

f = 2 2 2

1 1 2 2 1 2

1 1 1m u m u m m v

2 2 2

= 21 21 2

1 2

m m1u u

2 m m

SOLVED EXAMPLE

Example-19

Two identical balls are approaching towards each otheron a straight line with velocity 2 m/s and 4 m/srespectively. Find the final velocities, after elasticcollision between them.

mm 4m/s2m/s

Sol. The two velocities will be exchanged and the finalmotion is reverse of initial motion for both.

m m4m/s 2m/s

Example-20

Two elastic bodies P and Q having equal masses are

moving along the same line with velocities of 16 m/s

and 10m/s respectively. Their velocities after the elastic

collision will be in m/s -

(A) 0 and 25

(B) 5 and 20

(C) 10 and 16

(D) 20 and 5

Sol. (C)

Since collision is elastic so from linear momentum

conservation

m × 26 = m (v1 + v

2); v

1 + v

2 = 26 ......(i)

From K.E. conservation

2

1m (162 + 102) =

2

1 m )vv( 2

221 ;

356 = v1

2 + v22 ......(ii)

From equation (i) & (ii)v

1 = 10 m/s

v2 = 16 m/s & option (C) is correct

Example-21

Two solid balls of rubber A and B whose masses are

200gm and 400gm respectively, are moving in mutually

opposite directions. If the velocity A is 0.3 m/s and

both the balls come to rest after collision, then the

velocity of ball B is-

(A) 0.15 ms–1 (B) –0.15 ms–1

(C) 1.5 ms–1 (D) none of theseSol. (B)

From momentum conservationP

1 + P

2 = 0 [Given : m

1 = 200 gm m

2 = 400 gm]

m (0.3) + 2 m.VB = 0

VB = –

2

3.0 = – 0.15 ms–1 so correct options is (B)

Example-22

As shown in figure A, B and C are identical balls B and

C are at rest and, the ball A is moving with velocity v

collides elastically with ball B, then after collision –

A v B C

(A) All the three balls move with velocity v/2

(B) A comes to rest and (B + C) moves with velocity

v 2

(C) A moves with velocity v and (B + C) moves withvelocity v

(D) A and B come to rest and C moves with velocity v

Sol. (D)

From Question figure :

A v B C

It should be follow momentum conservationp

A = p

B’ + p

C’

from option (A), mv = m

2

v

2

v

2

v=

2

3mv

mv 2

3mv so its wrong

Option (B), mv = m × 0 + m

2

v

2

v

mv 2 mv & its wrong

Option (C), mv = mv + 2 mv = 3 mv (wrong)

Option (D), mv = m × 0 + m × 0 + mv mv = mv; or p

i = p

f

So option (D) is correct option

Example-23A sphere of mass m moving with velocity u hits another

stationary sphere of same mass. If e is the coefficient

of restitution, what is the ratio of velocities of two

spheres after the collision?

Sol. By definition of coefficient of restitution

2 1 2 1

1 2

v v v ve

u u u 0

i.e. v2 – v

1 = eu (i)

And by conservation of momentum

mu = mv1 + mv

2

or v2 + v

1 = u (ii)

Solving equations (i) and (ii) for v1 and v

2, we get

v1 = u(1 – e)/2 and v

2 = u(1 + e)/2

so1

2

v 1 e

v 1 e

Example-24A moving particle of mass m makes a head on elasticcollision with a particle of mass 2m which is initially atrest. Show that the colliding particle losses (8/9)th ofits energy after collision.

Sol. Let u be the initial velocity of body of mass m and v1

and v2 be the velocities of bodies of mass m and 2m

respectively after the collision. As in collisionmomentum is always conserved

mu = mv1 + 2mv

2

i.e. v1 + 2v

2 = u or (u – v

1) = 2v

2...(i)

And as the collision in elastic

2 2 21 2

1 1 1mu mv 2m v

2 2 2

i.e. 2 2 21 2v 2v u or 2 2 2

1 2u v 2v ...(ii)

Dividing equation (ii) by (i), we getu + v

1 = v

2

Substituting this value of v2 in equation (i),

u – v1 = 2(u + v

1), i.e., v

1 = – (u/3)

So 2

2 2f 1 i

1 1 u 1 1 1K mv m mu K

2 2 3 2 9 9

[as Ki =

21mu

2]

i f

i

K K 8

K 9

COLLISION IN TWO DIMENSIONS

In two or three dimensions (except for a completely

inelastic collision) the conservation laws alone cannot

tell us the motion of particles after a collision, if we

know the motion before the collision. For example, for

a two-dimensional elastic collision, which is simplest

case, we have four unknowns, namely the two

components of velocity for each of two particles after

collision, but we have only three known relations

between them, one for the conservation of kinetic

energy and a conservation of momentum relation for

each of the two dimensions. Hence we need more

information than just the initial conditions. When we

do not know the actual forces of interaction, as is often

the case, the additional information must be obtained

from experiment. It is simplest to specify the angle of

recoil of one of the colliding particles.

A typical situation is shown in figure. The distance b

between the initial line of motion and a line parallel to

it through the center of the target particle is called the

impact parameter. This is a measure of the directness

of the collision, b = 0, corresponding to a head-on

collision. The direction of motion of the incident particle

m1 after collision makes an angle

1 with the initial

direction, and the target projectile m2, initially at rest,

moves in a direction after collision making an angle 2

with the initial direction of the incident projectile.

m1

u1 m1 m2

Before

b

m2

v1

v2

1

2 x

y

O

After

Applying the conservation of momentum, which is a

vector relation, we obtain two scalar equations; for

the x-component of motion we have

1 1 1 1 1 2 2 2m u m v cos m v cos

and for the y-component

0 = 1 1 1 2 2 2m v sin m v sin

If the collision is an elastic one we have,

2 2 21 1 1 1 2 2

1 1 1m u m v m v

2 2 2

We have four unknowns v1, v

2,

1 and

2 and only

three equations relating them.

We can determine the motion after collision only if we

specify a value for one of these quantities such as 1.

SOLVED EXAMPLE

Example-25A gas molecule having a speed of 300 m/s collideselastically with another molecule of the same masswhich is initially at rest. After the collision the firstmolecule moves at an angle of 30o to its initial direction.Find the speed of each molecule after collision and theangle made with the incident direction by the recoilingtarget molecule.

Sol. This application corresponds exactly to the situationjust discussed, with m

1 = m

2, the equations become

u1 = v

1cos

1 + v

2cos

2(1)

v1sin

1 = v

2sin

2(2)

and 2 2 21 1 2u v v (3)

Squaring and adding the first two equations, we get2 2 21 1 1 1 1 2u v 2u v cos v

Combining this with the third equation, we obtain21 1 1 12v 2u v cos

or (since u1 0) 1 1 1v u cos = 260 m/s

From the third equation, v2 = 150 m/s

and finally from the second equation,

2 = 60o.

The molecules move apart at right angles (1 +

2 =

90o).The student should be able to show that in an elasticcollision between particles of equal mass, one of whichis initially at rest, the recoiling particles always moveoff at right angles to one another.

SYSTEM OF VARIABLE MASSSo far we have dealt with the dynamics of a systemwhose mass is constant. We now discuss the dynamicsof a system, such as a rocket, whose mass varies. Onemay try to apply the following equation in such a case.

extF

= dP dv dM

M vdt dt dt

However, this equation is correct in only a few veryspecial cases. This approach is not correct when massactually enters or leaves the system.

Let us examine the motion of a body of mass M, movingwith velocity v. Another small body of mass M

approaches with velocity u along the same line. Wee

assume u > v

and that after the collision the two

bodies stick together and move at velocity ( v + v

),

as shown in the figure.

Mu

M v

M+M vv

Before After

By defining the system comprising of both the bodies

we can use extF

= dP

dt

, where P

is the total momentum

of the system of constant mass, M + M. The change

in momentum of the system in time t is

P M M v v Mv- Mu M v- u-v M

Now relu v v

is the velocity of M relative to M

before the collision.

We divide both sides of this equation by t and take

the limit as t 0

extF

= rel

dP dv dMM v

dt dt dt

It is convenient to rewrite it in the form

dvM

dt

=

extF

+ relv dM

dt

The last term in the above equation rel

dMv

dt

, is the

rate at which momentum is being transferred into (or

out of) the system by the mass that system has

collected (or ejected). It can be interpreted as the force

exerted on the system by the mass that leaves or joins

it. This force is referred to as the reaction force.

SOLVED EXAMPLE

Example-26A hopper releases grain at a rate dm/dt onto a conveyorbelt that moves at a constant speed v. What is thepower of the motor driving the belt?

Sol.v

Let the system has some arbitrary length of belt whosemass we can call M. This mass of the system increasesat the same rate as that the grain falls,so dM/dt = dm/dt.

Since the grain falls vertically, u = 0 and relv

= u

–

v = v

. Since the speed is constant,

dv/dt = 0. Thus from equation for the system of variablemass

0

= extF

v

dt

dm

dt

dmvFext

where extF

is the force needed to maintain constant

speed because the mass is increasing. The power

required (P = F

. v ) is

dt

dmvP 2

It is interesting to compare this with the rate at whichthe kinetic energy of the grain increases.

dt

dK =

dt

d

2

2

1mv =

dt

dmv 2

2

1

This is only half the power input. The other half isdissipated as heat when the grain lands on the beltand slips relative to it.

Example-27A uniform chain of mass m and length l hangs on athread and touches the surface of a table by its lowerend. Find the force exerted by the chain on the surfacewhen half of its length has fallen on the table. Thefallen part does not form heap.

Sol. Let N be the normal reaction acting on the chain.

It consists of two parts

N = Nwt

+ Nthrust

Nwt

= m

xgl

Nthrust

= vrel

dm

dt

If v be the instantaneous velocity of the particle fallingon the floor, then

vrel

= v – 0 = v

dm dm dx

dt dx dt or

dm mv

dt l

Thus Nth =

2 mv l

Since v = 2gx

Nth =

2mgx

l

Total reaction is

N = mgx 2mgx

l l

N = 3mgx

l

for x = l/2

N = 3

mg2

ROCKET PROPULSION

Consider a rocket of mass M with fuel of mass m.

Their common velocity is v relative to some inertial

frame. When the rocket engines are fired, the gases

are expelled backward with an exhaust velocity

exv

= vex i relative to the rocket. This is a fixed

quantity determined by the design of the engine and

the type of fuel. If the rocket’s velocity changes to

v v relative to the inertial frame, then the velocity

of gas with respect to the frame will be

gas exv v v v

= (vex

+ v + v) i

Applying the law of conservation of momentum, we

get

(M + m) v

= M(v + v) + m(-vex

+ v + v)

After some cancellation we find

0 = Mv + m(-vex

+ v)

v

+ v

v

M + m

M v

ex+ v

+ v

m

If both v and m are small quantities relative to v and

M, respectively, their product , vm, is negligible in

comparison with the other terms, and we are left with

v = vex

m

M

Since an increase in the mass of the expelled gases

corresponds exactly to the loss in mass of the rocket

system, we have m = –M. In the limit as M 0 the

above equation becomes

dv = –vex

dM

M

On integrating both sides,

f f

i i

v M

ex

v M

dMdv v

M

We find

vf – v

i = v

ex ln

i

f

M

M

SOLVED EXAMPLE

Example-28The mass of a rocket is 2.8 106 kg at the launch time.Of this, 2 106 kg is fuel. The exhaust speed is2500 m/s and the fuel is ejected at the rate of 1.4 104

kg/s.(a) Find the thrust of the rocket(b) What is its initial acceleration at launch time?Ignore air resistance.

Sol. (a) The magnitude of the thrust is given by

Thrust = vex

dM

dt = (2500)(1.4 104) = 3.5 107 N

(b) Using equation

ext rel

d dMM

dt dt

vF v

We get the acceleration

a = ext relF vdv dM

= +dt M M dt

dv

dt = – g +

1

Mv

exdM

dt = –9.8 + 12.5 = +2.7 m/s2

ROTATIONAL KINEMATICS

In the previous chapters we have studied the

translatory motion. In this chapter we will study the

rotational motion of a rigid body about a fixed axis. A

rigid body is defined as an object that has fixed size

and shape. In other words, the relative positions of its

constituent particles remain constant.

Actually a rigid body does not exist – it is an useful

idealization. By the term fixed axis, we mean that the

axis must be fixed relative to the body and fixed in

direction relative to an inertial position. In figure (a)

below, translatory motion of a rigid body is shown. In

figure (b) rotation about an axis which is fixed both in

space and direction is shown. In figure (c), combined

translation and rotation is shown.

1 1

2 2

c.m. c.m.

(a)

1

2

1

2

c.m.

(b)

1

2

c.m.

(c)

1

2

c.m.

VARIABLES OF ROTATIONAL MOTION

Consider a body of arbitrary shape rotating about a

fixed axis 'O' as shown in figure. In a given interval all

the particles lying on the line OA move to their

corresponding positions lying on OB.

Although the particles of the body have different

linear displacements, they all have the same angular

displacement , which is given by =s

r

A

B s

r

O

The average angular velocity of the body for a finite

time interval is given by

av = f i

f i

θ - θΔθ=

Δt t - t

The unit of angular velocity is radian per second

(rad/s). The instantaneous angular velocity is defined

as

=Δt 0

Δθ dθLim =

Δt dt

It is the rate of change of the angular position with

respect to time. It is a vector quantity.The direction of angular velocity is given by the right-

hand rule. We hold the right hand such that when the

fingers of the right hand curve in the sense of rotation,

the thumb points in the direction of .

x

y

z

ROTATIONAL MOTION

PERIOD AND FREQUENCY OF REVOLUTION

The period T is the time for one revolution and the

frequency f is the number of revolutions per second

(rev/s). The period and frequency are related as

f = 1

T

If the angular velocity is constant, the instantaneous

and average angular velocities are equal. In terms of

period T and frequency f the angular velocity is given

by = 2π

= 2π fT

The relation between linear speed and angular speed

is obtained by differentiating equation

= rs

with respect to time. dθ 1 ds

=dt r dt

or = rv

or v = r

Although all particles have the same angular velocity,

their speeds increase linearly with distance from the

axis of rotation.

The average angular acceleration is defined as

av

= ω

t

and the instantaneous angular acceleration is defined

as

= 0t

dLim

t dt

Angular acceleration is a vector quantity measured in

rad/s2.

Rotation with Constant Angular Acceleration

When the angular acceleration is constant, we can

find the change in angular velocity by integrating

equation = dω

dt

d = dt

or o

ω t

ω 0

dω = αdt

or – o = at or oω = ω + αt

Substituting the value of in equation = dθ

dt,

we get

d = dt = (o + t)dt

On integrating o

θ t

o

θ 0

dθ = ω + αt dt

– o =

ot +

21αt

2

or2

o o

1θ = θ + ω t + αt

2

On eliminating t from the above equations, we get

2 2o oω = ω + 2α θ -θ

TOTAL ACCELERATION IN ROTATIONAL MOTION

A particle moving in a circular path with speed v has a

centripetal (or radial) acceleration at = 2

2v= ω r

r

ar

at

If there is angular acceleration, the speed of the particle

changes and thus we can find the tangential

acceleration

at =

dv dω= r = αr

dt dt

The net linear acceleration is

tr aaa

The magnitude of the net acceleration is given by

a = 22tr aa

Analogy Between Rotational Kinematics and Linear Kinematics

Quantity Linear Rotational 1. Position x 2. Displacement x 3. Velocity v =

dtdx

dtd

4. Acceleration

a = dtdv

dtd

v = cadt = cdt

x = 'cvdt 'cdt

5. Equations of Kinematics for constant linear acceleration / angular acceleration

v = vo + at t

x = xo + vot + 2

21

at t21t

v2 vo2 = 2ax 2

SOLVED EXAMPLE

Example-29A disc starts rotating with constant angular

acceleration of p rad/s2 about a fixed axis perpendicular

to its plane and through its centre at t = 0.

(a) Find the angular velocity of the disc at t = 4 s.

(b) Find the angular displacement of the disc in 4 s

and

(c) Find number of turns accomplished by the disc in

4 s.

Sol. Here a = p rad/sec2

0 = 0

t = 4 sec

(a) = 0 + ( rad/s2) 4 sec = 4 rad/s.

(b) = 0 + 2

1( rad/s2) (16 s2) = 8 rad.

(c) Let the number of turns be n

n (2 ) rad = 8 rad n = 4

Example-30A wheel rotates with an angular acceleration given by

= 4at3 – 3bt2 , where t is the time and a and b are

constants. If the wheel has initial angular speed 0,

write the equations for the:

(i) angular speed

(ii) angular displacement.Sol. (i) We know

d = dt

Integrating both sides, we get

340

23

btat + =

dt )3bt(4at

ωω

= = t

or

dtdt

o 00

(ii) We know

d = dt

On integrating both the sides, we get

4

bt-

5

at+t =

dt)bt-at+(

45

0

340

ωθ

ω

=

or

dtdt

o

t

00

SOLVED EXAMPLE

Example-31A flywheel of radius 20 cm starts from rest, and has a

constant angular acceleration of 60 rad/s2. Find

(a) the magnitude of the net linear acceleration of a

point on the rim after 0.15 s

(b) the number of revolutions completed in 0.25 s

Sol. (a) The tangential acceleration is constant and given

by

at = r = (60 rad/s2)(0.2 m) = 12 m/s2

In order to calculate the radial acceleration we first

need to find the angular velocity at the given time. We

have

= o + t = 0 + (60 rad/s2)(0.15 s) = 9 rad/s

ar = 2r = (81 rad2/s2)(0.2 m) = 16.2 m/s2

The magnitude of the net linear acceleration is

a = 2.2022 tr aa m/s2

(b) Finally

= radssradt 88.125.0/602

1

2

1 222

This corresponds to (1.88 rad) (1 rev/2 rad) = 0.3 rev

MOMENT OF INERTIA

Like the centre of mass, the moment of inertia is a

property of an object that is related to its mass

distribution. The moment of inertia (denoted by I) is an

important quantity in the study of system of particles

that are rotating. The role of the moment of inertia in

the study of rotational motion is analogous to that of

mass in the study of linear motion. Moment of inertia

gives a measurement of the resistance of a body to a

change in its rotaional motion.

If a body is at rest, the larger the moment of inertia of a

body the more difficuilt it is to put that body into

rotational motion. Similarly, the larger the moment of

inertia of a body, the more difficult to stop its rotational

motion. The moment of inertia is calculated about some

axis (usually the rotational axis).

Moment of inertia depends on :

(i) density of the material of body

(ii) shape & size of body

(iii) axis of rotation

In totality we can say that it depends upon distribution

of mass relative to axis of rotation.

Note

Moment of inertia does not change if the mass :

(i) is shifted parallel to the axis of the rotation

(ii) is rotated with constant radius about axis of

rotation

Moment of Inertia of a Single Particle

For a very simple case the moment of inertia of a single

particle about an axis is given by,

r

I = mr2 ...(i)

Here, m is the mass of the particle and r its distance

from the axis under consideration.

MOMENT OF INERTIA OF A SYSTEM OF PARTICLES

The moment of inertia of a system of particles about an

axis is given by,

I = 2i i

i

m r ...(ii)

m1

m2

m3r3

r2

r1

where ri is the perpendicular distance from the axis to

the ith particle, which has a mass mi.

SOLVED EXAMPLE

Example-32Two heavy particles having masses m

1 & m

2 are situated

in a plane perpendicular to line AB at a distance of r1

and r2 respectively.

r1 r2

A

B

m1 m2

C

D

E F

(i) What is the moment of inertia of the system about

axis AB?

(ii) What is the moment of inertia of the system about

an axis passing through m1 and perpendicular to

the line joining m1 and m

2 ?

(iii) What is the moment of inertia of the system about

an axis passing through m1 and m

2?

Sol. (i) Moment of inertia of particle on left is I1 = m

1r

12.

Moment of Inertia of particle on right is I2 = m

2r

22.

Moment of Inertia of the system about AB is

I = I1+ I

2 = m

1r

12 + m

2r

22

(ii) Moment of inertia of particle on left is I1 = 0

Moment of Inertia of the system about CD is

I = I1 + I

2 = 0 + m

2(r

1 + r

2)2

(iii) Moment of inertia of particle on left is I1 = 0

Moment of inertia of particle on right is I2 = 0

Moment of Inertia of the system about EF is

I = I1 + I

2 = 0 + 0

Example-33Three light rods, each of length 2, are joined together

to form a triangle. Three particles A, B, C of masses m,

2m, 3m are fixed to the vertices of the triangle. Find the

moment of inertia of the resulting body about

(a) an axis through A perpendicular to the plane ABC,

(b) an axis passing through A and the midpoint of BC.

Sol.

B

AX

Y

2l

C3m2m

m

2l

(a) B is at a distant 2 from the axis XY so the moment

of inertia of B (IB) about XY is 2 m (2)2 + 3m (2)2

= 20 m2

The moment of inertia of the body about X' Y' is

m(0)2 + 2m()2 + 3m()2 = 5 m2

C3m2m

A m

X'

B

Y'

Example-34Four particles each of mass m are kept at the four corners

of a square of edge a. Find the moment of inertia of the

system about a line perpendicular to the plane of the

square and passing through the centre of the square.

Sol. The perpendicular distance of every particle from the

given line is a / 2 . The moment of inertia of one

particle is, therefore, 2m(a / 2) =21

ma2

. The moment

of inertia of the system is, therefore, 214 ma

2 = 2 ma2.

a/

2

mm

mm

MOMENT OF INERTIA OF RIGID BODIES

For a continuous mass distribution such as found in a

rigid body, we replace the summation of 2

i ii

I m r by

an integral. If the system is divided into infinitesimal

element of mass dm and if r is the distance from a mass

element to the axis of rotation, the moment of inertia is,

I = 2r dm

r

(A) Uniform rod about a perpendicular bisector

Consider a uniform rod of mass M and length l figure

and suppose the moment of inertia is to be calculated

about the bisector AB. Take the origin at the middle

point O of the rod. Consider the element of the rod

between a distance x and x + dx from the origin. As the

rod is uniform,

Mass per unit length of the rod = M/l

so that the mass of the element = (M/l)dx.

x dx

B

A

0

The perpendicular distance of the element from the line

AB is x. The moment of inertia of this element about AB is

2MdI dx x

l.

When x = – l/2, the element is at the left end of the rod.

As x is changed from – l/2 to l/2, the elements cover the

whole rod.

Thus, the moment of inertia of the entire rod about AB is

/ 2/ 2 32

/ 2 / 2

M M x MI x dx

3 12

ll

l –l

l

l l

2

(B) Moment of inertia of a rectangular plate about a

line parallel to an edge and passing through the

centre

The situation is shown in figure. Draw a line parallel to

AB at a distance x from it and another at a distance

x + dx. We can take the strip enclosed between the two

lines as the small element.x

dx

b

B

A

lIt is “small” because the perpendiculars from different

points of the strip to AB differ by not more than dx. As

the plate is uniform,

its mass per unit area = M

bl

Mass of the strip = M M

b dx dxb

l l

.

The perpendicular distance of the strip from AB = x.

The moment of inertia of the strip about

AB = dI = 2M

dx xl

. The moment of inertia of the given

plate is, therefore,

/ 22

/ 2

M MI x dx

12

l

l

l

l

2

The moment of inertia of the plate about the line parallel

to the other edge and passing through the centre may

be obtained from the above formula by replacing l by b

and thus,

2MbI

12 .

(C) Moment of inertia of a circular ring about its axis

(the line perpendicular to the plane of the ring

through its centre)

Suppose the radius of the ring is R and its mass is M.

As all the elements of the ring are at the same

perpendicular distance R from the axis, the moment of

inertia of the ring is

2 2 2 2I r dm R dm R dm MR .

(D) Moment of inertia of a uniform circular plate about

its axis

Let the mass of the plate be M and its radius R. The

centre is at O and the axis OX is perpendicular to the

plane of the plate.

R

x0

X

dx

Draw two concentric circles of radii x and x + dx, both

centred at O and consider the area of the plate in

between the two circles.

This part of the plate may be considered to be a circular

ring of radius x. As the periphery of the ring is 2 x and

its width is dx, the area of this elementary ring is 2xdx.

The area of the plate is R2. As the plate is uniform,

Its mass per unit area = 2

M

R

Mass of the ring 2 2

M 2 M x dx2 x dx

R R

Using the result obtained above for a circular ring, the

moment of inertia of the elementary ring about OX is

22

2Mx dxdI x

R

.

The moment of inertia of the plate about OX is

R 23

20

2 M MRI x dx

2R .

(E) Moment of inertia of a hollow cylinder about its

axis

Suppose the radius of the cylinder is R and its mass is

M. As every element of this cylinder is at the same

perpendicular distance R from the axis, the moment of

inertia of the hollow cylinder about its axis is

2 2 2I r dm R dm MR

(F) Moment of inertia of a uniform solid cylinder about

its axis

Let the mass of the cylinder be M and its radius R.

Draw two cylindrical surface of radii x and x + dx coaxial

with the given cylinder. Consider the part of the cylinder

in between the two surface. This part of the cylinder

may be considered to be a hollow cylinder of radius x.

The area of crosssection of the wall of this hollow

cylinder is 2 x dx. If the length of the cylinder is l, the

volume of the material of this elementary hollow cylinder

is 2 x dxl.

The volume of the solid cylinder is R2 l and it is

uniform, hence its mass per unit volume is

dx

x

2

M

R

l

The mass of the hollow cylinder considered is

2 2

M 2M2 x dx x dx

R R

l

l.

As its radius is x, its moment of inertia about the given

axis is

22

2MdI x dx x

R

.

The moment of inertia of the solid cylinder is, therefore,

R 23

20

2M MRI x dx

2R .

Note that the formula does not depend on the length of

the cylinder.

(G) Moment of inertia of a uniform hollow sphere about

a diameter

Let M and R be the mass and the radius of the sphere,

O its centre and OX the given axis (figure). The mass is

spread over the surface of the sphere and the inside is

hollow.

Let us consider a radius OA of the sphere at an angle with the axis OX and rotate this radius about OX. The

point A traces a circle on the sphere. Now change to

+ d and get another circle of somewhat larger radius

on the sphere. The part of the sphere between these

two circles, shown in the figure, forms a ring of radius

R sin. The width of this ring is Rd and its periphery

is 2R sin.

Hence, the area of the ring = (2R sin) (Rd).

Mass per unit area of the sphere 2

M

4 R

.

The mass of the ring

2

M M(2 R sin )(Rd ) sin d .

24 R

R sinA

Rd

R

d

x

0

The moment of inertia of this elemental ring about OX is

2Md I sin d . (R sin )

2

2 3MR sin d

2

As increases from 0 to , the elemental rings cover

the whole spherical surface. The moment of inertia of

the hollow sphere is, therefore,

22 3 2

0 0

M MRI R sin d (1 cos )sin d

2 2

22

0

MR(1 cos )d (cos )

2

2 32

0

MR cos 2cos MR

2 3 3

(H) Moment of inertia of a uniform solid sphere about

a diameter

Let M and R be the mass and radius of the given solid

sphere. Let O be centre and OX the given axis. Draw

two spheres of radii x and x + dx concentric with the

given solid sphere. The thin spherical shell trapped

between these spheres may be treated as a hollow

sphere of radius x.

The mass per unit volume of the solid sphere

x

dx

0

x

= 3

3

M 3M4 4 RR3

The thin hollow sphere considered above has a surface

area 4x2 and thickness dx. Its volume is 4 x2 dx and

hence its mass is

= 23

3M(4 x dx)

4 R

= 23

3Mx dx

R

Its moment of inertia about the diameter OX is, therefore,

dl = 23

2 3Mx dx

3 R

x2 = 4

3

2Mx dx

R

If x = 0, the shell is formed at the centre of the solid

sphere. As x increases from 0 to R, the shells cover the

whole solid sphere.

The moment of inertia of the solid sphere about OX is,

therefore,

I =

R4

30

2Mx dx

R = 22

MR5

.

THEOREMS OF MOMENT OF INERTIA

There are two important theorems on moment of inertia,

which, in some cases enable the moment of inertia of a

body to be determined about an axis, if its moment of

inertia about some other axis is known. Let us now

discuss both of them.

Theorem of parallel axes

A very useful theorem, called the parallel axes theorem

relates the moment of inertia of a rigid body about two

parallel axes, one of which passes through the centre

of mass.

COM

r

Two such axes are shown in figure for a body of mass

M. If r is the distance between the axes and ICOM

and I

are the respective moments of inertia about them, these

moments are related by,

I = ICOM

+ Mr2

Theorem of parallel axis is applicable for any type of

rigid body whether it is a two dimensional or three

dimensional

SOLVED EXAMPLE

Example-35Three rods each of mass m and length l are joined

together to form an equilateral triangle as shown in

figure. Find the moment of inertia of the system about

an axis passing through its centre of mass and

perpendicular to the plane of triangle.

A

B C

COM

Sol. Moment of inertia of rod BC about an axis perpendicular

to plane of triangle ABC and passing through the mid-

point of rod BC (i.e., D) is

I1 =

2m

12

l

From theorem of parallel axes, moment of inertia of this

rod about the asked axis is

I2 = I

1 + mr2

tan 30° = r

BD =

r

L/ 2

r = L

2 3

=

22 2m mm

12 62 3

l l l

A

B C30°

r

COM

D

Moment of inertia of all the three rod is2 2

2

m mI 3I 3

6 2

l l

Example-36Find the moment of inertia of a solid sphere of mass Mand radius R about an axis XX shown in figure.

x

x

Sol.

x

x

COM

r=R

From theorem of parallel axis,I

XX = I

COM + Mr2

= 2 22

MR MR5

= 27

MR5

Example-37Consider a uniform thin wire of mass m and length 2l

with two particles of mass m each at its ends. Let AB be

a line perpendicular to the length of the rod passing

through its centre. Find the moment of inertia of the

system about AB.

Sol.

A

I I

m m

B

IAB

= Irod

+ Iboth particles

2

2m(2 )2(m )

12

ll

27m

3 l Ans.

THEOREM OF PERPENDICULAR AXES

The theorem states that the moment of inertia of a

plane lamina about an axis perpendicular to the plane

of the lamina is equal to the sum of the moments of

inertia of the lamina about two axes perpendicular to

each other, in its own plane and intersecting each other,

at the point where the perpendicular axis passes

through it.

Let x and y axes be chosen in the plane of the body

and z-axis perpendicular, to this plane, three axes being

mutually perpendicular, then the theorem states that.z

y

xO

xi

yiri

P

Iz = I

x + I

y

Important point in perpendicular axis theorem

(i) This theorem is applicable only for the plane bodies

(two dimensional).

(ii) In theorem of perpendicular axes, all the three axes

(x, y and z) intersect each other and this point may

be any point on the plane of the body (it may even

lie outside the body).

(iii) Intersection point may or may not be the centre of

mass of the body.

SOLVED EXAMPLE

Example-38Find the moment of Inertia of a cuboid along the axis as

shown in the figure.

I

b

c

a

Sol. After compressing the cuboid parallel to the axis

I = 2 2M(a b )

12

Example-39Find the moment of inertia of uniform ring of mass M

and radius R about a diameter.

Z

0

A

DC

B

Sol. Let AB and CD be two mutually perpendiculardiameters of the ring. Take them as X and Y-axes andthe line perpendicular to the plane of the ring throughthe centre as the Z-axis. The moment of inertia of thering about the Z-axis is I = MR2. As the ring is uniform,all of its diameter equivalent and so I

x = I

y, From

perpendicular axes theorem,

Iz = I

x + I

y Hence I

x = zI

2 =

2MR

2

Similarly, the moment of inertia of a uniform disc abouta diameter is MR2/4

Example-40

Two uniform identical rods each of mass M and length

are joined perpendicularly to form a cross as shown

in figure. Find the moment of inertia of the cross about

a bisector as shown dotted in the figure.

Sol. Consider the line perpendicular to the plane of the figure

through the centre of the cross. The moment of inertia

of each rod about this line is 2M

12

and hence the

moment of inertia of the cross is 2M

6

. The moment of

inertia of the cross about the two bisector are equal by

symmetry and according to the theorem of

perpendicular axes, the moment of inertia of the cross

about the bisector is 2M

12

.

Example-41In the figure shown find moment of inertia of a plate

having mass M, length and width b about axis 1, 2, 3

and 4. Assume that C is centre and mass is uniformly

distributed

b

3

14 2

C

Sol. Moment of inertia of the plate about axis 1 (by taking

rods perpendicular to axis 1)

I1 =

2mb

12 + n

2b

2

= 2mb

3

Moment of inertia of the plate about axis 2 (by taking


I2 = M2/12

Moment of inertia of the plate about axis 3 (by taking


2

3

MbI

12

Moment of inertia of the plate about axis 4(by taking


I4 =

2m

12

l+ m

2

2

l =

2m

3

l

MOMENT OF INERTIA OF COMPOUND BODIES

Consider two bodies A and B, rigidly joined together.

The moment of inertia of this compound

body, about an axis XY, is required. If IA is the moment

of inertia of body A about XY. IB is the moment of

inertia of body B about XY.Then, moment of Inertia of

compound body I = IA + I

B

Extending this argument to cover any number of bodies

rigidly joined together, we see that the moment of inertia

of the compound body, about a specified axis, is the

sum of the moments of inertia of the separate parts of

the body about the same axis.

Y

B

AX

SOLVED EXAMPLE

Example-42

Two rods each having length l and mass m joined

together at point B as shown in figure.Then findout

moment of inertia about axis passing thorugh A and

perpendicular to the plane of page as shown in figure.

× A

B

C

Sol. We find the resultant moment of inertia I by dividing in

two parts such as

I = M.I of rod AB about A +

M.I of rod BC about A

I = I1 + I

2... (1)

first calculate I1 :

B A×

I1 =

m

3

2

...(2)

Calculation of I2 :

Use parallel axis theorem

I2 = I

CM + md2

=

22m

m12 4

2

= 2m 5

m12 4

2

...(3)

×

×

/ 2

COMd

Put value from eq. (2) & (3) into (1)

I = 2 2m m 5 m

3 12 4

2

I = m

(4 1 15)12

2

I = 25m

3

MOMENT OF FORCE OR TORQUE :

Rotational analogue of force:

Let us take the example of opening or closing of a door.

A door is a rigid body which can rotate about a fixed

vertical axis passing through the hinges. What makes

the door rotate? It is clear that unless a force is applied

the door does not rotate. But any force does not do the

job. A force applied to the hinge line cannot produce

any rotation at all, whereas a force of given magnitude

applied at right angles to the door at its outer edge is

most effective in producing rotation. It is not the force

alone, but how and where the force is applied is impor-

tant in rotational motion.The quantitative measure of

the tendency of a force to cause or change the rota-

tional motion of a body is called torque. Consider an

example to understand this.

In the figure below, the wrench is trying to open the

nut. Now the ability of wrernch to open the nut will

depend not only on the applied force, but the distance

at which force is applied. This gives birth to a new

physical quantity called torque.

Example-43A uniform disc having radius 2R and mass density as

shown in figure. If a small disc of radius R is cut from

the disc as shown. Then find out the moment of inertia

of remaining disc around the axis that passes through

O and is perpendicular to the plane of the page.

2R O R

Sol. We assume that in remaning part a disc of radius R and

mass density ± is placed. Then

2R O R

2R

× I1

M R122 ( )

when is taken

+ × RI2

M R22 –

–when is takes

Total Moment of Inertia I = I1 + I

2

I1 =

21M (2R)

2

I1 =

2 24R .4R

2

= 8 R4

To calculate I2 we use parallel axis theorem.

I2 = I

CM + M

2R2

I2 =

22M R

2 + M

2R2

I2 =

22

3M R

2 =

2 23(– R )R

2

I2 =

43– R

2

Now I = I1 + I

2

I = 4 43

8 R – R2

I = 413

R2

Example-44A uniform disc of radius R has a round disc of radius R/

3 cut as shown in Fig. The mass of the remaining

(shaded) portion of the disc equals M. Find the moment

of inertia of such a disc relative to the axis passing

through geometrical centre of original disc and

perpendicular to the plane of the disc.

ORR

/3

Sol. Let the mass per unit area of the material of disc be .

Now the empty space can be considered as having

density – and .

Now I0 = I + I

–

(R2)R2/2 = M.I of about O

I– =

2 22 2– (R / 3) (R / 3)

[– (R / 3) ](2R / 3)2

= M.I of – about 0

I0 =

44R

9

where = 22

M

RR

9

= 2

9M

8 R

If only radial force Fr were present, the nut could not be

turned. Thus the force causing the rotation is tangen-

tial force FT only. The magnitude of the torque about

an axis due to a force is given by

= (Force causing the rotation)×(distance of

point of application of force from the axis)

= (F sin ) r

we may also write = F (r sin) = F ( r )

= rFsin = (r sin)F

= rFT = r

F

where r is known as moment arm (lever arm)

Thus, if a force acts on a single particle at a point P

whose position with respect to the origin O is given by

the position vector r , (see figure), the moment of force

(torque) acting on the particle with respect to the ori-

gin O is defined as the vector product ;

x

Fy

O

r

P

FrFT

Fr

Direction of torque is found by sliding the force vector

at the axis of rotation and using right hand thumb rule.

×

F3

F2

F1

F4

l2

l1

Talk about F1, F

2, F

3 , F

4 and F

5 and their moment arm or

force arm

×

l5

F5

(Torque) 1 = F

1 l

1

2 = F

2 l

2

3 = 0

4 = 0

5 = F

5 l

5

Important Note

(1) The SI unit of torque is newton-metre. This is also the

unit of work and energy. But torque is not work or

energy. Usually, the unit of work or energy is written as

"joule". But the torque should not be expressed in

"joule". However, both have same dimensions ML2T–2.

(2) Torque is always defined with reference to a given point

(or given line). On changing the referenece point (or

line) the torque may change.

(3) Torque is a vector quantity, whose direction is

perpendicular to the plane of force and position

vector and its direction is given by right hand screw

rule.

(4) If the torque rotates the body in anticlock wise

direction, the torque is positive and if the torque

rotates the body in clock-wire direction, the torque

will be negative.

(5) As a shortcut the magnitude of torque can be calcu-

lated as the product of the force and the length of line

perpendicular to the force starting from the point about

which the torque is to be calculated.

(6) = 0 if r = 0, if F = 0, or if = 0° or 180°. Hence if the line

of action of the force passes through the reference

point, the torque vanishes.

(7) The more is the value of r, the more will be torque and

easier to rotate the body.

(i) The handle of screw driver is taken thick.

(ii) In villages the handle of flour-mill is placed near

the circumference.

(iii) The handle of handpump is kept-long.

(iv) The rinch used for opening the tap, is kept-

long.

(8) If a body is acted upon by more than one force, the

total torque is the vector sum of each torque.

=

1

+ 2 +

2 +..........+

n

Thus, torque also follows superposition principle.

(9) There can be non-zero torque on body even when net

force is zero (couple) and there can be force producing

zero torque.

×

F

F

(10) Force couple :

A pair of forces each of same magnitude and in op-

posite direction is called a force couple. Torque due

to a couple about any point is same and its magni-

tude = Magnitude of one force X distance between

their lines of action

A PRACTICAL EXAMPLE OF TORQUE :

In judo, a weaker and smaller fighter who

understands physics can defeat a stronger and larger

fighter who does not. This fact is demonstrated by

the basic "hip throw," in which a fighter rotates the

fighter's opponent around his hip and if the throw

is successful onto the mat. Without the proper use

of physics, the throw requires considerable strength

and can easily fail.

SOLVED EXAMPLE

Example-45

Given that, r = 2 i + 3 j and

F = 2 i + 6 k . The

magnitude of torque will be-

(A) 405 N.m (B) 410 N.m

(C) 504 N.m (D) 510 N.m

Sol. We know that, =

r ×

F

= (2 i + 3 j ) × ( 2 i + 6 k )

= 12(– j ) + 6(– k ) + 18 i

= – 12 j – 6 k + 18 i

[Note : i × i = 0, i × j = k , j × i = – k etc]

Now, | | = 222 )18()6()12(

= 32436144 = 504

Example-46A particle is falling freely along line x = d. Find torque

on this particle due to gravity, about origin when it

(i) Crosses x axis (ii) is at y = d

Sol. (i)

d

O = mgd mg

y

x

(Clockwise)

(ii)

mg

d

d

= mgd

y

O x

(clockwise)

SOLVED EXAMPLE

Example-47Find torque due to gravity at any time t about pt. of

projection, if a body is projected with velocity u at an

angle .

Sol.

u cos t

mg

u

O

y

x

At every instant force mg is in downward direction and

perpendicular distance will be displacement along x-

axis i.e. x = (u cos)t

0 = mg u cos t it is increasing with time

Example-48

30°(2)

(3)

2m

10N

(1)

O

30M

Find net torque about axis

of rotation passing through O.

Sol. 1 = 10 × 2 = 20

2 = 20 × 2 sin 30° = 20

3 = 30 × 2 = 60

–––––––––––––

= 20

ROTATION ABOUT A FIXED AXIS

If Hinge

= moment of inertia about the axis of rotation

(since this axis passes through the hinge, hence the

name Hinge

).

ext

= resultant external torque

acting on the body about

axis of rotation xHinge

Fixed axis of Rotation = angular acceleration of the body.

ext Hinge

= Hinge

Rotational Kinetic Energy = 21

. .2

CMP M v

;external CMF Ma

Net external force acting on the body has two

component tangential and centripetal.

FC = ma

C =

2

CM

vm

r = m2 r

CM

Ft = ma

t = m r

CM

PULLEY BLOCK SYSTEM

If there is friction between pulley and string and pulley

have some mass then tension is different on two sides

of the pulley.

Reason : To understand this concept we take a pulley

block system as shown in figure.

M m

RB

C

D a M>mT1

aA

Let us assume that tension induced in part AB of the

string is T1 and block M move downward. If friction is

present between pulley and string then it opposes the

relative slipping between pulley and string, take two

point e and f on pulley and string respectively. If friction

is there then due to this, both points wants to move

together. So friction force act on e and d in the direction

as shown is figure

This friction force f acting on point d increases the

tension T1 by a small amount dT.

Then T1 = T

2 + dT

or we can say T2 = T

1 – f

C

T1

T2

f

f

ed

In this way the tension on two side of pulley is different

If there is no relative slipping between pulley and string

then ta

R =

a

R

Example-49The pulley shown in figure has a moment of inertia about its axis and its radius is r. Calculate the magnitude

of the acceleration of the two blocks. Assume that the

string is light and does not slip on the pulley.

Sol. Suppose the tension in the left string is T1 and that in

the right string in T2. Suppose the block of mass m

1

goes down with an acceleration and the other block

moves up with the same acceleration a. This is also the

tangential acceleration of the rim of the wheel as the

string does not slip over the rim. The angular

acceleration of the wheel is, therefore, = a/r. The

equations of motion for the mass m1, the mass m

2 and

the pulley are as follows :

m1g – T

1 = m

1a .........(i)

T2 – m

2g = m

2a .........(ii)

T1r – T

2r = I = Ia /r .........(iii)

Putting T1 and T

2 from (i) and (ii) into (iii),

[(m1g – a) – m

2(g + a)] r =

a

r

which gives a = 2

1 22

1 2

(m m )gr

(m m )r

.

Example-50A uniform rod of mass m and length can rotate in

vertical plane about a smooth horizontal axis hinged at

point H.

(i) Find angular acceleration of the rod just after it is

released from initial horizontal position from rest ?

(ii) Calculate the acceleration (tangential and radial) of

point A at this moment.

(iii) Calculate net hinge force acting at this moment.

(iv) Find and when rod becomes vertical.

(v) Find hinge force when rod become vertical.

Sol. (i) H = I

H

mg.2

=

2m

3

=

3g

2

(ii) atA

= = 3g

2 . = 3g

2

aCA

= 2 r = 0. = 0 ( = 0 just after release)

(iii) Suppose hinge exerts normal reaction in

component form as shown

mg

N2

N1

at

ac

In vertical direction

Fext

= maCM

mg – N1 = m.

3g

4

(we get the value of aCM

from previous example)

N1 =

mg

4

In horizontal direction

Fext

= maCM

N2 = 0

( aCM

in horizontal = 0 as = 0 just after release).

(vi) Torque = 0 when rod becomes vertical.

so = 0

using energy conservation

2mg 1

2 2

2m

3

=

3g

(v) When rod becomes vertical

= 0, =

3g

FH – mg =

2m

2

FH =

5mg

2

Example-51The pulley shown in figure has moment of inertia l about

its axis and radius R. Find the acceleration of the two

blocks. Assume that the string is light and does not

slip on the pulley.

Sol. Suppose the tension in the left string is T1 and that in

the right string is T2. Suppose the block of mass M

goes down with an acceleration a and the other block

moves up with the same acceleration. This is also the

tangential acceleration of the rim of the wheel as the

string does not slip over the rim.

The angular acceleration of the wheel = a

R.

The equations of motion for the mass M,

the mass m and the pulley are as follows ;

M

R

m

Mg – T1 = Ma ...(i)

T2 – mg = ma ...(ii)

T1R – T

2R = I =

Ia

R...(iii)

Substituting for T1 and T

2 from equations (i) and (ii) in

equation (iii)

[M(g – a) – m (g + a)]R = Ia

R

Solving, we get, a = 2

2

(M – m)gR

I (M m)R

KINETIC ENERGY OF A RIGID BODY ROTATING ABOUT

A FIXED AXIS.

Suppose a rigid body is rotating about a fixed axis with

angular speed .

Then, kinetic energy of the rigid body will be :

Here, is the resultant torque acting on the body..

Further, since

d L

dt

dt d L

or2

1

t

tdt

= angular impulse =

2 1L – L

K = 2

i ii

1m v

2 = 2

i ii

1m ( r )

2

ri

mi

= 2 2

i ii

1m r

2 =

21I

2 (as

2i i

i

m r I )

Thus, KE = 21

I2

Sometimes it is called the rotational kinetic energy.

ANGULAR MOMENTUM

Consider a particle that has linear momentum p

and is

located at position r relative to an origin O, as shown

in the figure. Its angular momentum about the origin is

defined as

O

r

x

r

p

y

L = r × p

The magnitude of momentum is given by

L = rp sinThe direction of angular momentum is given by the

right hand rule.

The SI unit of angular momentum is kg m2/s.

SOLVED EXAMPLE

Example-52What is the angular momentum of a particle of mass

m = 2 kg that is located 15 m from the origin in the

direction 30º south of west and has a velocity v = 10

m/s in the direction 30º east of north?

Sol. In the figure (a), the x - axis points east. We know r = 15

m; p = mv = 20 kg m/s. The angle between r and p is

(180º – 30º) = 150º

30O

30O

P

x

y

r

r

O

150o

Thus,

L = rp sin = (15)(20)sin150º = 150 kg m2/s

In unit vector notation,

r = –15 cos30º i – 15 sin30º j m

p = 20 sin30º i + 20 cos30º j kg m/s.

Therefore,

L

= (15 3

–2 i – y

15

2 j ) (10 i + 10 3 j )

= –150 k kg m2/s

Note that angular momentum is defined always with

respect to a point

The total angular momentum L

of a system of par-

ticles relative to a given origin is the sum of the angu-

lar momentum of the particles.

L

= i ir ×p

Consider an object rotating about a fixed axis as shown

in the figure.

Since ir

and ip

are perpendicular, therefore,

ri

mi

L = rip

i

Also pi = m

iv

i = m

ir

i

or L = mir

i2 or L = I

where I = mir

i2

SOLVED EXAMPLE

Example-53A disc of mass M and radius R is rotating at angularvelocity about an axis perpendicular to its plane at adistance R/2 from the center, as shown in the figure.What is its angular momentum? The moment of inertia

of a disc about the central axis is 21MR

2.

Sol. The moment of inertia of the disc about the given axismay be found from the parallel axes theorem, equationI = I

cm + Mh2, where h is the distance between the

given axis and a parallel axis through the center ofmass.

R/2

R

Here h = R

2, therefore,

221 R

MR + M2 2

= 23

MR4

The angular momentum is

L = I = 2

4

3MR

TORQUE AND RATE OF CHANGE OF ANGULARMOMENTUM

As force changes the linear momentum of a particle,torque changes the angular momentum of a particle.The rate ofchange of angular momentum with time isgiven by

dL d= r × mv

dt dt

or dL dr dv

= × mv + r × mdt dt dt

or dL

= v× mv + r × Fdt

or dL

= r × F = τdt

because v×mv

is zero

Thus,dL

τ =dt

The above equation is the rotational analog of

equation dP

F =dt

SOLVED EXAMPLE

Example-54Show that the above equation can be applied to the

motion of a projectile.

Sol. In figure, we take the initial point as the origin. At a

later time, r = x i + y j .

Since the force on the particle is F

= -mg j , the

gravitational torque on it is

τ

= (x i + y j ) × (-mg j ) = -mgx kThe rate of change of the angular momentum

L

r × p

= is

v r

y

x mg

O

d L d p d v= r × = m r ×

dt dt dt

But the acceleration is dv

dt = -g j . Thus

d L d v ˆ ˆ ˆ= mr × = m xi + yj × -g jdt dt

= – mgx k

Hence the equation d L

τ =dt

is applicable.

SOLVED EXAMPLE

Example-55Two blocks with masses m

1 = 3 kg and m

2 = 1 kg are

connected by a rope that passes over a pulley of radius

R = 0.2 m and mass M = 4 kg. The moment of inertia of

the pulley about its center is I = 21

MR2

. Use the

concept of angular momentum to find the linear

acceleration of the blocks. There is no friction. Assume

that the c.m. of the block of mass m2 is at a distance R

above the center of the pulley.

m2

m1

m1g

v

R

v

Sol. If we take the origin at the center of the pulley, theangular momenta of the blocks are m

1vR and m

2vR and

that of the pulley is I. Therefore, the total angularmomentum isL = m

1vR + m

2vR + I ....(i)

If the rope does not slip, then v = R.

L = (m1 + m

2)vR +

MRv

2

The net external torque about the center of the pulleyis due to the weight of m

1text

= r^F = R(m

1g) .....(ii)

Applying equation ext

d Lτ =

dt

We obtain

Rm1g = (m

1 + m

2)Ra +

MRa

2

or a = 1

1 2

m g

m + m + M / 2

putting m1 = 3 kg; m

2 = 1 kg; M = 4 kg; R = 0.2 m

a = 3 10

3 +1+ 4 / 2 = 5 m/s2

CONSERVATION OF ANGULAR MOMENTUM

If the net external torque on a system is zero, the total

angular momentum is constant in magnitude and

direction.

That is, if ext

dLτ = 0 = 0

dt

Thus, L

= constant

For a rigid body rotating about a fixed axis,

Lf = Li

or Iff = Iii

SOLVED EXAMPLE

Example-56A disc of moment of inertia 4 kg m2 is spinning freely at

3 rad/s. A second disc of moment of inertia 2 kg m2

slides down the spindle and they rotate together.

(a) What is the angular velocity of the combination?

(b) What is the change in kinetic energy of the system?

Sol. (a) Since there are no external torques acting, we may

apply the conservation of angular momentum.

f

f = I

i

i

(6 kg m2)f = (4 kg m2)(3 rad/s)

Thus, f = 2 rad/s

(b) The kinetic energies before and after the collision

are

Ki =

2i i

1I ω = 18

2J;

Kf =

2f f

1I ω = 12

2J

The change is

K = Kf – K

i = -6 J.

In order for the two discs to spin together at the samerate, there had to be friction between them. The lostkinetic energy is converted to thermal energy.

A

B

ANGULAR IMPULSE

In complete analogy with the linear momentum,angular impulse is defined as

extJ = τ dt

Using Newton’s second law for rotation motion,

ext

d Lτ =

dt

f iJ = ΔL = L - L

The net angular impulse acting on a rigid body is equal

to the change in angular momentum of the body. This

is called the impulse – momentum theorem for

rotational dynamics.

SPIN AND ORBITAL ANGULAR MOMENTUM

For a rigid body undergoing linear and rotational

motion, the total angular momentum may be split into

two parts – the orbital angular momentum and the

spin angular momentum. The orbital angular

momentum oL is the angular momentum of the center

of mass motion about an origin O in an inertial frame.

The spin angular momentum cmL

is the angular

momentum relative to the center of mass.

The orbital term treats the system as a point particle at

the center of mass, whereas the spin term is the sum of

the angular momenta of the particles relative to the

center of mass. The total angular momentum relative

to the origin O in an inertial frame is the sum:

o cmL = L + L

Example-57A solid sphere of mass M and radius R rolls without

slipping on a horizontal surface as shown in the figure.

Find the total angular momentum of the sphere with

respect to the origin O fixed on the ground.

Sol. Let us assume the clockwise sense of rotation positive.

Orbital angular momentum about O isL

o = MvR

Spin angular momentum about c.m. is

Lc.m.

= I = 22

MR ω5

The total angular momentum is

L= Lo + L

cm = MvR +

2

5MR2

For pure rolling, v = R, therefore, L = 7

MvR5

Finally, we present an analogy between rotational dynamics and linear dynamics.

Quantity Linear Rotational

1. Inertia m I = 2

iirm

or I = dmr 2

2. Newton’s Second Law Fext = ma

dtdp

Fext

ext = I

dtdL

τext

3. Work Wlin = s.F

d Wrot = θ.dτ

4. Kinetic Energy Klin = 2

21

mv Krot = 2

21I

5. Work Energy Theorem Wlin = Klin Wrot = Krot

6. Impulse dtextFI dtext

J

7. Momentum p = mv L = I8. Impulse Momentum Theorem pI

LJ

9. Power P = vF

. ωτ

.P

SOLVED EXAMPLE

Example-58A uniform rod of mass m and length can rotate freely

on a smooth horizontal plane about a vertical axis hinged

at point H. A point mass having same mass m coming

with an initial speed u perpendicular to the rod, strikes

the rod in-elastically at its free end. Find out the angular

velocity of the rod just after collision?

×H

m

u

m

Sol. Angular momentum is conserved about H because no

external force is present in horizontal plane which is

producing torque about H.

mu = 2m

m3

2

= 3u

4

Example-59A person of mass m stands at the edge of a circularplatform of radius R and moment of inertia I. A platformis at rest initially. But the platform rotate when the personjumps off from the platform tangentially with velocity uwith respect to platfrom. Determine the angular velocityof the platform.

Sol. Let the angular velocity of platform is .Then the velocity of person with respect to ground v.

R

M

vmD

= vmG

– VDG

u = vm + R

vm = u – R

Now from angular momentum conservationL

i = L

f

0 = mvmR – I

I = m (u – R) . R

R

u

= 2

muR

I mR

ROLLING - COMBINATION OF TRANSLATIONAL ANDROTATIONAL MOTION :

Rolling Motion :

A photograph of a rolling bicycle wheel. The spokes near the top of the wheel are more blurred than those near the bottom

of the wheel because they are moving faster

Pure rolling means no sliding. Now, the motion of anybody can be divided into pure translation & purerotation. And we can see rotation about any axis. So;for a wheel rolling on a horizontal surface, we are takingits COM as reference point to study its motion. If C isthe reference point, then the wheel can be consideredrotating about C. And the point C would be translatingwith velocity v

c.

Sliding refers to the condition under which two bodiesin contact have relative velocity. And under pure rolling,the relative velocity at point of contact should be zero.

Now for the wheel shown, point P is in contact with theground. The point P on the ground has zero velocity.Thus, the point P on the wheel should also have zerovelocity.

Rolling with sliping :

Case-I Vcm > R

Vcm

AV – Rcm fk

ground

NOTE : The friction will be kinetic in nature & its magnitude

can be determined using fk = kN. Direction will be

opposite to Vcm (because pt. of contact A is moving

forward w.r.t. ground)

(This is how brakes stop a car)

Case -2 Vcm < R

Vcm

A

R – Vcmfk

ground

(This is how a car accelerate because of friction)

Case -3 Vcm = r

Known as perfect rolling

Maximum problems we come across this situation

NOTE : 1. The friction will be be static in nature. Its magnitude

be determined, it will very from O to sN.

2. Even its direction can not be perdicted

3. The total work done by static friction is zero. Thus

mechanical energy of the system will remain conserved.

Kinetic energy of a rolling Body :

Since the rolling motion is a combination of linear

velocity of the center and rotational motion about the

center. Therefore, the total kinetic energy of a rolling

body is given by

K = 2

c2c I

2

1m

2

1 .....(i)

Where 2cm

2

1 is the translational kinetic energy and

2cI

2

1 is the rotational kinetic energy about the center

of mass

In pure rolling motion, c = R

K = 2

1m(R)2 +

2

1I

c2

or K = (Ic + mR2) 2

Using parallel axes theorem, the term Ic + mR2 gives the

moment of inertia about the point of contact, therefore,

I0 = I

c + mR2

and K = 2

1I

02 ....(ii)

Note that equation (ii) give the rotational kinetic energy

of the wheel about the point of contact.

Application of Newton’s Second Law in Rolling Motion

1. Write Fnet = M acm for the object as if it were a point-

mass, that is, ignoring ratation.

2. Write =Icm as if the object were only rotating

about the centre of mass, that is ignoring

translation.

3. Use of no-slip condition

4. Solve the resulting equations simultaneously for

any unknown.

Caution :

• In general, it is not the case that f = N

• Be certain that the sign convention of forces and

torques are consistent.

ROLLING ON THE INCLINED PLANE :

(a) Velocity : When a rigid body of mass M rolls on an

inclined plane without slipping, the friction force

'f ' is static, and no work is done by the friction force.

Hence from the conservation of mechanical energy,

//////////////////////////////////////////////

f

mgcosmgmgsinvcm

N

R

h

Mgh = 222

cm2

cm2cm MK

2

1Mv

2

1I

2

1Mv

2

1

(where K is the radius of gyration)

= 222

22cm MR

R

K

2

1Mv

2

1

= 2cm2

2

vR

K1.M

2

1

, [where v

cm = R for pure rolling]

or,

2

2cm

RK

1

gh2v

,

[where 2

2

R

K is a pure number between 0 and 1 that

depends on the shape of the body]

It is important to note that the velocity of a rolling

body is independent of its mass (M) and radius (R). All

uniform solid spheres have the same speed at the bottom

even if their masses and radii are different because they

have same

5

2

R

K2

2

. All solid cylinders have the

same speed

2

1

R

K2

2

, all hollow cylinders have the

same speed

1

R

K2

2

and so on. Smaller the value

of 2

2

R

K, faster the body is moving.

(b) Acceleration : Let the linear acceleration of the rolling

body be acm

. Then for the linear motion, the net force =

mg sin – f = macm

...(i)

For the rotational motion, the net torque

f R = Icm

. = MK2 . R

acm

or, cm2

2

aR

MKf ...(ii)

solving (i) and (ii) for acm

, we get

2

2cm

RK

1

singa

It is also independent of mass (M) and radius R.

(c) Time of descent (t) :

sin

ht.a

2

1 2cm

or,

sina

h2t

cm

g

RK

1h2

sin

1t

2

2

It is also independent of mass and radius. Clearly, a

body of smaller 2

2

R

K, takes less time.

SOLVED EXAMPLE

Example-60A solid cylinder of mass m and radius r starts rolling

down an inclined plane of inclination . Friction is

enough to prevent slipping. Find the speed of tis centre

of mass when, its centre of mass has fallen a height h.

Sol. Consider the two shown positions of the cylinder. As it

does not slip, total mechanical energy will be conserved.

Energy at position 1 is E1 = mgh

h

1

2

Energy at position 2 is E2 =

1

2mv2

c.m. +

1

2I

c.m. 2

2

c.m.c.m.

V mrω, and I

r 2

E2 =

3

2mv2

c.m.

From COE, E1 = E

2

Vc.m.

= 3

gh2

Example-61A ball of radius R and mass m is rolling without slipping

on a horizontal surface with velocity of its centre of

mass v. It then rolls without sloping up a hill to a height

h before momentarily coming to rets. Find h.

Sol.2 21 1

mv Iω mgh2 2

h

v

v = R , I = 2

5mR2

h = g10

v7 2CM

Example-62Figure shows a solid sphere rolling without slipping

on a horizontal surface. Its radius is r, mass is m and the

velocity of its cenre is v. Find the angular momentum

about any point O on the horizontal surface along the

line parallel to its velocity v

Sol. The angular momentum about O is

v

rO

L = Lcm

+ Icm

= cmcm vrm

+Icm

= mvr + 5

2mr2 ·

5

7

r

v mvr

Example-63A spherical ball rolls on a table without slipping.

Then the fraction of its total energy associated with

rotation is-

(A) 2/5 (B) 3/5

(C) 2/7 (D) 3/7

Sol. Total energy,

E = (1/2) I2 + (1/2) mv2

= (1/2) (2/5 mr2) 2 + (1/2) mr22

=(1/5) mr22 + (1/2) mr22 = (7/10) mr22

Rotational energy = (1/5) mr22

Rotational energy

Total energy =

2 2

2 2

1mr

57

mr10

=

2

7

CENTER OF MASSCalculation of COMQ.1 The centre of mass of a body :

(1) Lies always at the geometrical centre(2) Lies always inside the boy(3) Lies always outside the body(4) Lies within or outside the body

Q.2 A body has its centre of mass at the origin. Thex-coordinates of the particles(1) may be all positive(2) may be all negative(3) must be all non-negative(4) may be positive for some particles and negativein other particles

Q.3 All the particles of a body are situated at a distanceR from the origin. The distance of the centre of massof the body from the origin is(1) = R (2) R(3) > R (4) R

Q.4 Where will be the centre of mass on combining twomasses m and M(M > m) :(1) towards m (2) towards M(3) between m and M (4) anywhere

Q.5 A uniform solid cone of height 40 cm is shown infigure. The distance of centre of mass of the conefrom point B (centre of the base) is :

40 cm

30°

B

(1) 20 cm (2) 10/3 cm(3) 20/3 cm (4) 10 cm

Q.6 In the HCl molecule, the separation between the nucleiof the two atoms is about 1.27 Å (1 Å = 10–10 m). Theapproximate location of the centre of mass of themolecule, distance from hydrogen atom assuming thechlorine atom to be about 35.5 times massive ashydrogen is(1) 1Å (2) 2.5 Å(3) 1.24 Å (4) 1.5 Å

Q.7 Centre of mass is a point(1) Which is geometric centre of a body(2) From which distance of particles are same(3) Where the whole mass of the body is supposed to

concentrated(4) Which is the origin of reference frame

Q.8 Choose the correct statement about the centre of mass(CM) of a system of two particles(1) The CM lies on the line joining the two particles

midway between them(2) The CM lies on the line joining them at a point

whose distance form each particle is inverselyproportional to the mass of that particle

(3) The CM lies on the line joining them at a pointwhose distance from each particle is proportionalto the square of the mass of that particle

(4) The CM is on the line joining them at a point whosedistance from each particle is proportional to themass of that particle

Q.9 The centre of mass of triangle shown in figure hascoordinates

h

b

y

x

(1) x = h

2, y =

b

2(2) x =

b

2, y =

h

2

(3) x = b

3, y =

h

3(4) x =

h

3, y =

b

3

EXERCISE-I

Q.10 Three identical spheres, each of mass 1 kg are kept as

shown in figure, touching each other, with their centres

on a straight line. If their centres are marked P, Q, R

respectively, the distance of centre of mass of the

system from P (origin) is

P Q Rx

y

(1) PQ PR QR

3

(2)

PQ PR

3

(3) PQ QR

3

(4)

PR QR

3

Q.11 A uniform square plate ABCD has a mass of 10 kg. If

two point masses of 3 kg each are placed at the corners

C and D as shown in the adjoining figure, then the

centre of mass shifts to the point which is lie on -

x

y

y'

x'O

D C(1) OC (2) OD (3) OY (4) OX

Displacement, velocity, Acceleration of COM

Q.12 A bomb travelling in a parabolic path under the effect

of gravity, explodes in mid air. The centre of mass of

fragments will:

(1) Move vertically upwards and then downwards

(2) Move vertically downwards

(3) Move in irregular path

(4) Move in the parabolic path which the unexploded

bomb would have travelled.

Q.13 A body at rest may have(1) Energy (2) Momentum(3) Speed (4) Velocity

Q.14 Internal forces can change :

(1) the linear momentum but not the kinetic energy

of the system.

(2) the kinetic energy but not the linear momentum

of the system.

(3) linear momentum as well as kinetic energy of the

system.

(4) neither the linear momentum nor the kinetic

energy of the system.

Q.15 Two balls are thrown in air. The acceleration of the

centre of mass of the two balls while in air (neglect

air resistance)

(1) depends on the direction of the motion of the

balls

(2) depends on the masses of the two balls

(3) depends on the speeds of the two balls

(4) is equal to g

Q.16 The motion of the centre of mass of a system of two

particles is unaffected by their internal forces :

(1) irrespective of the actual directions of the internal

forces

(2) only if they are along the line joining the particles

(3) only if they are at right angles to the line joining the

particles

(4) only if they are obliquely inclined to the line joining

the particles.

Q.17 Two objects of masses 200 gm and 500 gm posses

velocities ˆ10i m/s and ˆ ˆ3i 5j m/s respectively. The

velocity of their centre of mass in m/s is :

(1) ˆ ˆ5i 25j (2) 5 ˆ î 25j7

(3) 25ˆ ˆ5i j7

(4) 5ˆ ˆ25i j7

Q.18 2 bodies of different masses of 2kg and 4kg are movingwith velocities 20m/s and 10m/s towards each otherdue to mutual gravitational attraction. What is thevelocity of their centre of mass ?(1) 5 m/s (2) 6 m/s (3) 8 m/s (4) zero

Q.19 Two bodies of masses 2 kg and 4 kg are moving withvelocities 2 m/s and 10m/s respectively along samedirection. Then the velocity of their centre of mass willbe(1) 8.1 m/s (2) 7.3 m/s (3) 6.4 m/s (4) 5.3 m/s

Q.20 Two particle of masses m1 and m

2 initially at rest start

moving towards each other under their mutual force ofattraction. The speed of the centre of mass at any timet, when they are at a distance r apart, is

(1) Zero (2) 1 22

1

m m 1G . t

r m

(3) 1 22

2

m m 1G . t

r m

(4) 1 22

1 2

m m 1G . t

r m m

Q.21 The two particles X and Y, initially at rest, start movingtowards each other under mutual attraction. If at anyinstant the velocity of X is V and that of Y is 2V, thevelocity of their centre of mass will be(1) Zero (2) V (3) 2V (4) V/2

Q.22 Two particles whose masses are 10 kg and 30 kg and

their position vectors are ˆ ˆ î j k and ˆ ˆ î j k

respectively would have the centre of mass at -

(1) ˆ ˆ ˆ(i j k)

2

(2)

ˆ ˆ ˆ(i j k)

2

(3) ˆ ˆ ˆ(i j k)

4

(4)

ˆ ˆ ˆ(i j k)

4

Q.23 Two balls A and B of masses 100 gm and 250 gmrespectively are connected by a stretched spring ofnegligible mass and placed on a smooth table. Whenthe balls are released simultaneously the initialacceleration of B is 10 cm/sec2 west ward. What is themagnitude and direction of initial acceleration of theball A -(1) 25 cm/sec2 East ward(2) 25 cm/sec2 North ward(3) 25 cm/sec2 West ward(4) 25 cm/sec2 South ward

Law of conservation of momentumQ.24 Two bodies of masses m

1 and m

2 have equal kinetic

energies. If p1 and p

2 are their respective momentum,

then ratio p1 : p

2 is equal to

(1) m1 : m

2(2) m

2 : m

1

(3) 1 2m : m (4) 2 21 2m : m

Q.25 A bullet of mass m is being fired from a stationarygun of mass M. If the velocity of the bullet is v,the velocity of the gun is-

(1) Mv

m M (2) mv

M

(3) (M m)v

M

(4)

M m

Mv

Q.26 Two identical blocks A and B, each of mass ‘m’ restingon smooth floor are connected by a light spring ofnatural length L and spring constant K, with the springat its natural length. A third identical block ‘C’ (massm) moving with a speed v along the line joining A andB collides with A. the maximum compression in thespring is (consider all collision are elastic)

(1) m

v2k

(2) v

m2k

(3) mv

k(4)

mv

2k

Q.27 A ball of mass 3 kg collides with a wall with velocity 10

m/sec at an angle of 30° and after collision reflects at

the same angle with the same speed. The change in

momentum of ball in MKS unit is-

(1) 20 (2) 30 (3) 15 (4) 45

Q.28 A bomb at rest has mass 60 kg. It explodes and a

fragment of 40 kg has kinetic energy 96 joule. Then

kinetic energy of other fragment is-

(1) 180 J (2) 190 J

(3) 182 J (4) 192 J

Idea of impulsive / Non impulsive force

Q.29 A force of 50 dynes is acted on a body of mass

5gm which is at rest for an interval of 3 sec, then

impulse is-

(1) 0.16 × 10–3 N-S (2) 0.98 × 10–3 N-S

(3) 1.5 × 10–3 N-S (4) 2.5 × 10–3 N-S

Q.30 The area of F-t curve is A, where 'F' is the force on

one mass due to the other. If one of the colliding

bodies of mass M is at rest initially, its speed just

after the collision is :

(1) A/M (2) M/A (3) AM (4) 2A

M

Q.31 If two balls, each of mass 0.06 kg, moving in opposite

directions with speed of 4m/s, collide and rebound with

the same speed, then the impulse imparted to each ball

due to other (in kg-m/s) is :

(1) 0.48 (2) 0.53 (3) 0.81 (4) 0.92

Collisions (Oblique & head on)

Q.32 A block moving in air explodes in two parts then

just after explosion

(1) the total momentum must be conserved

(2) the total kinetic energy of two parts must be

same as that of block before explosion.

(3) the total momentum must change

(4) the total kinetic energy must not be increased

Q.33 In head on elastic collision of two bodies of equal

masses, it is not possible :

(1) the velocities are interchanged

(2) the speeds are interchanged

(3) the momenta are interchanged

(4) the faster body speeds up and the slower body

slows down

Q.34 A ball of mass 'm', moving with uniform speed, collideselastically with another stationary ball. The incidentball will lose maximum kinetic energy when the massof the stationary ball is(1) m (2) 2m (3) 4m (4) infinity

Q.35 The coefficient of restitution e for a perfectly elasticcollision is(1) 1 (2) 0 (3) (4) – 1

Q.36 A lead ball strikes a wall and falls down, a tennis ballhaving the same mass and velocity strikes the walland bounces back. Check the correct statement(1) The momentum of the lead ball is greater than that

of the tennis ball(2) The lead ball suffers a greater change in momentum

compared with the tennis ball(3) The tennis ball suffers a greater change in

momentum as compared with the lead ball(4) Both suffer an equal change in momentum

Q.37 In the figure shown the block A collides head onwith another block B at rest. Mass of B is twice themass of A. The block A stops after collision. Theco-efficient of restitution is :

A B

m 2m

(1) 0.5 (2) 1(3) 0.25 (4) it is not possible

Q.38 Two identical smooth spheres A and B are movingwith same velocity and collides with similar spheresC and D, then after collision- (Consider one dimensionalcollision and all collision are elastic)

A B C D

V

(1) D will move with greater speed(2) C and D will move with same velocity v(3) C will stop and D will move with velocity v(4) All spheres A, B, C & D will move with velocityv/2

Q.39 A body of mass m having an initial velocity v, makeshead on collision with a stationary body of mass M.After the collision, the body of mass m comes to restand only the body having mass M moves. This willhappen only when (consider all collision are elastic)

(1) m >> M (2) m < < M (3) m = M (4) m 1

2 M

Q.40 Two equal masses m1 and m

2 moving along the same

straight line with velocities + 3 m/s and – 5 m/srespectively collide elastically. Their velocities afterthe collision will be respectively(1) + 4 m/s for both (2) – 3 m/s and +5 m/s(3) – 4 m/s and + 4 m/s (4) – 5 m/s and + 3 m/s

Q.41 A body falls on a surface of coefficient of restitution0.6 from a height of 1 m. Then the body rebounds to aheight of

(1) 0.6 m (2) 0.4 m (3) 1 m (4) 0.36 m

Q.42 A ball of mass m falls vertically to the ground from a

height h1 and rebound to a height h

2. The change in

momentum of the ball on striking the ground is

(1) 1 2mg(h h ) (2) 1 2m( 2gh 2gh )

(3) 1 2m 2g(h h ) (4) 1 2m 2g(h h )

Q.43 Two spheres approaching each other collides

elastically. Before collision the speed of A is 5m/s

and that of B is 10m/s. Their masses are 1kg and

0.5kg. After collision velocities of A and B are

respectively-

(1) 5 m/s –10 m/s (2) 10 m/s, –5 m/s

(3) –10 m/s, –5 m/s (4) –5 m/s, 10 m/s

Q.44 There are hundred identical sliders equally spaced

on a frictionless track as shown in the figure. Initially

all the sliders are at rest. Slider 1 is pushed with

velocity v towards slider 2. In a collision the sliders

stick together. The final velocity of the set of

hundred stucked sliders will be :

1v

2 3 100

(1) v

99(2)

v

100(3) zero (4) v

Q.45 A space craft of mass M is moving with velocity V andsuddenly explodes into two pieces. A part of it of mass

m becomes at rest, then the velocity of other part willbe

(1) MV

M m(2)

MV

M m

(3) mV

M m(4)

(M m)V

m

Q.46 The bob A of a simple pendulum is released when the

string makes an angle of 45°with the vertical. It hits

another bob B of the same material and same mass

kept at rest on the table. If the collision is elastic

B

A

45°

O

(1) Both A and B rise to the same height(2) Both A and B come to rest at B(3) Both A and B move with the same velocity of A

(4) A comes to rest and B moves with the velocity of A

Q.47 A body of mass M1 collides elastically with another

mass M2 at rest. There is maximum transfer of energy

when

(1) M1 > M

2

(2) M1 < M

2

(3) M1 = M

2

(4) Same for all values of M1 and M

2

Q.48 A mass ‘m’ moves with a velocity ‘v’ and collidesinelastically with another identical mass. After collision

the 1st mass moves with velocity v

3 in a direction

perpendicular to the initial direction of motion. Findthe speed of the 2nd mass after collision

before collision After collision

m v

m At rest

(1) (2) (3) v (4)

Q.49 A completely inelastic collision is one in which thetwo colliding particles(1) Are separated after collision(2) Remain together after collision(3) Split into small fragments flying in all directions(4) None of the above

Q.50 A body of mass 2kg is moving with velocity 10 m/stowards east. Another body of same mass and samevelocity moving towards north collides with formerand coalsces and moves towards north-east. Itsvelocity is(1) 10 m/s (2) 5 m/s

(3) 2.5 m/s (4) 5 2 m /s

Q.51 A body of mass m1 is moving with a velocity V. It

collides with another stationary body of mass m2 . They

get embedded. At the point of collision, the velocity

of the system(1) Increases(2) Decreases but does not become zero(3) Remains same(4) Become zero

Variable Mass system

Q.52 A rocket with a life-off mass 3.5 × 104 kg is blasted

upwards with an initial acceleration of 10 m/s2. The

initial thrust of the blast is-

(1) 14.0 × 105 N (2) 1.76 × 105 N

(3) 3.5 × 105 N (4) 7.0 × 105 N

Q.53 Fuel is consumed at the rate of 100 kg/sec. in a rocket.

The exhaust gases are ejected as a speed of 4.5 ×

104 m/s. What is the thrust experienced by the rocket-

(1) 3 × 106 N (2) 4.5 × 106 N

(3) 6 × 106 N (4) 9 × 106 N

Q.54 A 6000 kg rocket is set for vertical firing. If the exhaust

speed is 1000 m/sec. How much gas must be ejected

each second to supply the thrust needed to give

the rocket an initial upward acceleration of 20 m/sec²-

(consider g = 9.8 msec–2 acceleration due to gravity )

(1) 92.4 kg/sec (2) 178.8 kg/sec

(3) 143.2 kg/sec (4) 47.2 kg/sec

ROTATIONAL MOTION

Angular displacement, velocity & acceleration

Q.55 The shaft of a motor rotates at a constant angular

velocity of 3000 rpm. The radians it has turned in 1 sec

are

(1) 1000 (2) 100 (3) (4) 10

Q.56 The angular speed of second hand of a clock is

(1) (1/60) rad/s (2) (/60) rad/s

(3) (2/60) rad/s (4) (360/60) rad/s

Q.57 Two bodies of mass 10 kg and 5 kg moving in concentricorbits of radius r

1 and r

2 such that their periods are

same. The ratio of centripetal accelerations is(1) r

1/r

2(2) r

2/r

1(3) (r

1/r

2)3 (4) (r

2/r

1)2

Q.58 A wheel starts rotating from rest and attains an angularvelocity of 60 rad/sec in 5 seconds. The total angulardisplacement in radians will be-(1) 60 (2) 80 (3) 100 (4) 150

Q.59 Figure shows a small wheel fixed coaxially on a biggerone of double the radius. The system rotates uniformlyabout the common axis. The strings supporting A andB do not slip on the wheels. If x and y be the distancestravelled by A and B in the same time interval, then-

AB

(1) x = 2y (2) x = y (3) y = 2x (4) None of these

Q.60 The linear and angular acceleration of a particle are10 m/sec2 and 5 rad/sec2 respectively. It will be at adistance from the axis of rotation.

(1) 50 m (2) 1

2m (3) 1 m (4) 2 m

Q.61 A particle is moving with a constant angular velocityabout an exterior axis. Its linear velocity will dependupon -(1) perpendicular distance of the particle form the axis(2) the mass of particle(3) angular acceleration of the particle(4) the linear acceleration of particle

Q.62 A body is in pure rotation. The linear speed of aparticle, the distance r of the particle from the axisand the angular velocity of the body are related as

= r

v. Thus-

(1) r

1(2) r

(3) = 0 (4) is independent of r.

Moment of Inertia

Q.63 Four masses are fixed on a massless rod as shown in

fig. The moment of inertia about the axis P is nearly

0.2 m 0.2 m 0.2 m 0.2 m

2 kg5 kg5 kg2 kg

P

(1) 2 kg m2 (2) 1 kg m2

(3) 0.5 kg m2 (4) 0.3 kg m2

Q.64 By the theorem of perpendicular axes, if a body be

in X-Z-plane then :-

(1) Ix – I

y = I

z(2) I

x + I

z = I

y

(3) Ix + I

y = I

z(4) I

y + I

z = I

x

Q.65 The axis X and Z in the plane of a disc are mutually

perpendicular and Y-axis is perpendicular to the plane

of the disc. If the moment of inertia of the body about

X and Y axes is respectively 30 kg m2 and 40 kg m2

then M.I. about Z-axis in kg m2 will be :-

(1) 70 (2) 50

(3) 10 (4) Zero

Q.66 Two rods each of mass m and length are joined at

the centre to form a cross. The moment of inertia of

this cross about an axis passing through the common

centre of the rods and perpendicular to the plane

formed by them, is :

(1) 2m

12

(2) 2m

6

(3) 2m

3

(4) 2m

2

Q.67 For the same total mass which of the following will

have the largest moment of inertia about an axis

passing through the centre of mass and perpendicular

to the plane of the body

(1) A disc of radius a

(2) A ring of radius a

(3) A square lamina of side 2a

(4) Four rods forming a square of side 2a

Q.68 The moment of inertia of a thin uniform circular disc

about one of the diameters is I. Its moment of inertia

about an axis perpendicular to the .plane of disc and

passing through its centre is

(1) 2 I (2) 2I (3) I/2 (4) I / 2

Q.69 The moment of inertia of a uniformsemicircular wire of

mass M and radius r about a line perpendicular to the

plane of the wire through the centre is

(1) Mr2 (2) 21

Mr2

(3) 21

Mr4

(4) 22

Mr5

Q.70 The density of a rod AB increases linearly from A to B.

Its midpoint is O and its centre of mass is at C. Four

axes pass through A, B, O and C, all perpendicular to

the length of the rod. The moments of inertia of the

rod about these axes are IA, I

B, I

O and I

C respectively.

(1) IA > I

B(2) I

A < I

B(3) I

O = I

C(4) I

O < I

C

Q.71 A stone of mass 4 kg is whirled in a horizontal circle

of radius 1m and makes 2 rev/sec. The moment of

inertia of the stone about the axis of rotation is-

(1) 64 kg × m2 (2) 4 kg × m2

(3) 16 kg × m2 (4) 1 kg × m2

Q.72 In an arrangement four particles, each of mass 2 gram

are situated at the coordinate points (3, 2, 0),

(1, –1, 0), (0, 0, 0) and (–1, 1, 0). The moment of inertia

of this arrangement about the Z-axis will be-

(1) 8 units (2) 16 units (3) 43 units (4) 34 units

Q.73 A solid sphere and a hollow sphere of the same mass

have the same M.I. about their respective diameters.

The ratio of their radii will be :

(1) 1 : 2 (2) 3 : 5 (3) 5 : 3 (4) 5 : 4

Q.74 Three solid spheres of mass M and radius R are

shown in the figure. The moment of inertia of the

system about XX' axis will be :-X

X'

(1) 7

2MR2 (2)

14

5MR2 (3)

16

5MR2 (4)

21

5MR2

Q.75 The moment of inertia of a square lamina about the

perpendicular axis through its centre of mass is 20 kg-

m2. Then, its moment of inertia about an axis touching

its side and in the plane of the lamina will be :-

(1) 10 kg-m2 (2) 30 kg-m2

(3) 40 kg-m2 (4) 25 kg-m2

Q.76 Three rings, each of mass P and radius Q are arranged

as shown in the figure. The moment of inertia of the

arrangement about YY’ axis will be-

Q QPP

Y1 2

Q

P3Y'

(1) 2

7PQ2 (2)

7

2 PQ2 (3)

5

2 PQ2 (4)

2

5 PQ2

Q.77 The moment of inertia of a rod of mass M and length

L about an axis passing through one edge and

perpendicular to its length will be :–

(1) 2ML

12(2)

2ML

6(3)

2ML

3(4) ML2

Q.78 Three thin uniform rods each of mass M and length L

and placed along the three axis of a Cartesian

coordinate system with one end of each rod at the

origin. The M.I. of the system about z-axis is-

(1) 3

ML2

(2) 3

ML2 2

(3) 6

ML2

(4) ML2

Q.79 Four particles each of mass m are placed at the

corners of a square of side length .The radius of

gyration of the system about an axis perpendicular

to the square and passing through centre is :–

(1) 2

(2)

2

(3) (4) 2

Q.80 The M.I. of a thin rod of length about the

perpendicular axis through its centre is I. The M.I. of

the square structure made by four such rods about

a perpendicular axis to the plane and through the

centre will be :-

(1) 4 I (2) 8 I (3) 12 I (4) 16 I

Q.81 The moment of inertia of a ring of mass M and radiusR about PQ axis will be :

R

MD

O

D'

O'

P

Q

(1) MR2 (2) 2MR

2(3)

3

2 MR2 (4) 2MR

Q.82 Four point masses (each of mass m) are arranged itthe X-Y plane the moment of inertia of this array ofmasses about Y-axis is

Y

O(0,0)

(a,–a)

(2a,0)

(a,a)

a

aa

a

X

(1) ma2 (2) 2ma2 (3) 4ma2 (4) 6ma2

Q.83 An equilateral traingular wire frame is made from 3 rodsof equal mass and length each. The frame is rotatedabout an axis perpendicular to the plane of the frameand passing through its end. What is the radius ofgyration of the frame ?

(1) 2

(2) (3)

2

(4)

32

Fixed axis rotation + TopplingQ.84 For a system to be in equilibrium, the torques acting

on it must balance. This is true only if the torques aretaken about(1) the centre of the system(2) the centre of mass of the system(3) any point on the system(4) any point on the system or outside it

Q.85 A rectangular block has a square base measuring a × a,and its height is h. It moves on a horizontal surface ina direction perpendicular to one of the edges h beingvertical. The coefficient of friction is µ. It will topple if

(1) h

a (2)

a

h (3)

2a

h (4)

a

2h

Q.86 A force of )k2j4i2( Newton acts at a point

)k4j2i3( metre from the origin. The magnitude

of torque is -(1) zero (2) 24.4 N-m(3) 0.244 N-m (4) 2.444 N-m

Q.87 Rate of change of angular momentum with respect totime is proportional to :(1) angular velocity (2) angular acceleration(3) moment of inertia (4) torque

Q.88 When constant torque is acting on a body then :–(1) body maintain its state or moves in straight linewith same velocity(2) acquire linear acceleration(3) acquire angular acceleration(4) rotates with a constant angular velocity

Q.89 If I = 50 kg-m2, then how much torque will be applied tostop it in 10 sec. Its initial angular speed is 20 rad/sec.:(1) 100 N-m (2) 150 N-m(3) 200 N-m (4) 250 N-m

Q.90 A particle is at a distance r from the axis of rotation. Agiven torque produces some angular acceleration init. If the mass of the particle is doubled and its distancefrom the axis is halved, the value of torque to producethe same angular acceleration is -(1) /2 (2) (3) 2 (4) 4

Energy analysisQ.91 A ring of radius r and mass m rotates about an axis

passing through its centre and perpendicular to itsplane with angular velocity . Its kinetic energy is

(1) mr (2) 21

mr2

(3) mr22 (4) 2 21

mr2

Q.92 A rod of length L is hinged at one end. It is brought toa horizontal position and released. The angular velocityof the rod when it is in vertical position is

(1) 2g / L (2) 3g / L (3) g / 2L (4) g / L

Q.93 Two bodies A and B having same angular momentumand I

A > I

B, then the relation between (K.E.)

A and (K.E.)

B

will be :–

(1) (K.E.)A > (K.E.)

B(2) (K.E.)

A = (K.E.)

B

(3) (K.E.)A < (K.E.)

B(4) (K.E.)

A (K.E.)

B

Angular MomentumQ.94 A thin circular ring of mass M and radius r is rotating

about its axis with a constant angular velocity . Twoobjects, each of mass m are attached gently to theopposite ends of a diameter of the ring. The wheelnow rotates with an angular velocity

(1) M

M m

(2)

M 2m

M 2m

(3) M

M 2m

(4)

M 2m

M

Q.95 A rotating table completes one rotation is 10 sec. andits moment of ineratia is 100 kg-m2. A person of 50kg. mass stands at the centre of the rotating table.If the person moves 2m. from the centre, the angularvelocity of the rotating table in rad/sec. will be:

(1) 2

30

(2)

20

30

(3)

2

3

(4) 2

Q.96 A circular hoop of mass m, and radius R rests flat on ahorizontal frictionless surface. A bullet, also of massm, and moving with a velocity v, strikes the hoop andgets embedded in it. The thickness of the hoop is muchsmaller than R. The angular velocity with which thesystem rotates after the bullet strikes the hoop is

R

m

m v

(1) V

4R(2)

V

3R(3)

2V

3R(4)

3V

4R

Q.97 A uniform heavy disc is rotating at constant angularvelocity () about a vertical axis through its centre O.Some wax W is dropped gently on the disc. The angularvelocity of the disc-(1) does not change (2) increases(3) decreases (4) becomes zero

Q.98 A girl sits near the edge of a rotating circular platform.If the girl moves from circumference towards the centreof the platform then the angular velocity of the platformwill-(1) decrease (2) increase(3) remain same (4) becomes zero

Q.99 Two wheels P and Q are mounted on the same axle.

The moment of inertia of P is 6 kg-m2 and it is rotating

at 600 rotations per minute and Q is at rest. If the two

are joined by means of a clutch then they combined

and rotate at 400 rotations per minute. The moment of

inertia of Q will be -

(1) 3 kg-m2 (2) 4 kg-m2 (3) 5 kg-m2 (4) 8 kg-m2

Combined Rotation and translation

Q.100 If a spherical ball rolls on a table without slipping, the

fraction of its total kinetic energy associated with

rotation is

(1) 3/5 (2) 2/7 (3) 2/5 (4) 3/7

Q.101 The speed of a homogeneous solid sphere after rolling

down an inclined plane of vertical height h, from rest

without sliding is

(1) gh (2) g / 5 gh

(3) 4 / 3 gh (4) (10 / 7)gh

Q.102 The rotational kinetic energy of a body is E. In the

absence of external torque, if mass of the body is

halved and radius of gyration doubled, then its

rotational kinetic energy will be :-

(1) 0.5E (2) 0.25E

(3) E (4) 2E

Q.103 A ring is rolling without slipping. Its energy of

translation is E. Its total kinetic energy will be :-

(1) E (2) 2E

(3) 3E (4) 4E

Q.104 One hollow and one solid cylinder of the same outer

radius rolls down on a rough inclined plane. The foot

of the inclined plane is reached by

(1) solid cylinder earlier

(2) hollow cylinder earlier

(3) simultaneously

(4) the heavier earlier irrespective of being solid or

hollow

Q.105 If a solid sphere, disc and cylinder are allowed to roll

down an inclined plane from the same height

(1) cylinder will reach the bottom first

(2) disc will reach the bottom first

(3) sphere will reach the bottom first

(4) all will reach the bottom at the same time

Q.106 A solid sphere rolls down without slipping twodifferent inclined planes of the same height but ofdifferent inclinations(1) in both cases the speeds and time of descend willbe same(2) the speeds will be same but time of descend will bedifferent(3) the speeds will be different but time of descend willbe same(4) speeds and time of descend both will be different

Q.107 A solid homogeneous sphere is moving on a roughhorizontal surface, partly rolling and partly sliding.During this kind of motion of the sphere(1) total kinetic energy is conserved(2) angular momentum of the sphere about the point ofcontact with the plane is conserved(3) only the rotational kinetic energy about the centreof mass is conserved(4) angular momentum about centre of mass isconversed

Q.108 A ring of mass M is kept on a horizontal rough surface.A force F is applied tangentially at its rim as shown.The coefficient of friction between the ring and thesurface is . Then

F

f(1) friction will act in the forward direction(2) friction will act in the backward direction(3) frictional force will not act(4) frictional force will be Mg.

Q.109 A disc is rolling on an inclined plane without slippingthen what fraction of its total energy will be in formof rotational kinetic energy :(1) 1 : 3 (2) 1 : 2 (3) 2 : 7 (4) 2 : 5

Q.110 A ring takes time t1 and t

2 for sliding down and rolling

down an inclined plane of length L respectively forreaching the bottom. The ratio of t

1 and t

2 is :-

(1) 2 : 1 (2) 1 : 2 (3) 1 : 2 (4) 2 : 1

Q.111 A ladder rests against a frictionless vertical wall, with itsupper end 6m above the ground and the lower end 4maway from the wall. The weight of the ladder is 500 N andits C. G. at 1/3rd distance from the lower end. Wall’sreaction will be, (in Newton)(1) 111 (2) 333 (3) 222 (4) 129

EXERCISE-II

CENTRE OF MASSQ.1 A thin uniform wire is bent to form the two equal sides

AB and AC of triangle ABC, where AB = AC = 5 cm.The third side BC, of length 6cm, is made from uniformwire of twice the density of the first. The distance ofcentre of mass from A is :

(1) 34

11cm (2)

11

34 cm (3)

34

9cm (4)

11

45 cm

Q.2 The centre of mass of a system of particles is at theorigin. From this we conclude that(1) the number of particles on positive x-axis is equal tothe number of particles on negative x-axis(2) the total mass of the particles on positive x-axis issame as the total mass on negative x-axis(3) the number of particles on X-axis may be equal tothe number of particles on Y-axis.(4) if there is a particle on the positive X-axis, theremust be at least one particle on the negative X-axis.

Q.3 A body of mass 1 kg moving in the x-direction,

suddenly explodes into two fragments of mass 1

8kg

and 7

8kg. An instant later, the smaller fragment is 0.14

m above the x-axis. The position of the heavier fragmentis -

(1) 1

50 m above x-axis (2)

1

50 m below x-axis

(3) 7

50 m below x-axis (4)

7

50 m above x-axis

Q.4 A man of mass M stands at one end of a plank of lengthL which lies at rest on a frictionless surface. The manwalks to other end of the plank. If the mass of the plank

is M

3, then the distance that the man moves relative to

ground is :

(1) 3L

4(2)

L

4(3)

4L

5(4)

L

3

Q.5 A particle of mass 3m is projected from the ground atsome angle with horizontal. The horizontal range is R.At the highest point of its path it breaks into two piecesm and 2m. The smaller mass comes to rest and largermass finally falls at a distance x from the point ofprojection where x is equal to

(1) 3R

4(2)

3R

2(3)

5R

4(4) 3R

Q.6 Two particles of mass 1 kg and 0.5 kg are moving in thesame direction with speed of 2m/s and 6m/s respectivelyon a smooth horizontal surface. The speed of centre ofmass of the system is :

(1) 10

3 m/s (2)

10

7 m/s (3)

11

2 m/s (4)

12

3 m/s

Q.7 Two particles having mass ratio n : 1 are interconnectedby a light inextensible string that passes over a smoothpulley. If the system is released, then the accelerationof the centre of mass of the system is :

(1) (n – 1)2 g (2)

2n 1

gn 1

(3)

2n 1

gn 1

(4) n 1

gn 1

Q.8 Internal forces in a system can change(1) linear momentum only(2) kinetic energy only(3) both kinetic energy and linear momentum(4) neither the linear momentum nor the kinetic energyof the system.

Q.9 A man of mass 'm' climbs on a rope of length Lsuspended below a balloon of mass M. The balloon isstationary with respect to ground. If the man begins toclimb up the rope at a speed vrel (relative to rope). Inwhat direction and with what speed (relative to ground)will the balloon move?

(1) downwards, relmv

m M (2) upwards, relMv

m M

(3) downwards, relmv

M(4) downwards, rel(M m)v

M

Q.10 There are some passengers inside a stationary railwaycompartment. The track is frictionless. The centre ofmass of the compartment itself (without the passengers)is C

1, while the centre of mass of the compartment plus

passengers system is C2. If the passengers move about

inside the compartment along the track.(1) both C

1 and C

2 will move with respect to the ground

(2) neither C1 nor C

2 will move with respect to the ground

(3) C1 will move but C

2 will be stationary with respect to

the ground(4) C

2 will move but C

1 will be stationary with respect to

the ground

Q.11 A shell is fired from a canon with a velocity V at anangle with the horizontal direction. At the highestpoint in its path, it explodes into two pieces of equalmasses. One of the pieces come to rest. The speed ofthe other piece immediately after the explosion is(1) 3V cos (2) 2V cos

(3) 3

2 V cos (4) V cos

Q.12 A small sphere is moving at a constant speed in a verticalcircle. Below is a list of quantities that could be used todescribe some aspect of the motion of the sphereI - kinetic energyII - gravitational potential energyIII - momentumWhich of these quantities will change as this spheremoves around the circle ?(1) I and II only (2) I and III only(3) III only (4) II and III only

Q.13 A bomb at rest explodes into two parts of masses m1and m2 . If the momentums of the two parts be p1 andp2, then their kinetic energies will be in the ratio of-(1) m1 / m2 (2) m2 / m1 (3) p1 / p2 (4) p2 / p1

Q.14 A body of mass m collides against a wall with thevelocity and rebounds with the same speed. Itschange of momentum is-(1) 2 m (2) m (3) – m (4) 0

Q.15 A bomb initially at rest explodes by it self into threeequal mass fragments. The velocities of two fragments

are (3 i +2 j ) m/s and (– i – 4j ) m/s. The velocity of

the third fragment is (in m/s) -

(1) 2 i + 2 j (2) 2 i – 2 j

(3) – 2 i + 2 j (4) –2 i – 2 j

Q.16 A stone of mass m1 moving with a uniform speed vsuddenly explodes on its own into two fragments. Ifthe fragment of mass m2 is at rest, the speed of theother fragment is-

(1) 1

1 2

m v

(m m ) (2) 2

1 2

m v

(m m )

(3) 1

1 2

m v

(m m ) (4) 1

2

m v

m

Q.17 A nucleus of mass number A originally at rest emits -particle with speed v. The recoil speed of daughternucleus is :

(1) 4v

A 4(2)

4v

A 4(3)

v

A 4(4)

v

A 4

Q.18 A super-ball is to bounce elastically back and forthbetween two rigid walls at a distance d from each other.Neglecting gravity and assuming the velocity of super-ball to be v

0 horizontally, the average force being exerted

by the super-ball on each wall is :

(1) 20mv1

2 d(2)

20mv

d(3)

202mv

d(4)

204mv

d

Q.19 A force exerts an impulse I on a particle changing itsspeed from u to 2u. The applied force and the initialvelocity are oppositely directed along the same line.The work done by the force is

(1) 3

I u2

(2) 1

I u2

(3) Iu (4) 2Iu

Q.20 A particle of mass 4m which is at rest explodes intothree fragments. Two of the fragments each of mass mare found to move with a speed 'v' each in mutuallyperpendicular directions. The minimum energy releasedin the process of explosion is :(1) (2/3) mv2 (2) (3/2) mv2

(3) (4/3) mv2 (4) (3/4) mv2

Q.21 A bullet of mass m moving vertically upwardsinstantaneously with a velocity 'u' hits the hangingblock of mass 'm' and gets embedded in it. As shown inthe figure the height through which block rises afterthe collision, assume sufficient space above block :

m

mu

(1) u2/2g (2) u2/g (3) u2/8g (D) u2/4g

Q.22 In an inelastic collision-(1) momentum is conserved but kinetic energy is not(2) momentum is not conserved but kinetic energy isconserved(3) neighter momentum nor kinetic energy is conserved(4) both the momentum and kinetic energy are conserved

Q.23 In the arrangement shown, the pendulum on the left ispulled aside. It is then released and allowed to collidewith other pendulum which is at rest. A perfectly inelasticcollision occurs and the system rises to a height h/4 .The ratio of the masses (m1 / m2) of the pendulum is :

h

m1

m2

(1) 1 (2) 2 (3) 3 (4) 4

Q.24 Two perfectly elastic balls of same mass m are movingwith velocities u

1 and u

2. They collide elastically n

times. The kinetic energy of the system finally is :

(1) 21

1 mu

2 u(2)

2 21 2

1 m(u u )

2 u

(3) 2 21 2

1m(u u )

2 (4)

2 21 2

1mn(u u )

2

Q.25 A ball hits the floor and rebounds after an inelasticcollision . In this case-(1) the momentum of the ball just after the collision isthe same as that just before the collision(2) the mechanical energy of the ball remains the samein the collision(3) the total momentum of the ball and the earth isconserved.(4) the total energy of the ball and the earth isconserved

Q.26 Six steel balls of identical size are lined up along astraight frictionless groove. Two similar balls movingwith a speed V along the groove collide with this rowon the extreme left hand then (all collision are ellastic)-

v

Frictionless Groove

(1) all the balls will start moving to the right withspeed 1/8 each(2) all the six balls initially at rest will move on withspeed V/6 each and two identical balls will come torest(3) two balls from the extreme right end will move onwith speed V each and the remaining balls will remainat rest(4) one ball from the right end will move on with speed2V, the remaining balls will be at rest.

Q.27 A particle of mass m moves with velocity v0 = 20 m/sec

towards a large wall that is moving with velocity v = 5 m/sec. towards a particle as shown. If the particle collideswith the wall elastically, the speed of the particle justafter the collision is :

V0

V

(1) 30 m/s (2) 20 m/s(3) 25 m/s (4) 22 m/s

Q.28 A sphere of mass m moving with a constant velocityhits another stationary sphere of the same mass. If e isthe coefficient of restitution, then ratio of speed of thefirst sphere to the speed of the second sphere aftercollision will be :

(1) 1 e

1 e

(2) 1 e

1 e

(3) e 1

e 1

(4) e 1

e 1

Q.29 Three blocks are initially placed as shown in the figure.Block A has mass m and initial velocity v to the right.Block B with mass m and block C with mass 4 m are bothinitially at rest. Neglect friction. All collisions are elastic.The final velocity of block A is

4mmmA B C

V

(1) 0.60 v to the left (2) 1.4 v to the left(3) v to the left (4) 0.4 v to the left

Q.30 Two billiard balls undergo a head-on collision. Ball 1 istwice as heavy as ball 2. Initially, ball 1 moves with aspeed v towards ball 2 which is at rest. Immediatelyafter the collision, ball 1 travels at a speed of v/3 in thesame direction. What type of collision has occured ?(1) inelastic(2) elastic(3) completely inelastic(4) Cannot be determined from the information given

Q.31 An object of mass 5 kg and speed 10 ms-1 explodesinto two pieces of equal mass. One piece comes torest. The kinetic energy added to the system duringthe explosion is-(1) Zero. (2) 50 J.(3) 250 J. (4) 500 J.

Q.32 A block of mass m starts from rest and slides down africtionless semi-circular track from a height h as shown.When it reaches the lowest point of the track, it collideswith a stationary piece of putty also having mass m. Ifthe block and the putty stick together and continue toslide, the maximum height that the block-putty systemcould reach is

h

(1) h/4 (2) h/2(3) h (4) independent of h

ROTATIONAL MOTIONQ.33 A fan is running at 3000 rpm. It is switched off. It comes

to rest by uniformly decreasing its angular speed in 10seconds. The total number of revolutions in thisperiod.(1) 150 (2) 250 (3) 350 (4) 300

Q.34 The moment of inertia of a non-uniform semicircularwire having mass m and radius r about a lineperpendicular to the plane of the wire through thecentre is

(1) mr2 (2) 1

2mr2 (3)

1

4mr2 (4)

2

5mr2

Q.35 Let A and

B be the moments of inertia of two solid

cylinders of identical geometrical shape and size abouttheir axes, the first made of aluminium and the secondof iron.(1)

A <

B

(2) A =

B

(3) A >

B

(4) relation between A and

B depends on the actual

shapes of the bodies.

Q.36 A unifrom thin rod of length L and mass M is bent atthe middle point O as shown in figure. Consider anaxis passing through its middle point O andperpendicular to the plane of the bent rod. Thenmoment of inertia about this axis is :

O

(1) 2

3 mL2 (2)

1

3 mL2

(3) 1

12 mL2 (4) dependent on

Q.37 Two spheres of same mass and radius are in contactwith each other. If the moment of inertia of a sphereabout its diameter is I, then the moment of inertia ofboth the spheres about the tangent at their commonpoint would be -(1) 3I (2) 7I (3) 4I (4) 5I

Q.38 The M.I. of a disc about its diameter is 2 units. ItsM.I. about axis through a point on its rim and in theplane of the disc is(1) 4 units. (2) 6 units (3) 8 units (4) 10 units

Q.39 Moment of inertia of a thin semicircular disc (mass =M & radius = R) about an axis through point O andperpendicular to plane of disc, is given by :

R

O

(1) 21

MR4

(2) 21

MR2

(3) 21

MR8

(4) MR2

Q.40 A rigid body can be hinged about any point on thex-axis. When it is hinged such that the hinge is at x, themoment of inertia is given by I = 2x2 – 12x + 27 Thex-coordinate of centre of mass is(1) x = 2 (2) x = 0 (3) x = 1 (4) x = 3

Q.41 On applying a constant torque on a body-(1) linear velocity may be increases(2) angular velocity may be increases(3) it will rotate with constant angular velocity(4) it will move with constant velocity

Q.42 A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100 radian/sec in 15seconds. If moment of inertia is 10kg-m2, then appliedtorque (in newton-metre) is(1) 900 (2) 100 (3) 90 (4) 60

Q.43 A torque of 2 newton-m produces an angularacceleration of 2 rad/sec2 a body. If its radius ofgyration is 2m, its mass will be:(1) 2kg (2) 4 kg (3) 1/2 kg (4) 1/4 kg

Q.44 A uniform circular disc A of radius r is made from ametal plate of thickness t and another uniform circulardisc B of radius 4r is made from the same metal plate ofthickness t/4. If equal torques act on the discs A andB, initially both being at rest. At a later instant, theangular speeds of a point on the rim of A and anotherpoint on the rim of B are

A and

B respectively. We

have(1)

A >

B

(2) A =

B

(3) A <

B

(4) the relation depends on the actual magnitude ofthe torques.

Q.45 A force F

= ˆ ˆ4i 10 j acts on a body at a point having

position vector ˆ ˆ5i 3 j relative to origin of co-ordinates on the axis of rotation . The torque acting onthe body about the origin is :

(1) 38 k (2) 25 k (3) 62 k (4) none of these

Q.46 Four equal and parallel forces are acting on a rod (asshown in figure) at distances of 20 cm, 40 cm, 60 cmand 80 cm respectively from one end of the rod. Underthe influence of these forces the rod :

0 20

F

40

F

60 80

F

F

(1) is at rest(2) experiences a torque(3) experiences a linear motion(4) experiences a torque and also a linear motion

Q.47 The uniform rod of mass 20 kg and length 1.6 m ispivoted at its end and swings freely in the verticalplane. Angular acceleration of rod just after the rod isreleased from rest in the horizontal position as shownin figure is

(1) 15g

16 (2)

17g

16(3)

16g

15(4)

g

15

Q.48 Two men support a uniform horizontal rod at its twoends. If one of them suddenly lets go, the force exertedby the rod on the other man will:(1) remain unaffected(2) increase(3) decrease(4) become unequal to the force exerted by him on thebeam.

Q.49 A Cubical bloc of mass M and edge a slides down arough inclined plane of inclination with a uniformvelocity. The torque of the normal force on the blockabout its centre has a magnitude.(1) zero (2) Mga

(3) Mga sin (4) 1

Mga sin2

Q.50 A uniform cube of side a and mass m rests on a roughhorizontal table. A horizontal force 'F' is applied normalto one of the faces at a point that is directly above the

centre of the face, at a height 3a

4 above the base. The

minimum value of 'F' for which the cube begins to tiltabout the edge is (assume that the cube does not slide).

(1) 2

mg3

(2) 4

mg3

(3) 5

mg4

(4) 1

mg2

Q.51 The moment of inertia and rotational kinetic energyof a fly wheel are 20kg-m2 and 1000 joule respectively.Its angular frequency per minute would be -

(1) 600

(2) 2

25

(3)

5

(4)

300

Q.52 A circular ring of wire of mass M and radius R ismaking n revolutions/sec about an axis passingthrough a point on its rim and perpendicular to itsplane. The kinetic energy of rotation of the ring isgiven by-(1) 42MR2n2 (2) 22MR2n2

(3) 1

22MR2n2 (4) 82MR2n2

Q.53 Rotational kinetic energy of a disc of constantmoment of inertia is -(1) directly proportional to angular velocity(2) inversely proportional to angular velocity(3) inversely proportional to square of angularvelocity(4) directly proportional to square of angular velocity

Q.54 A constant torque acting on a uniform circular wheelchanges its angular momentum from A

0 to 4A

0 in 4 sec.

the magnitude of this torque is :(1) 4A

0(2) A

0(3) 3A

0 /4 (4) 12A

0

Q.55 A particle moves with a constant velocity parallel tothe Y-axis. Its angular momentum about the origin.(1) is zero(2) remains constant(3) goes on increasing(4) goes on decreasing.

Q.56 A boy sitting firmly over a rotating stool has his armsfolded. If he stretches his arms, his angular momentumabout the axis of rotation(1) increases (2) decreases(3) remains unchanged (4) doubles

Q.57 The rotational kinetic energy of a rigid body ofmoment of inertia 5 kg-m2 is 10 joules. The angularmomentum about the axis of rotation would be -(1) 100 joule-sec (2) 50 joule-sec(3) 10 joule-sec (4) 2 joule -sec

Q.58 The angular velocity of a body changes from onerevolution per 9 second to 1 revolution per secondwithout applying any torque. The ratio of its radiusof gyration in the two cases is(1) 1 : 9 (2) 3 : 1 (3) 9 : 1 (4) 1 : 3

Q.59 A disc rolls on a table. The ratio of its K.E. of rotationto the total K.E. is -(1) 2/5 (2) 1/3(3) 5/6 (4) 2/3

Q.60 A thin string is wrapped several times around a cylinderkept on a rough horizontal surface. A boy standing ata distance from the cylinder draws the string towardshim as shown in figure. The cylinder rolls withoutslipping. The length of the string passed through thehand of the boy while the cylinder reaches his hand is

(1) (2) 2 (3) 3 (4) 4

Q.61 A solid sphere, a hollow sphere and a ring, all havingequal mass and radius, are placed at the top of anincline and released. The friction coefficients betweenthe objects and the incline are equal and but notsufficient to allow pure rolling. The greastest kineticenergy at the bottom of the incline will be achieved by(1) the solid sphere(2) the hollow sphere(3) the ring(4) all will achieve same kinetic energy.

Q.62 A body is given translational velocity and kept on asurface that has sufficient friction. Then:(1) body will move forward before pure rolling(2) body will move backward before pure rolling(3) body will start pure rolling immediately(4) none of these

Q.63 A disk and a ring of the same mass are rolling tohave the same kinetic energy. What is ratio of theirvelocities of centre of mass(1) (4:3)1/2 (2) (3 : 4)1/2

(3) (2)1/2 : (3) 1/2(4) (3)1/2 : (2)1/2

Q.64 A body kept on a smooth horizontal surface is pulledby a constant horizontal force applied at the top pointof the body. If the body rolls purely on the surface, itsshape can be :(1) thin pipe (2) uniform cylinder(3) uniform sphere (4) thin spherical shell

Q.65 A solid sphere with a velocity (of centre of mass) vand angular velocity is gently placed on a roughhorizontal surface. The frictional force on the sphere :(1) must be forward (in direction of v)(2) must be backward (opposite to v)(3) cannot be zero(4) none of the above

Q.66 A cylinder is pure rolling up an incline plane. It stopsmomentarily and then rolls back. The force of friction.(1) on the cylinder is zero throughout the journey(2) is directed opposite to the velocity of the centre ofmass throughout the journey(3) is directed up the plane throughout the journey(4) is directed down the plane throughout the journey

Q.67 A sphere is released on a smooth inclined plane fromthe top. When it moves down its angular momentumis:(1) conserved about every point(2) conserved about the point of contact only(3) conserved about the centre of the sphere only(4) conserved about any point on a fixed line parallelto the inclined plane and passing throughthe centre of the ball.

JEE-ADVANCEDCOMPREHENSION/STATEMENT/MATCHING/MCQ ]

CENTRE OF MASSQ.1 A system of particles has its centre of mass at the origin.

The x-coordinates of all the particles(A) may be positive (B) may be negative(C) may be non-negative (D) may be non-positive

Q.2 An object comprises of a uniform ring of radius R andits uniform chord AB (not necessarily made of the samematerial) as shown. Which of the following can not bethe centre of mass of the object.

Ax

y

B

(A) (R/3, R/3) (B) (R3, R/2)

(C) (R/4, R/4) (D) ( / , / )R R2 2

Q.3 If the net external forces acting on a system is zero,then the centre of mass(A) must not move (B) must not accelerate(C) may move (D) may accelerate

Q.4 An external force F (F 0)

acts on a system of

particles. The velocity and the acceleration of the centreof mass are found to be v

cm and a

cm , then it is possible

that(A) v

cm = 0, a

cm = 0 (B) v

cm = 0, a

cm 0

(C) vcm

0, acm

= 0 (D) vcm

0, acm

0Q.5 Two blocks A and B each of mass ‘m’ are connected by

a massless spring of natural length L and spring constantk. The blocks are initially resting on a smooth horizontalplank block C also of mass m moves on the floor with aspeed ‘v’ along the line joining A and B and collideselastically with A then which of the following is/arecorrect :(A) K

E of the AB system at maximum compression of

the spring is zero(B) The KE of AB system at maximum compression is(1/4) mv2

(C) The maximum compression of spring is v m / k

(D) The maximum compression of spring is v m / 2k

Q.6 In an elastic collision in absence of external force, whichof the following is/are correct :(A) The linear momentum is conserved(B) The potential energy is conserved in collision(C) The final kinetic energy is less than the initial kineticenergy(D) The final kinetic energy is equal to the initial kineticenergy

Q.7 A block moving in air explodes in two parts then justafter explosion (neglect change in momentum duet togravity)(A) the total momentum of two parts must be same tothe momentum of the block before explosion.(B) the total kinetic energy of two parts must be sameas that of block before explosion.(C) the total momentum must change(D) the total kinetic energy must increase

Q.8 A set of n-identical cubical blocks lie at rest parallel toeach other along a line on a smooth horizontal surface.The separation between the near surfaces of any twoadjacent blocks is L. The block at one end is given aspeed V towards the next one at time t = 0. All collisionsare completely inelastic, then

(A) The last block starts moving at t = n (n – 1)

L

2V

(B) The last block starts moving at t = (n – 1)

L

V(C) The centre of mass of the system will have a finalspeed v/n(D) The centre of mass of the system will have a finalspeed v

Q.9 The fig. shows a string of equally spaced beads ofmass m, separated by distance d. The beads are free toslide without friction on a thin wire. A constant force Facts on the first bead initially at rest till it makes collisionwith the second bead. The second bead then collideswith the third and so on. Suppose all collisions areelastic, then :

(A) speed of the first bead immediately before andimmediately after its collision with the second bead is

m

Fd2 and zero respectively..

EXERCISE-III

(B) speed of the first bead immediately before andimmediately after its collision with the second bead is

m

Fd2 and

1

22Fd

m respectively..

(C) speed of the second bead immediately after itscollision with third bead is zero.

(D) the average speed of the first bead is 1

22Fd

m.

Q.10 A shell explodes in a region of negligible gravitationalfield, giving out n fragments of equal mass m. Then itstotal(A) Kinetic energy is smaller than that before the explosion(B) Kinetic energy is greater than that before the explosion(C) Momentum and kinetic energy depend on n(D) Momentum is equal to that before the explosion.

Q.11 Two identical balls are interconnected with a masslessand inextensible thread. The system is in gravity freespace with the thread just taut. Each ball is imparted avelocity v, one towards the other ball and the otherperpendicular to the first, at t = 0. Then,(A) the thread will become taut at t = (L/v)(B) the thread will become taut at some time t < (L/v).(C) the thread will always remain taut for t > (L/v)(D) the kinetic energy of the system will always remainmv2.

Q.12 A particle moving with kinetic energy = 3 joule makesan elastic head on collision with a stationary particlewhen has twice its mass during the impact.(A) The minimum kinetic energy of the system is 1 joule(B) The maximum elastic potential energy of thesystem is 2 joule.(C) Momentum and total kinetic energy of the systemare conserved at every instant.(D) The ratio of kinetic energy to potential energy ofthe system first decreases and then increases.

Q.13 In an inelastic collision,(A) the velocity of both the particles may be same aftercollision.(B) kinetic energy is not conserved(C) linear momentum of the system is conserved.(D) velocity of separation will be less than velocity ofapproach.

ROTATIONAL MOTION Q.14 The moment of inertia of a thin uniform square plate

ABCD of uniform thickness about an axis passingthrough the centre O and perpendicular to the plate is

O3

B1A

CD 2

4

where 1 ,

2 ,

3 , and

4 are respectively the moments of

inertia about axes 1, 2, 3, and 4 which are in the planeof the plate.(A)

1 +

2 (B)

3 +

4

(C) 1 +

3(D)

1 +

2 +

3 +

4

Q.15 A block with a square base measuring a a and heighth, is placed on an inclined plane. The coefficient offriction is . The angle of inclination () of the plane isgradually increased. The block will:

(A) topple before sliding if > a

h

(B) topple before sliding if < a

h

(C) slide before toppling if > a

h

(D) slide before toppling if < a

h

Q.16 A rod of weight w is supported by two parallel knifeedges A and B and is in equilibrium in a horizontalposition. The knives are at a distance d from each other.The centre of mass of the rod is at a distance x from A.

(A) the normal reaction at A is wx

d

(B) the normal reaction at A is w d x

d

( )

(C) the normal reaction at B is wx

d

(D) the normal reaction at B is w d x

d

( )

Q.17 A body is in equilibrium under the influence of a numberof forces. Each force has a different line of action. Theminimum number of forces required is(A) 2, if their lines of action pass through the centre ofmass of the body(B) 3, if their lines of action are not parallel(C) 3, if their lines of action are parallel(D) 4, if their lines of action are parallel and all theforces have the same magnitude

Q.18 Four point masses are fastened to the corners of aframe of negligible mass lying in the xy plane. Let w bethe angular speed of rotation. Then

y-axis

m

Mm

M x-axis

z-axis a

b

(A) rotational kinetic energy associated with a givenangular speed depends on the axis of rotation.(B) rotational kinetic energy about y-axis isindependent of m and its value is Ma22

(C) rotational kinetic energy about z-axis depends onm and its value is (Ma2 + mb2)2

(D) rotational kinetic energy about z-axis isindependent of m and its value is Mb22

Q.19 In the given figure a ball strikes a rod elastically androd is hinged at point A. Then which of thestatement(s) is/are correct for the collision?

(A) linear momentum of system (ball + rod) isconserved(B) angular momentum of system about hinged pointA is conserved(C) initial KE of the system is equal to final KE of thesystem(D) linear momentum of ball is conserved.

Q.20 A particle falls freely near the surface of the earth.Consider a fixed point O (not vertically below theparticle) on the ground.(A) Angular momentum of the particle about O isincreasing(B) Torque of the gravitational force on the particleabout O is decreasing(C) The moment of inertia of the particle about O isdecreasing(D) The angular velocity of the particle about O isincreasing

Q.21 A man spinning in free space changes the shape of hisbody, eg. by spreading his arms or curling up. By doingthis, he can change his(A) moment of inertia (B) angular momentum(C) angular velocity (D) rotational kinetic energy

Q.22 When a bicycle is in motion (accelerating) on a roughplane, the force of friction exerted by the plane on thetwo wheels is such that it acts :

(A) In the backward direction on the front wheel andin the forward direction on the rear wheel(B) In the forward direction on the front wheel and inthe backward direction on the rear wheel(C) In the backward direction on both front and therear wheels(D) In the forward direction on both the front and therear wheels

Q.23 A ring rolls without slipping on the ground. Its centreC moves with a constant speed u. P is any point on thering. The speed of P with respect to the ground is v.(A) 0 v 2u(B) v = u, if CP is horizontal(C) v = u, if CP makes an angle of 30º with the horizontaland P is below the horizontal level of C

(D) v u 2 , if CP is horizontal

Q.24 Consider a sphere of mass ‘m’ radius ‘R’ doing pure

rolling motion on a rough surface having velocity v0

as shown in the Figure. It makes an elastic impact withthe smooth wall and moves back and starts pure rollingafter some time again.

v0

O

(A) Change in angular momentum about ‘O’ in theentire motion equals 2mv

0 R in magnitude.

(B) Moment of impulse provided by wall during impactabout O equals 2mv

0R in magnitude

(C) Final velocity of ball will be 3

7 0

v

(D) Final velocity of ball will be 3

7 0

v

CENTRE OF MASS Comprehension # 1 (Q. No. 25 to 27)

Two blocks of equal mass m are connected by anunstretched spring ànd the system is kept at rest on africtionless horizontal surface. A constant force F isapplied on the first block pulling it away from the otheras shown in figure.

Q.25 Then the displacement of the centre of mass at time tis

(A)

2Ft

2m (B)

2Ft

3m (C) 2Ft

4m(D)

2Ft

m

Q.26 If the extension of the spring is x0 at time t, then the

displacement of the right block at this instant is :

(A) 1

2

2

0

Ftx

2m

(B) –

1

2

2

0

Ftx

2m

(C) 1

2

2

0

Ftx

2m

(D)

2

0

Ftx

2m

Q.27 If the extension of the spring is x0 at time t, then the

displacement of the left block at this instant is :

(A)

2

0

Ftx

2m

(B)

1

2

2

0

Ftx

2m

(C) 1

2

2

0

2Ftx

m

(D)

1

2

2

0

Ftx

2m

Comprehension # 2 (Q. No. 28 to 34)A particle of mass m moving horizontal with v

0 strikes a

smooth wedge of mass M, as shown in figure. Aftercollision, the ball starts moving up the inclined face ofthe wedge and rises to a height h.

m v0

h

M

Q.28 The final velocity of the wedge v2 is

(A) 0mv

M(B) 0mv

M m(C) v

0(D) insufficient data

Q.29 When the particle has risen to a height h on the wedge,then choose the correct alternative(s)(A) The particle is stationary with respect to ground(B) Both are stationary with respect to the centre ofmass(C) The kinetic energy of the centre of mass remainsconstant(D) The kinetic energy with respect to centre of mass isconverted into potential energy

Q .30 The maximum height h attained by the particle is

(A)

20vm

m M 2g

(B)

20vm

M 2g

(C)

20vM

m M 2g

(D) none of these.

Q.31 Identify the correct statement(s) related to the situationwhen the particle starts moving downward.(A) The centre of mass of the system remains stationary(B) Both the particle and the wedge remain stationarywith respect to centre of mass(C) When the particle reaches the horizontal surface itsvelocity relative to the wedge is v

0

(D) None of these

Q.32 Suppose the particle when reaches the horizontalsurfaces, its velocity with respect to ground is v

1 and

that of wedge is v2. Choose the correct statement (s)

mv1 M

v2

(A) mv1 = Mv

2(B) Mv

2 – mv

1 = mv

0

(C) v1 + v

2 = v

0(D) v

1 + v

2 < v

0

Q.33 Choose the correct statement(s) related to particle m

(A) Its kinetic energy is f

mMK gh

m M

(B) 1 0

M mv v

M m

(C) The ratio of its final kinetic energy to its initial kinetic

energy is

2

f

i

K M

K m M

(D) It moves opposite to its initial direction of motion

Q.34 Choose the correct statement related to the wedge M

(A) Its kinetic energy is 2

f

4mK gh

m M

(B) 2 0

2mv v

m M

(C) Its gain in kinetic energy is

202

4mM 1K mv

2(m M)

(D) Its velocity is more that the velocity of centre ofmass

Comprehension # 3 (Q. No. 35 to 38)A small ball B of mass m is suspended with light inelasticstring of length L from a block A of same mass m whichcan move on smooth horizontal surface as shown inthe figure. The ball is displaced by angle fromequilibrium position & then released.

L

L

B

u=0

A

Q.35 The displacement of block when ball reaches theequilibrium position is

(A) Lsin

2

(B) L sin

(C) L (D) none of these

Q.36 Tension in string when it is vertical, is(A) mg (B) mg(2 – cos)(C) mg (3 – 2 cos) (D) none of these

Q.37 Maximum velocity of block during subsequent motionof the system after release of ball is(A) [gl(1 – cos )]1/2

(B) [2gl(1 – cos )]1/2

(C) [glcos]1/2

(D) informations are insufficient to decide

Q.38 The displacement of centre of mass of A + B system tillthe string becomes vertical is

(A) zero (B) L

(1 – cos )2

(C) L

(1– sin )2

(D) none of these

ROTATIONAL MOTIONComprehension # 04 (Q. No. 39 to 41)

A uniform disc of mass ‘m’ and radius R is free to rotatein horizontal plane about a vertical smooth fixed axispassing through its centre. There is a smooth groovealong the diameter of the disc and two small balls of

mass 2

m each are placed in it on either side of the centre

of the disc as shown in fig. The disc is given initialangular velocity

0 and released.

Q.39 The angular speed of the disc when the balls reach theend of the disc is :

(A) 20 (B)

30

(C) 3

2 0(D)

40

Q.40 The speed of each ball relative to ground just afterthey leave the disc is :

(A) 3

R 0(B)

2

R 0

(C) 3

R2 0 (D)

3

R 0

Q.41 The net work done by forces exerted by disc on one ofthe ball for the duration ball remains on the disc is

(A) 9

mR2 20

2(B)

18

mR 20

2

(C) 6

mR 20

2(D)

9

mR 20

2

Comprehension # 05 (Q. No. 42 to 44)A uniform disc of mass M and radius R initially standsvertically on the right end of a horizontal plank of massM and length L, as shown in the figure.The plank rests on smooth horizontal floor and frictionbetween disc and plank is sufficiently high such thatdisc rolls on plank without slipping. The plank is pulledto right with a constant horizontal force of magnitudeF.

Q.42 The magnitude of acceleration of plank is -

(A) M8

F(B)

M4

F

(C) M2

F3(D)

M4

F3

Q.43 The magnitude of angular acceleration of the disc is -

(A) mR4

F(B)

mR8

F

(C) mR2

F(D)

mR2

F3

Q.44 The distance travelled by centre of disc from its initialposition till the left end of plank comes vertically belowthe centre of disc is

(A) 2

L(B)

4

L

(C) 8

L(D) L

Comprehension # 06 (Q. No. 45 to 48)The figure shows an isosceles triangular plate of massM and base L. The angle at the apex is 90°. The apexlies at the origin and the base is parallel to X - axis.

Y

M

X

L

Q.45 The moment of inertia of the plate about the z-axis is

(A) ML2

12(B)

ML2

24

(C) ML2

6(D) none of these

Q.46 The moment of inertia of the plate about the x-axis is

(A) ML2

8(B)

ML2

32

(C) ML2

24(D)

ML2

6Q.47 The moment of inertia of the plate about its base parallel

to the x-axis is

(A) ML2

18(B)

ML2

36

(C) ML2

24(D) none of these

Q.48 The moment of inertia of the plate about the y-axis is

(A) ML2

6(B)

ML2

8

(C) ML2

24(D) none of these

CENTRE OF MASSQ.49 Two blocks A and B of mass m and 2m respectively are connected by a massless spring of spring constant K. This system lies

over a smooth horizontal surface. At t = 0 the block A has velocity u towards right as shown while the speed of block B is zero,and the length of spring is equal to its natural length at that instant. In each situation of column I, certain statements are givenand corresponding results are given in column II. Match the statements in column corresponding results in column .

Column I Column II(A) The velocity of block A (p) can never be zero(B) The velocity of block B (q) may be zero at certain instants of time(C) The kinetic energy of system of two blocks (r) is minimum at maximum compression of spring(D) The potential energy of spring (s) is maximum at maximum extension of spring

ROTATIONAL MOTIONQ.50 In each situation of column-I, a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface as shown in

the figure. At t=0 ( initially) the angular velocity of disc is and velocity of centre of mass of disc is v(in horizontaldirection). The relation between v and for each situation and also initial sense of rotation is given for each situation incolumn-I . Then match the statements in column-I with the corresponding results in column-II .

Column-I Column-II

(A) (v > R ) (p) The angular momentum of disc about point A

(as shown in figure) remains conserved.

(B) (v > R ) (q) The kinetic energy of disc after it starts rolling

without slipping is less than its initial kinetic energy.

(C) (v < R ) (r) In the duration disc rolls with slipping, the

friction acts on disc towards left.

(D) (v < R ) (s) In the duration disc rolls with slipping, the friction acts on disc

for some time towards right and for some time towards left.

Q.51 A uniform disc rolls without slipping on a rough horizontal surface with uniform angular velocity. Point O is the centre ofdisc and P is a point on disc as shown in the figure. In each situation of column a statement is given and the correspondingresults are given in column-II. Match the statements in column-I with the results in column-II.

Column Column (A) The velocity of point P on disc (p) Changes in magnitude with time.(B) The acceleration of point P on disc (q) is always directed from that point (the point on disc given in column-

I) towards centre of disc.(C) The tangential acceleration of point (r) is always zero.

P on disc(D) The acceleration of point on disc which (s) is non-zero and remains constant is in contact with rough horizontal surface in magnitude.

NUMERICAL BASED QUESTIONSCENTER OF MASSQ.52 A 2kg mass and 3 kg mass are used to compress oppoite

ends of a spring (k = 750 N/m) by a distance of 40 cmfrom natural length and released from rest. If the speeds

of the two masses as they leave the spring are v2 andv3, find v2 + v3 (in m/s).

2kg 3kg

Smooth

Q.53 For years together, people thought that hell is locatedinside earth. Assume that hell is a spherical bubblecreated inside earth. It is completely evacuated to makesinful people feel suffocated. It has a radius R/4 and islocated as shown. (C is earth's centre). What is thedistance of centre of mass of the resulting body from C(in km). Assume that earth's radius is 6300 km.

RR/4

C

Q.54 A bullet of mass m and velocity v passes through apendulum bob of mass M and emerges with velocity v/2 (figure). The pendulum bob is at the end of a string oflength l. What is the minimum value of v (in m/s) suchthat the pendulum bob will swing through a completecircle ? (Take : l = 2m, M = 1kg, m = 10 gm.)

v v/2

O

l

Q.55 Two particles of masses m and 4m, moving in vacuum

at right angles to each other experience same force F

for time T simultaneously. Consequently the particle mmoves with velocity 4v in its original direction. Findthe new magnitude of the velocity v' (in m/s) of theparticle 4m. Given v = 100 m/s.

mv

m4v

4m

v

vBefore

After

y

x

Q.56 A block of mass 2kg is sliding on a smooth surface. At

t = 0, its speed is v0 = 2 m/s. At t = 0, a time - varying

force starts acting on the block in the direction opposite

to v0. Find the speed of object (in m/s) at t = 10

2.0 kgv0

0 2 4 6 8 10036

Forc

e (N

)

Time (s)

F

ROTATIONAL MOTIONQ.57 A merry-go-round is a common piece of playground

equipment. A 4 m diameter merry-go-round with amoment of inertia of 500 kg-m2 is spinning at 5.6 rad/s.John runs tangent to the merry-go-round at 5 m/s, inthe same direction that it is turning, and jumps onto theouter edge. John's mass is 30 kg. What is the merry-go-round's angular velocity, (in rad/s), after John jumpson ?

Q.58 A solid Hemisphere rests in equilibrium on a roughground and against a smooth wall. The curved surfacetouches the wall and the ground. The angle ofinclination of the circular base to the horizontal is 30°,Find minimum coefficient of friction required betweenground and hemisphere. Fill 32µ in OMR sheet.

Q.59 The figure shown is in equilibrium. Find out theextension in the spring (in cm).

0.3m

0.6m

K=100 N/m

40Kg

70Kg A

B

Q.60 A wheel of radius 0.20 m is mounted on a frictionlesshorizontal axis. The rotational inertia of the wheel aboutthe axis is 0.07 kgm2. A massless cord wrapped aroundthe wheel is attached to a 2.0 kg block that slides on ahorizontal frictionless surface. If a horizontal force ofmagnitude P = 3.0 N is applied to the block as shown infigure, what is the magnitude of the angular acceleration(in r/s2) of the wheel? Assume the string does not slipon the wheel.

Q.61 A force of 100 N is applied on a disc along the tangentfor 10 sec. During this time, the disc attains an angularvelocity of 10 rad/s. There is constant frictional torqueat the axis which opposes the rotation of the disc. Nowthe force is removed. After what time (in sec) from theremoval of force will the disc come to rest.

m=10kgR=10m

Q.62 The drawing shows the top view of two doors. Thedoors are uniform and identical. Door A rotates aboutan axis through its left edge, while door B rotates aboutan axis through the center. The same force F is appliedperpendicular to each door at its right edge, and theforce remains perpendicular as the door turns. Startingfrom rest, door A rotates through a certain angle in 3 s.How long does it take door B to rotate through thesame angle ? Round off to nearest integer.

F

F

Door A

Door B

Axis

Axis

Q.63 A yo-yo-shaped device mounted on a horizontalfrictionless axis is used to lift a 30kg box as shown infigure. The outer radius R of the device is 0.50 m, an theradius r of the hub is 0.20 m. When a constant horizontalforce of magnitude 152 N is applied in the left directionto a rope wrapped around the outside of the device,the box, which is suspended from a rope wrappedaround the hub, has an upward acceleration ofmagnitude 0.80 m/s2. What is the rotational inertia I (inkg-m2) of the device about its axis of rotation?

r

R

Q.64 Seven pennies are arranged in a hexagonal, planarpattern so as to touch each neighbor, as shown inthe figure below. Each penny is a uniform disk ofmass m = 2kg and radius r = 1m. What is the momentof inertia of the system (in kg-m2) of seven penniesabout an axis that passes through the center of thecentral penny and is normal to the plane of thepennies ?

Q.65 A ring of mass m = 1 kg and radius R = 1.25 m is kept ona rough horizontal ground. A small body of mass m isstruck to the top of the ring. When it was given a slightpush forward, the ring started rolling purely on theground. What is the maximum speed of the centre ofthe ring (in m/s)?

EXERCISE-IV

JEE-MAINPREVIOUS YEAR'SCENTRE OF MASSQ.1 It is found that if a neutron suffers an elastic collinear

collision with deuterium at rest, fractional loss of itsenergy is p

d : while for its similar collision with carbon

nucleus at rest, fractional loss of energy is pc . The

values of pd and p

c are respectively :

[JEE Main-2018](1) (·28, ·89) (2) (0, 0)(3) (0, 1) (4) (·89, ·28)

Q.2 In a collinear collision, a particle with an initial speed v0

strikes a stationary particle of the same mass. If thefinal total kinetic energy is 50 % greater than the original

kinetic energy, the magnitude of the relative velocitybetween the two particles, after collision, is

[JEE Main-2018]

(1) 02 v (2) 0v

2(3)

0v

2(4) 0v

4

Q.3 The mass of a hydrogen molecule is 3.32 × 10–27 kg. If1023 hydrogen molecules strike, per second, a fixed wallof area 2 cm2 at an angle of 45° to the normal, andrebound elastically with a speed of 103 m/s, then thepressure on the wall is nearly :

[JEE Main-2018](1) 4.70 × 103 N/m2 (2) 2.35 × 102 N/m2

(3) 4.70 × 102 N/m2 (4) 2.35 × 103 N/m2

Q.4 A particle of mass m is moving in a straight line withmomentum p. Starting at time t = 0, a force F = kt acts inthe same direction on the moving particle during timeinterval T so that its momentum changes from p to 3p.Here k is a constant. The value of T is :

[JEE Main - 2019 (January)]

(1) k

2p (2)

p2

k(3)

2k

p (4) 2p

k

Q.5 Three blocks A, B and C are lying on a smoothhorizontal surface, as shown in the figure. A and Bhave equal masses, m while C has mass M. Block A isgiven an initial speed v towards B due to which itcollides with B perfectly inelastically. The combined

mass collides with C, also perfectly inelastically if 5

6th

of the initial kinetic energy is lost in whole process.What is value of M/m ? [JEE Main - 2019 (January)]

m m MA B C

(1) 5 (2) 2 (3) 4 (4) 3

Q.6 A piece of wood of mass 0.03 kg is dropped from thetop of a 100 m height building. At the same time, abullet of mass 0.02 kg is fired vertically upward, with avelocity 100 ms–1, from the ground. The bullet getsembedded in the wood. Then the maximum height towhich the combined system reaches above the top ofthe building before falling below is (g = 10 ms–2)

[JEE Main-2019 (January)](1) 20 m (2) 30 m (3) 40 m (4) 10 m

Q.7 A particle of mass ‘m’ is moving with speed ‘2v’ andcollides with a mass ‘2m’ moving with speed ‘v’ in thesame direction. After collision, the first mass is stoppedcompletely while the second one splits into twoparticles each of mass ‘m’, which move at angle 45ºwith respect to the original direction. The speed of eachof the moving particle will be :

[JEE Main-2019 (April)]

(1) v / 2 2 (2) 2 2v

(3) 2v (4) v / 2

Q.8 A wedge of mass M = 4m lies on a frictionless plane. Aparticle of mass m approaches the wedge with speed v.There is no friction between the particle and the planeor between the particle and the wedge. The maximumheight climbed by the particle on the wedge is given by:- [JEE Main-2019 (April)]

(1) 22v

7g(2)

2v

g(3)

22v

5g(4)

2v

2g

Q.9 Four particles A, B, C and D with masses mA= m, m

B =

2m, mC = 3m and m

D = 4m are at the corners of a

square. They have accelerations of equal magnitudewith directions as shown. The acceleration of thecentre of mass of the particles is :


A

B C

D

Y

X

aa

a

a

(1) a ˆ î j5

(2) a ˆ î j5

(3) Zero (4) ˆ â i j

Q.10 If 1022 gas molecules each of mass 10–26 kg colide witha surface (perpendicular to it) elastically per secondover an area 1 m2 with a speed 104 m/s, the pressureexerted by the gas molecules will be of the order of:

[JEE Main-2019 (April)](1) 108 N/m2 (2) 104 N/m2

(3) 103 N/m2 (4) 1016 N/m2

Q.11 A uniform rectangular thin sheet ABCD of mass M haslength a and breadth b, as shown in the figure. If theshaded portion HBGO is cut-off, the coordinates of thecentre of mass of the remaining portion will be :-


a2

b2

D(0, 0)

C(a, 0)F

(0, b)A

H (a, b)B

GO

E

(1) 2a 2b

,3 3

(2)

5a 5b,

3 3

(3) 3a 3b

,4 4

(4)

5a 5b,

12 12

Q.12 A body of mass m1 moving with an unknown velocity

of v1 i , undergoes a collinear collision with a body of

mass m2 moving with a velocity v

2 i , After colision, m1

and m2 move with velocities of v

3 i and v4 i ,

respectively. If m2 = 0.5 m

1 and v

3 = 0.5 v

1, then v

1 is :-


(1) v4 –

2v

4(2) v

4 –

2v

2(3) v

4 – v

2(4) v

4 + v

2

Q.13 A body of mass 2 kg makes an eleastic collision with asecond body at rest and continues to move in theoriginal direction but with one fourth of its originalspeed. What is the mass of the second body ?

[JEE Main-2019 (April)](1) 1.8 kg (2) 1.2 kg (3) 1.5 kg (4) 1.0 kg

Q.14 A ball is thrown vertically up (taken as +z-axis) from theground. The correct momentum-height (p-h) diagramis : [JEE Main-2019 (April)]

(1) O

p

h(2) O

p

h

(3) O

p

h(4) O

ph

Q.15 Three particles of masses 50 g, 100g and 150 g are placedat the vertices of an equilateral triangle of side 1 m (asshown in the figure). The (x, y) coordinates of the centreof mass will be :


m = 150g3

m = 100g2

1.0m0.5m50g = m1

0

Y

60º

(1) 7 3

m, m12 8

(2) 3 5

m, m4 12

(3) 7 3

m, m12 4

(4) 3 7

m, m8 12

Q.16 A man (mass = 50 kg) and his son (mass = 20 kg) arestanding on a frictionless surface facing each other.The man pushes his son so that he starts moving at aspeed of 0.70 ms–1 with respect to the man. The speed

of the man with respect to the surface is : [JEE Main-2019 (April)]

(1) 0.20 ms–1 (2) 0.14 ms–1

(3) 0.47 ms–1 (4) 0.28 ms–1

Q.17 Two particles, of masses M and 2M, moving as shown,with speeds of 10 m/s and 5 ms–1 collide elastically atthe origin. After the collision, they move along theindicated directions with speeds

1 and

2 respectively.

The values of and 2 are nearly :

[JEE Main-2019 (April)]M 2M

10 m/s

30º

v1

45º

v2

5 m/s

45º

30º

2MM

(1) 3.2 m/s and 6.3 m/s (2) 3.2 m/s and 12.6 m/s(3) 6.5 m/s and 6.3 m/s (4) 6.5 m/s and 3.2 m/s

Q.18 Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kgare placed at three corners of a right angle triangle ofsides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure.The center of mass of the system is at a point :

[JEE Main-2020 (January)]

4 cm

2.5 kg

5 cm

1.5 kg3 cm1.0 kg

(1) 0.6 cm right and 2.0 cm above 1kgmass(2) 2.0 cm right and 0.9 cm above 1 kg mass(3) 0.9 cm right and 2.0 cm above 1kg mass(4) 1.5 cm right and 1.2 cm above 1kg mass

Q.19 The coordinates of centre of mass of a uniform flagshaped lamina (thin flat plate) of mass 4 kg. (Thecoordinates of the same are shown in figure) are:


(0, 3) (2, 3)

(2, 2)(1,2)

(0, 0) (1, 0)

(1) (1.25 m, 1.50 m) (2) (0.75 m, 0.75 m)(3) (0.75 m, 1.75 m) (4) (1m, 1.75m)

Q.20 A body A, of mass m = 0.1 kg has an initial velocity of

1ˆ3i ms . It collides elastically with another body, B of

the same mass which has an initial velocity of 1ˆ5jms .

After collision, A moves with a velocity ˆ ˆv 4(i j)

.

The energy of B after collision is written as x

J10

. The

value of x is .[JEE Main-2020 (January)]

Q.21 A ball is dropped from the top of a 100 m high tower on

a planet. In the last 1

2s before hitting the ground, it

covers a distance of 19 m. Acceleration due to gravity(in ms–2) near the surface on that planet is _____


Q.22 As shown in figure. When a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphereof radius R (centred at C), the centre of mass ofremaining (shaded) part of sphere is at G, i.e on thesurface of the cavity. R can be determined by theequation :


R

1

OCG

(1) (R2 – R + 1) (2 – R) = 1(2) (R2 + R – 1) (2 – R) = 1(3) (R2 – R – 1) (2 – R) = 1(4) (R2 + R – 1) (2 – R) = 1

Q.23 A particle of mass m is dropped from a height h abovethe ground. At the same time another particle of thesame mass is thrown vertically upwards from the ground

with a speed of 2gh . If they collide head-on

completely in elastically, the time taken for the combined

mass to reach the ground, in units of h

is :g


(1) 3

4(2)

3

2(3)

1

2(4)

1

2

Q.24 Two particles of equal mas m have respective initial

velocities ûi and ˆ î j

u2

. They collide completely


(1) 23

mu4

(2) 22

mu3

(3) 21

mu3

(4) 21

mu8

Q.25 A body A of mass m is moving in a circular orbit of

radius R about a planet. Another body B of mass m

2

collides with A with a velocity which is half v

2

the

instantaneous velocity v of A. The collision is

completely inelastic. Then, the combined body:

[JEE Main-2020 (January)](1) Escapes from the Planet’s Gravitational field(2) Continues to move in a circular orbit(3) Falls vertically downwards towards the planet(4) Starts moving in an elliptical orbit around the planet

Q.26 A particle of mass m is projected with a speed u fromthe ground at an angle = /3 w.r.t. horizontal (x –axis). When it has reached its maximum height, it collidescompletely inelastically with another particle of the samemass and velocity u i . The horizontal distance coveredby the combined mass before reaching the ground is :


(1) 23 3 u

8 g(2)

2u2 2

g

(3) 25 u

8 g(4)

23 2 u

4 g

Q.27 A rod of length L has non-uniform linear mass density

given by (x) = a + b 2

x

L

, where a and b are

constants and 0 x L. The value of x for the centreof mass of the rod is at :


(1) 4 a b

L3 2a 3b

(2) 3 2a b

L4 3a b

(3) 3 2a b

L2 3a b

(4) 3 a b

L2 2a b

Q.28 A particle of mass m is moving along the x-axis with

initial velocity ûi . It collides elastically with a particle

of mass 10 m at rest and then moves with half its initial

kinetic energy(see figure). If sin1 = 2n sin , then

value of n is _________.

[JEE Main-2020 (September)]

Q.29 A square shaped hole of side I = a

2is carved out at a

distance d = a

2from the centre 'O' of a uniform circular

disk of radius a. If the distance of the centre of mass of

the remaining portion from O is a

x , value of X (to the

nearest integer) is __________.


a

O}

d }I=a/2

Q.30 A particle of mass m with an initial velocity ûi collides

perfectly elastically with a mass 3 m at rest. It moves

with a velocity vj after collision, then, v is given by


(1) v = 1

u6

(2) v = u

3

(3) v = 2

3u (4) v =

u

2

Q.31 A block of mass 1.9 kg is at rest at the edge of a table,of height 1m. A bullet of mass 0.1 kg collides with theblock and sticks to it. If the velocity of the bullet is 20m/s in the horizontal direction just before the collisionthen the kinetic energy just before the combined systemstrikes the floor, is [JEE Main-2020 (September)][Take g = 10 m/s2 Assume there is no rotational motionand loss of energy after the collision is negligable.](1) 19 J (2) 23 J(3) 20 J (4) 21 J

Q.32 Moment of inertia of a cylinder of mass M, length Land radius R about an axis passing through its centreand perpendicular to the axis of the cylinder is

I=M

2 2R L

4 12

. If such a cylinder is to be made for a

given mass of a material, the ratio L/R for it to haveminimum possible I is


(1) 2

3(2)

3

2

(3) 2

3(4)

3

2

Q.33 Blocks of masses m, 2m, 4m and 8m are arranged in aline on a frictionless floor. Another block of mass m,moving with speed v along the same line (see figure)collides with mass m in perfectly inelastic manner. Allthe subsequent collisions are also perfectly inelastic.By the time the last block of mass 8 m starts movingthe total energy loss is p% of the original energy.Value of ‘p’ is close to


m

v

m 2m 4m 8m(1) 37 (2) 77 (3) 87 (4) 94

Q.34 A spaceship in space sweeps stationary interplanetarydust. As a result, its mass increases at a rate

2dM(t)bv t ,

dt where v(t) is its instantaneous

velocity. The instantaneous acceleration of the satelliteis [JEE Main-2020 (September)]

(1) –bv3(t) (2) 32bv

M(t)

(3) 3bv

M(t) (4)

3bv

2M(t)

Q.35 The centre of mass of a solid hemisphere of radius 8 cmis x cm from the centre of the flat surface. Then value ofx is _________. [JEE Main-2020 (September)]

Q.36 The linear mass density of a thin rod AB of length L

varies from A to B as (x) = 0

x1

L

, where x is the

distance from A. If M is the mass of the rod then itsmoment of inertia about an axis passing through A andperpendicular to the rod is


(1) 22

ML5

(2) 25

ML12

(3) 23

ML7

(4) 27

ML18

Q.37 Particle A of mass m1 moving with velocity ˆ ˆ( 3i j) ms –1

collides with another particle B of mass m2 which is at

rest initially. Let 1V

and

2V

be the velocities of particles

A and B after collision respectively. If m1 = 2m

2 and

after collision 1V

= ˆ ˆ(i 3j) ms–1, the angle between

1V

and 2V

is [JEE Main-2020 (September)]

(1) – 45º (2) 60º (3) 15º (4) 105º

Q.38 Two bodies of the same mass are moving with the samespeed, but in different directions in a plane. They havea completely inelastic collision and move togetherthereafter with a final speed which is half of their initialspeed.The angle between the initial velocities of the twobodies (in degree) is __.[JEE Main-2020 (September)]

ROTATIONAL MOTION

Q.39 A particle of mass m is moving along the side of asquare of side ‘a’, with a uniform speed u in the x-y

plane as shown in the figure : [JEE Main-2016]

Which of the following statement is false for the

angular momentum L

about the origin ?

(1) m ˆL Rk

2

when the particle is moving from D

to A

(2) m ˆL Rk

2

when the particle is moving from AA

to B

(3) R ˆL m a k2

when the particle is moving

from C to D

(4) R ˆL m a k2

when the particle is moving

from B to C

Q.40 A roller is made by joining together two cones at theirvertices O. It is kept on two rails AB and CD which areplaced asymmetrically (see figure), with its axisperpendicular to CD and its centre O at the centre ofline joining AB and CD (see figure). It is given a lightpush so that it starts rolling with its centre O movingparallel to CD in the direction shown. As it moves, theroller will tend to :-

[JEE Main-2016]

(1) turn left and right alternately(2) turn left.(3) turn right.(4) go straight.

Q.41 The moment of inertia of a uniform cylinder of length and radius R about its perpendicular bisector is I. Whatis the ratio /R such that the moment of inertia isminimum ? [JEE Main-2017]

(1) 1 (2) 2

(3)

3

2(4)

3

2

Q.42 A slender uniform rod of mass M and length ispivoted at one end so that it can rotate in a verticalplane (see figure). There is negligible friction at thepivot. The free end is held vertically above the pivotand then released. The angular acceleration of the rodwhen it makes an angle with the vertical is :

[JEE Main-2017]y

x

(1) 3g

cos2

(2)

2gcos

3

(3) 3g

sin2

(4)

2gsin

3

Q.43 Seven identical circular planar disks, each of mass Mand radius R are welded symmetrically as shown. Themoment of inertia of the arrangement about the axisnormal to the plane and passing through the point P is:

[JEE Main-2018]

(1) 55

2MR2 (2)

73

2MR2 (3)

181

2MR2 (4)

19

2MR2

Q.44 A circular disc D1 of mass M and radius R has two

identical discs D2 and D

3 of the same mass M and radius

R attached rigidly as its opposite ends (see figure).The moment of inertia of the system about the axis OO',passing through the centre of D

1 as shown in the figure,

will : [JEE Main-2019 (January)]

O’

D2 D3O

D1

(1) MR2 (2) 3MR2 (3) 24

MR5

(4) 22

MR3

Q.45 The magnitude of torque on a particle of mass 1 kg is2.5 Nm about the origin. If the force acting on it is 1N,and the distance of the particle from the origin is 5m,the angle between the force and the position vector is(in radians) : [JEE Main - 2019 (January)]

(1) 6

(2)

3

(3)

8

(4)

4

Q.46 A string is wound around a hollow cylinder of mass 5kg and radius 0.5m. If the string is now pulled with ahorizontal force of 40 N. and the cylinder is rollingwithout slipping on a horizontal surface (see figure),then the angular acceleration of the cylinder will be(Neglect the mass and thickness of the string)


40 N

(1) 20 rad/s2 (2) 16 rad/s2

(3) 12 rad/s2 (4) 10 rad/s2

Q.47 An L-shaped object, made of thin rods of uniform massdensity, is suspended with a string as shown in figure.If AB = BC, and the angle made by AB with downwardvertical is , then: [JEE Main - 2019 (January)]

(1) 1

tan2 3

(2) 1

tan2

(3) 2

tan3

(4) 1

tan3

Q.48 A rod of length 50 cm is pivoted at one end. It is raisedsuch that if makes an angle of 30° for the horizontal asshown and released from rest. Its angular speed whenit passes through the horizontal (in rad s –1) will be (g =10 ms –2). [JEE Main - 2019 (January)]

30°

(1) 30

2(2) 30

(3) 20

2(4)

30

2

Q.49 A simple pendulum, made of a string of length and abob of mass m, is released from a small angle

0. It

strikes a block of mass M, kept on a horizontal surfaceat its lowest point of oscillations, elastically. It bouncesback and goes up to an angle

1. Then M is given by :


(1) 0 1

0 1

m

2

(2) 0 1

0 1

m

(3) 0 1

0 1

m–

(4) 0 1

0 1

m

2

Q.50 Two identical spherical balls of mass M and radius Reach are stuck on two ends of a rod of length 2R andmass M (see figure). The moment of inertia of the systemabout the axis passing perpendicularly through thecentre of the rod is : [JEE Main - 2019 (January)]

2RR R

(1) 137

15MR2 (2)

17

15MR2

(3) 209

15MR2 (4)

152

15MR2

Q.51 A particle of mass 20 g is released with an initial velocity5 m/s along the curve from the poitn A, as shwon in thefigure. The point A is at height h from point B. Theparticle slides along the frictionless surface. When theparticle reaches point B, its angular momentum aboutO will be : [Take g = 10 m/s2]

h=10m

a=10m

O

B

A

[JEE Main - 2019 (January)](1) 2 kg-m2/s (2) 8 kg-m2/s(3) 6 kg-m2/s (4) 3 kg-m2/s

Q.52 A rigid massless road of length 3l has two massesattached at each end as shown in the figure. The rod ispivoted at point P on the horizontal axis (see figure).When released form initial horizontal position, itsinstantaneous angular acceleration will be:


l 2l

P5M0 2M0

(1) g

13I(2)

g

3I(3)

g

2I(4)

7g

3I

Q.53 A homogeneous solid cylindrical roller of radius R andmass M is pulled on a cricket pitch by a horizontalforce. Assuming rolling without slipping, angularacceleration of the cylinder is :


(1) 3F

2mR(2)

F

3mR(3)

F

2mR(4)

2F

3mR

Q.54 The moment of inertia of a solid sphere, about an axisparallel to its diameter and at a distance of x from it, is'I(x)'. Which one of the graphs represents the variationof I(x) with x correctly ? [JEE Main - 2019 (January)]

(1)

I(x)

O x

(2)

I(x)

O x

(3)

I(x)

O x

(4)

I(x)

O x

Q.55 To mop–clean a floor, a cleaning machine presses acircular mop of radius R vertically down with a totalforce F and rotates it with a constant angular speedabout its axis. If the force F is distributed uniformlyover the mop and the floor is µ, the torque, applied bythe machine on the mop is :


(1) µFR/3 (2) µFR/6 (3) µFR/2 (4) 2

3µFR

Q.56 Moment of inertia of a body about a given axis is 1.5 kgm2. Initially the body is at rest. In order to produce arotational kinetic energy of 1200J, the angularacceleration of 20 rad/s2 must be applied about the axisfor a duration of :- [JEE Main-2019 (April)](1) 2 s (2) 5s(3) 2.5 s (4) 3 s

Q.57 A thin smooth rod of length L and mass M is rotatingfreely with angular speed

0 about an axis perpendicular

to the rod and passing through its center. Two beadsof mass m and negligible size are at the center of therod initially. The beads are free to slide along the rod.The angular speed of the system, when the beads reachthe opposite ends of the rod, will be :-


(1) 0M

M 3m

(2) 0M

M m

(3) 0M

M 2m

(4) 0M

M 6m

Q.58 A thin circular plate of mass M and radius R has itsdensity varying as (r) =

0r with

0 as constant and r

is the distance from its centre. The moment of inertia ofthe circular plate about an axis perpendicular to theplate and passing through its edge is I = aMR2. Thevalue of the coefficient a is :

(1) 3

2(2)

1

2

(3) 3

5(4)

8

5

Q.59 A metal coin of mass 5 g and radius 1 cm is fixed to athin stick AB of negligible mass as shown in the figure.The system is initially at rest. The constant torque,that will make the system rotate about AB at 25 rotationsper second in 5 s, is close to : [JEE Main-2019 (April)]

A

B

1cm

(1) 4.0 × 10–6Nm (2) 2.0 × 10–5Nm(3) 1.6 × 10–5Nm (4) 7.9 × 10–6Nm

Q.60 The time dependence of the position of a particle of

mass m = 2 is given by 2ˆ ˆr(t) 2ti 3t j

. Its angular

momentum, with respect to the origin, at time t = 2 is : [JEE Main-2019 (April)]

(1) 36 k (2) ˆ ˆ–34(k i)

(3) 48 ˆ ˆ(i j) (4) – 48 k

Q.61 A solid sphere of mass M and radius R is dividedinto two unequal parts. The first part has a mass of7M

8 and is converted into a uniform disc of radius

2R. The second part is converted into a uniform solidsphere. Let I1 be the moment of inertia of the discabout its axis and I2 be the moment of inertia of thenew sphere about its axis. The ratio I1/I2 is given by :

[JEE Main-2019 (April)](1) 185 (2) 65 (3) 285 (4) 140

Q.62 A solid sphere and solid cylinder of identical radiiapproach an incline with the same linear velocity (seefigure). Both roll without slipping all throughout. Thetwo climb maximum heights h

sph and h

cyl on the

incline. The ratio sph

cyl

h

h is given by :


(1) 14

15(2)

4

5

(3) 1 (4) 2

5

Q.63 A rectangular solid box of length 0.3 m is heldhorizontally, with one of its sides on the edge of aplatform of height 5m. When released, it slips off thetable in a very short time = 0.01s, remaining essentiallyhorizontal. The angle by which it would rotate when ithits the ground will be (in radians) close to :-


h

(1) 0.02 (2) 0.28 (3) 0.5 (4) 0.3

Q.64 The following bodies are made to roll up (withoutslipping) the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius

R

2and (iii) a solid sphere of radius

R

4. If in each case,

the speed of the centre of mass at the bottom of theincline is same, the ratio of the maximum height theyclimb is : [JEE Main-2019 (April)](1) 4 : 3 : 2 (2) 14 : 15 : 20(3) 10 : 15 : 7 (4) 2 : 3 : 4

Q.65 A stationary horizontal disc is free to rotate about itsaxis. When a torque is applied on it, its kinetic energyas a functionof , where is the angle by which it hasrotated, is given as k2. If its moment of inertia is I thenthe angular acceleration of the disc is :


(1) k

2I (2)

k

I (3)

k

4I (4)

2k

I

Q.66 A circular disc of radius b has a hole of radius a at itscentre (see figure). If the mass per unit area of the disc

varies as 0 ,

r

then the radius of gyration of the disc

about its axis passing through the centre is :[JEE Main-2019 (April)]

a

(1) a b

2

(2)

a b

3

(3) 2 2a b ab

2

(4)

2 2a b ab

3

Q.67 A thin disc of mass M and radius R has mass per unitarea (r) = kr2 where r is the distance from its centre. Itsmoment of inertia about an axis going through its centreof mass and perpendicular to its plane is :

(1) 2MR

6(2)

2MR

3(3)

22MR

3(4)

2MR

2

Q.68 Two coaxial discs, having moments of inertia I1 and

1I

2, are rotating with respective angular velocities

1

and 1

2

, about their common axis. They are brought in

contact with each other and thereafter they rotate witha common angular velocity. If E

f and E

1 are the final and

initial total energies, then (Ef – E

i) is :


(1) 2

1 1I

12

(2)

21 1

3I

8 (3)

21 1I

6

(4)

21 1I

24

Q.69 A particle of mass m is moving along a trajectory givenby [JEE Main-2019 (April)]x = x

0 + a cos

1t

y = y0 + b sin

2t

The torque, acting on the particle about the origin, at t= 0 is :

(1) 20 0 1

ˆm( x b y a) k

(2) + my0a 2

1 k

(3) – 2 20 2 0 1

ˆm(x b y a )k (4) zero

ROTATIONAL MOTION

Q.70

mr

m

As shown in the figure, a bob a mass m is tied by amassless string whose other end portion is woundon a fly wheel (disc) of radius r and mass m. Whenreleased from rest the bob starts falling vertically.When it has covered a distance of h, the angularspeed of the wheel will be :


(1) 3

r4gh (2)

1 2gh

r 3

(3) 3

r2gh (4)

1 4gh

r 3

Q.71 The radius of gyration of a uniform rod of length ,

about an axis passing through a point 4

away from

the centre of the rod, and perpendicular to it, is


(1) 7

48 (2)

1

8

(3) 3

8 (4)

1

4

Q.72 Consider uniform cubical box of side a on a rough floorthat is to be moved by applying minimum possible forceF at a point b above its centre of mass (see figure). Ifthe coefficient of friction is = 0.4, the maximum

possible value of b

100a

for box not to topple before

moving is [JEE Main-2020 (January)]

F

Q.73 Mass per unit area of a circular disc of radius a dependson the distance r from its centre as (r) = A + Br. Themoment of inertia of the disc about the axis,perpendicular to the plane passing through its centreis: [JEE Main-2020 (January)]

(1) 4 aA B

2 a4 5

(2) 4 A aB

2 a4 5

(3) 4 A aB

a4 5

(4) 4 A B

2 a4 5

Q.74 Consider a uniform rod of mass M = 4m and length pivoted about its centre. A mass m moving with

velocity v making angle = 4

to the rod's long axis

collides with one end of the rod and sticks to it. Theangular speed of the rod-mass system just after thecollision is: [JEE Main-2020 (January)]

(1) 4

7 v

(2) 3 2 v

7

(3) 3 v

7 (4) 3 v

7 2 Q.75 A uniform sphere of mass 500 g rolls without slipping

on a plane horizontal surface with its centre moving ata speed of 5.00 cm/s. Its kinetic energy is :

[JEE Main-2020 (January)](1) 8.75 × 10–4 J (2) 8.75 × 10–3 J(3) 1.13 × 10–3 J (4) 6.25 × 10–4 J

Q.76 Three solid spheres each of mass m and diameter d arestuck together such that the lines connecting thecentres form an equilateral triangle of side of length d.The ratio I

0/I

A of moment of inertial I

0 of the system

about an axis passing the centroid and about center ofany of the spheres I

A and perpendicular to the plane of

the triangle is : [JEE Main-2020 (January)]

(1) 23

13(2)

13

15(3)

13

23(4)

15

23

Q.77 One end of a straight uniform 1 m long bar is pivotedon horizontal table. It is released from rest when it makesan angle 30° from the horizontal (see figure). Its angular

speed when it hits the table is given as 1n s , where

n is an integer. The value of n is ….. .


Q.78 A uniformly thick wheel with moment of inertial I andradius R is free to rotate about its centre of mass (seefigure). A massless string is wrapped over its rim andtwo blocks of masses m

1 and m

2 (m

1 > m

2) are attached

to the ends of the string. The system is released fromrest. The angular speed of the wheel when m

1 descents

by a distance h is : [JEE Main-2020 (January)]

m2

m1

(1)

1

21 2

21 2

m mgh

(m m )R I

(2)

1

21 2

21 2

2 m m gh

(m m )R I

(3)

1

21 2

21 2

2 m m gh

(m m )R I

(4)

1

21 2

21 2

m mgh

(m m )R I

Q.79 Two uniform circular discs are rotating independentlyin the same direction around their common axis passingthrough their centres. The moment of inertia and angularvelocity of the first disc are 0.1 kg-m2 and 10 rad s–1

respectively while those for the second one are 0.2 kg-m2 and 5 rad s–1 respectively. At some instant they getstuck together and start rotating as a single systemabout their common axis with some angular speed. Thekinetic energy of the combined system is


(1) 20

J3

(2) 5

J3

(3) 10

J3

(4) 2

J3

Q.80 A B

75 10050250

2m

Shown in the figure is rigid and uniform one meter longrod AB held in horizontal position by two strings tiedto its ends and attached to the ceiling. The rod is ofmass ‘m’ and has another weight of mass 2 m hung at adistance of 75 cm from A. The tension in the string at Ais [JEE Main-2020 (September)](1) 0.5 mg (2) 2 mg(3) 0.75 mg (4) 1 mg

Q.81 A uniform cylinder of mass M and radius R is to bepulled over a step of height a (a < R) by applying aforce F at its centre ‘O’ perpendicular to the planethrough the axes of the cylinder on the edge of the step(see figure). The minimum value of F required is


R

F

a

O

(1) 2

R aMg 1

R

(2) 2

RMg 1

R a

(3) 2

2

aMg 1

R (4)

aMg

R

Q.82 An massless equilateral triangle EFG of side ‘a’ (Asshown in figure) has three particles of mass m situatedat its vertices. The moment of inertia of the systemabout the line EX perpendicular to EG in the plane of

EFG is 2N

ma20

where N is an integer. The value of N is

_________. [JEE Main-2020 (September)]

Q.83

Fv

FH

l

A uniform rod of length ‘l’ is pivoted at one of its endson a vertical shaft of negligible radius. When the shaftrotates at angular speed the rod makes an angle with it (see figure). To find equate the rate of changeof angular momentum (direction going into the paper)

22m

sin cos12

l

about the centre of mass (CM) to

the torque provided by the horizontal and vertical forcesF

H and F

V about the CM. The value of is then such

that [JEE Main-2020 (September)]

(1) 2

gcos

l(2) 2

2gcos

3

l

(3) 2

gcos

2

l(4) 2

3gcos

2

l

Q.84 A person of 80 kg mass is standing on the rim of acircular platform of mass 200 kg rotating about its axisat 5 revolutions per minute (rpm). The person now startsmoving towards the centre of the platform. What willbe the rotational speed (in rpm) of the platform whenthe person reaches its centre _________.


Q.85 A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides witha uniform vertical rod and sticks to it as shown. Therod is pivoted about O and swings as a result of thecollision making angle before momentarily coming torest. If the rod has mass M = 2 kg and length l = 1 m, thevalue of is approximately (take g = 10 m/s2)


m mv m

M,l

O

(1) 49° (2) 55° (3) 63° (4) 69°

Q.86

y

x60 cm

O

O'

80 cm

For a uniform rectangular sheet shown in the figure,the ratio of moments of inertia about the axesperpendicular to the sheet and passing through O (thecentre of mass) and O (corner point) is

[JEE Main-2020 (September)](1) 1/2 (2) 1/4(3) 1/8 (4) 2/3

Q.87 Consider two uniform discs of the same thickness anddifferent radii R

1 = R and R

2 = R made of the same

material. If the ratio of their moments of inertia I1 and

I2, respectively, about their axes is I

1 : I

2 = 1 : 16 then the

value of is


(1) 2 (2) 4 (3) 2 2 (4) 2

Q.88 A circular disc of mass M and radius R is rotatingabout its axis with angular speed

1. If another

stationary disc having radius R

2and same mass M is

dropped co-axially on to the rotating disc. Graduallyboth discs attain constant angular speed

2 . The energy

lost in the process is p% of the initial energy. Value ofp is _____. [JEE Main-2020 (September)]

Q.89 ABC is a plane lamina of the shape of an equilateraltriangle. D, E are mid points of AB, AC and G is thecentroid of the lamina. Moment of inertia of the laminaabout an axis passing through G and perpendicular tothe plane ABC is I

0 . If part ADE is removed, the moment

of inertia of the remaining part about the same axis is

0NI

16 where N is an integer. Value of N is ______.


A

D EG

B C

Q.90 A thin rod of mass 0.9 kg and length 1 m is suspended,at rest, from one end so that it can freely oscillate in thevertical plane. A particle of move 0.1 kg moving in astraight line with velocity 80 m/s hits the rod at itsbottom most point and sticks to it (see figure). Theangular speed (in rad/s) of the rod immediately afterthe collision will be –––––– .


Q.91 A force ˆ ˆ ˆF (i 2j 3k)N

acts at a point

ˆ ˆ ˆ(4i 3j k)m . Then the magnitude of torque about

the point ˆ ˆ ˆ(i 2j k)m will be x N – m. The value of

x is ––––. [JEE Main-2020 (September)]

Q.92 A wheel is rotating freely with an angular speed on ashaft. The moment of inertia of the wheel is I and themoment of inertia of the shaft is negligible. Anotherwheel of moment of inertia 3I initially at rest is suddenlycoupled to the same shaft. The resultant fractional lossin the kinetic energy of the system is


(1) 5

6(2)

1

4(3) 0 (4)

3

4

Q.93 Four point masses, each of mass m, are fixed at thecorners of a square of side . The square is rotatingwith angular frequency , about an axis passing throughone of the corners of the square and parallel to itsdiagonal, as shown in the figure. The angularmomentum of the square about this axis is


axis

(1) 3 m2 (2) 4 m2 (3) m2 (4) 2 m2

Q.94 Shown in the figure is a hollow icecream cone (it openat the top). If its mass is M, radius of its top, R andheight, H, then its moment of inertia about its axis is


(1) 2 2M(R H )

4

(2)

2MR

2

(3) 2MR

3(4)

2MH

3

JEE-ADVANCEDQ.1 A point mass of 1kg collides elastically with a stationary

point mass of 5 kg. After their collision, the 1 kg massreverses its direction and moves with a speed of 2 ms–

1. Which of the following statement(s) is/are correct forthe system of these two masses ?

[JEE-2010](A) Total momentum of the system is 3 kg ms–1

(B) Momentum of 5 kg mass after collision is 4 kg ms–1

(C) Kinetic energy of the centre of mass is 0.75 J(D) Total kinetic energy of the system is 4 J

Q.2 A ball of mass 0.2 kg rests on a vertical post of height 5m. A bullet of mass 0.01 kg, traveling with a velocity Vm/s in a horizontal direction, hits the centre of the ball.After the collision, the ball and bullet travelindependently. The ball hits the ground at a distanceof 20 m and the bullet at a distance of 100 m from thefoot of the post. The initial velocity V of the bullet is

[JEE-2011]

(A) 250 m/s (B) 250 2 m/s(C) 400 m/s (D) 500 m/s

Q.3 A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between theblock and the floor is 0.1. Initially the block is at restand the spring is un-stretched. An impulse is given tothe block as shown in the figure. The block slides adistance of 0.06 m and comes to rest for the first time.The initial velocity of the block in m/s isV= N/10. Then N is [JEE-2011]

Q.4 A bob of mass m, suspended by a string of length l1, is

given a minimum velocity required to complete a fullcircle in the vertical plane, At the highest point, itcollides elastically with another bob of mass msuspended by a string of length l

2, which is initially at

rest. Both the strings are mass-less and inextensible. Ifthe second bob, after collision acquires the minimumspeed required to complete a full circle in the vertical

plane, the ratio 1

2

l

l is : [JEE-2013]

Q.5 A particle of mass m is projected from the ground withan initial speed u

0 at an angle with the horizontal. At

the highest point of its trajectory, it makes a completelyinelastic collision with another identical particle, whichwas thrown vertically upward from the ground with thesame initial speed u

0. The angle that the composite

system makes with the horizontal immediately after thecollision is : [JEE -2013]

(A) 4

(B) a

4

(C) a

4

(D)

4

Q.6 A tennis ball is dropped on a horizontal smooth surface.It bounces back to its original position after hitting thesurface. The force on the ball during the collision isproportional to the length of compression of the ball.Which one of the following sketches describes thevariation of its kinetic energy K with time t mostappropriately ? The figures are only illustrative andnot to the scale. [JEE Advanced-2014]

(A)

(B)

t

K

(C)

(D)

Q.7 A block of mass M has a circular cut with a frictionlesssurface as shown. The block rests on the horizontalfrictionless surface of a fixed table. lnitially the rightedge of the block is at x = 0, in a coordinate systemfixed to the table. A point mass m is released from restat the topmost point of the path as shown and its slidesdown. When the mass loses contact with the block, itsposition is x and the velocity is v. At the instant, whichof the following options is/are correct?

[JEE Advanced-2017]

x = 0

R

R

m

M

y

x

(A) The velocity of the point mass m is :

2gRv

m1

M

(B) The x component of displacement of the center of

mass of the block M is : mR

M m

(C) The position of the point mass is

mRx 2

M m

(D) The velocity of the block M is :

mV 2gR

M

Q.8 A flat plate is moving normal to its plane through a gasunder the action of a constant force F. The gas is keptat very low pressure. The speed of the plate v is muchless than the average speed u of the gas molecules.Which of the following options is /are true?

[JEE Advanced -2017](A) The pressure difference between the leading and

trailing faces of the plate is proportional to uv.(B) At a later time the external force F balances the

resistive force(C) The resistive force experienced by the plate is

proportional to v(D) The plate will continue to move with constant non-

zero acceleration, at all times

Q.9 Consider regular polygone with number of sides n = 3,4, 5.............. as shown in the figure. The centre of massof all the polygons is at height h from the ground. Theyroll on a horizontal surface about the leading vertexwithout slipping and sliding as depicted. The maximumincrease in height of the locus of the center of mass foreach polygon is .Then depends on n and h as

[JEE Advanced -2017]

h

h

h

(A) 2

h sinn

(B) 2h tan

2n

(C) 2h sin

n

(D) 1

h 1cos

n

Q.10 A solid horizontal surface is covered with a thin layerof oil. A rectangular block of mass m = 0.4 kg is at reston this surface. An impulse of 1.0 N s is applied to theblock at time to t = 0 so that it starts moving along the

x-axis with a velocity t0v t e , where 0 is a

constant and = 4s. The displacement of the block, inmetres, at t = is ..... Take e–1 = 0.37 ?

[JEE Advanced - 2018]

Q.11 A small particle of mass m moving inside a heavy, hollow

and straight tube along the tube axis undergoes elastic

collision at two ends. The tube has no friction and it is

closed at one end by a flat surface while the other end

is fitted with a heavy movable flat piston as shown in

figure. When the distance of the piston from closed

end is L = L0 the particle speed is v = v

0. The piston is

moved inward at a very low speed V such that V <<

dL

Lv

0,where dL is the infinitesimal displacement of the

piston. Which of the following statemnt (s) is/are correct


L

v

(A) The rate at which the particle strikes the piston is

v/L

(B) After each collision with the piston, the particle

speed increases by 2V

(C) The particle's kinetic energy increases by a factor

of 4 when the piston is moved inward from L0 to

1

2

L0

(D) If the piston moves inward by dL, the particle speed

increases by 2vdL

L

ROTATIONAL MOTION

Q.12 A boy is pushing a ring of mass 2 kg and radius 0.5 m

with a stick as shown in the figure. The stick applies a

force of 2 N on the ring and rolls it without slipping

with an acceleration of 0.3 m/s2. The coefficient of

friction between the ground and ring is large enough

that rolling always occurs and the coefficient of friction

between the stick and the ring is (P/10). The value of P

is? [IIT JEE-2011]

Ground

stick

Q.13 A thin uniform rod, pivoted at O is rotating in thehorizontal plane with constant angular speed , asshown in the figure. At time t = 0, small insect startsfrom O and moves with constant speed with respectto the rod towards the other end. it reaches the end ofthe rod at t = T and stops. The angular speed of thesystem remains throughout. The magnitude of the

torque on the system about O, as a function of

time is best represented by which plot? [IIT JEE-2012]

O

Z

(A)

0 tT

(B)

0 tT

(C)

0 tT

(D)

0 tT

Q.14 A small mass m is attached to a massless string whoseother end is fixed at P as shown in the figure. The massis undergoing circular motion in the x-y plane withcentre at O and constant angular speed .If the angularmomentum of the system, calculated about O and P are

denoted by 0L

and

PL

respectively, then.[IIT JEE-2012]

m

P

O

z

(A) 0L

and

PL

do not vary with time

(B) 0L

varies with time while

PL

remains constant

(C) 0L

remains constant while

PL

varies with time

(D) 0L

and

PL

both vary with time.

Q.15 A lamina is made by removing a small disc of diameter2R from a bigger disc of uniform mass density andradius 2R, as shown in the figure. The moment of inertiaof this lamina about axes passing through O and P is Ioand IP, respectively. Both these axes are perpendicular

to the plane of the lamina. The ratio o

P

II

to the nearest

integer is [IIT JEE-2012]

Q.16 Consider a disc rotating in the horizontal plane with aconstant angular speed about its centre O. The dischas a shaded region on one side of the diameter andan unshaded region on the other side as shown in thefigure. When the disc is in the orientation as shown,two pebbles P and Q are simultaneously projected atan angle towards R. The velocity of projection is in they-z plane and is same for both pebbles with respect tothe disc. Assume that (i) they land back on the discbefore the disc has completed 1/8 rotation, (ii) theirrange is less than half the disc radius, and (iii) remainsconstant throughout. Then

Q

R

P

y

xO

(A) P lands in the shaded region and Q in the unshadedregion

(B) P lands in the unshaded region and Q in the shadedregion

(C) Both P and Q land in the unshaded region(D) Both P and Q land in the shaded region

Comprehension #1 (Q. No. 17 to 18)The general motion of a rigid body can be consideredto be a combination of (i) a motion of its centre of massabout an axis, and (ii) its motion about aninstantanneous axis passing through center of mass.These axes need not be stationary. Consider, for example,a thin uniform welded (rigidly fixed) horizontally at itsrim to a massless stick, as shown in the figure. Wheredisc-stick system is rotated about the origin onahorizontal frictionless plane with angular speed , themotion at any instant can be taken as a combination of(i) a rotation of the centre of mass the disc about the z-axis, and (ii) a rotation of the disc through aninstantaneous vertical axis passing through its centreof mass (as is seen from the changed orientation ofpoints P and Q). Both the motions have the sameangular speed in the case.

Z

Y

QP

X

PQ

Now consider two similar systems as shown in the

figure: case (a) the disc with its face vertical and parallel

to x-z plane; Case (b) the disc with its face making an

angle of 45° with x-y plane its horizontal diameter parallel

to x-axis. In both the cases, the disc is welded at point

P, and systems are rotated with constant angular speed

about the z-axis.

Z

Y

Q

P

X

Case (a)

Z

Y

Q

P

X

Case (b)

45°

Q.17 Which of the following statements about the

instantaneous axis (passing through the centre of mass)

is correct ? [IIT JEE-2012]

(A) It is vertical for both the cases (a) and (b).

(B) It is vertical for case (a); and is at 45o to the x-z plane

and lies in the plane of the disc for case (b).

(C) It is horizontal for case (a); and is at 45o to the x-z

plane and is normal to the plane of the disc for case

(b).

(D) It is vertical for case (a); and is at 45o to the x-z

plane and is normal to the plane of the disc for case

(b).

Q.18 Which of the following statements regarding the

angular speed about the instantaneous axis (passing

through the centre of mass) is correct [IIT JEE-2012]

(A) It is 2 for both the cases.

(B) It is for case (a); and / 2 for case (b).

(C) It is for case (a); and 2 for case (b).

(D) It is for both the cases.

Q.19 The figure shows a system consisting of (i) a ring of

outer radius 3R rolling clockwise without slipping on a

horizontal surface with angular speed and (ii) an inner

disc of radius 2R rotating anti-clockwise with angular

speed 2. The ring and disc are separated by

frictionless ball bearings. The system is in the x-z plane.

The point P on the inner disc is at a distance R from the

origin, where OP makes an angle of 30o with the

horizontal. Then with respect to the horizontal surface.

[IIT JEE-2012]

(A) the point O has a linear velocity iR3

(B) the point P has a linear velocity 11 3 ˆˆR i R k4 4

(C) the point P has a linear velocity kR43

iR413

(D) the point P has a linear velocity

kR41

iR43

3

Q.20 Two solid cylinders P and Q of same mass and sameradius start rolling down a fixed inclined plane fromthe same height at the same time. Cylinder P has mostof its mass concentrated near its surface, while Q hasmost of its mass concentrated near the axis. Whichstatement(s) is(are) correct? [IIT JEE-2012](A) Both cylinders P and Q reach the ground at the

same time.(B) Cylinder P has larger linear acceleration than

cylinder Q.(C) Both cylinders reach the ground with same

translational kinetic energy.(D) Cylinder Q reaches the ground with larger angular

speed.

Q.21 A uniform circular disc of mass 50 kg and radius 0.4 m is

rotating with an angular velocity of 10 rad s-1 about its

own axis, which is vertical. Two uniform circular rings,

each of mass 6.25 kg and radius 0.2 m, are gently placed

symmetrically on the disc in such a manner that they

are touching each other along the axis of the disc and

are horizontal. Assume that the friction is large enough

such that the rings are at rest relative to the disc and

the system rotates about the original axis. The new

angular velocity (in rad s-1) of the system is:

[JEE Advance 2013]

Q.22 A horizontal circular platform of radius 0.5 m and mass

0.45 kg is free to rotate about its axis. Two massless

spring toy-guns, each carrying a steel ball of mass 005kg

are attached to the platform at a distance 0.25 m from

the centre on its either sides along its diameter (see

figure). Each gun simultaneously fires the balls

horizontally and perpendicular to the diameter in

opposite directions. After leaving the platform the balls

have horizontal speed of 9 ms -1 with respect to the

ground. The rotational speed of the platform in rad s -1

after the balls leave the platform is [JEE Advance 2014]

Q.23 A uniform circular disc of mass 1.5 kg and radius 0.5 mis ini8tially at rest on a horizontal friction less surface.Three forces of equal magnitude F = 0.5 N are appliedsimultaneously along the three sides of an equilateraltriangle XYZ with its vertices on the perimeter of thedisc (see figure). One second after applying the forces,the angular speed of the disc in rad/s is

[JEE Advance 2014]

O

F

Y Z F

F

X

Q.24 Two identical uniform discs roll without slipping ontwo different surfaces AB and CD (see figure) startingat A and C with linear speeds v

1 and v

2, respectively,

and always remain in contact with the surfaces. If theyreach B and D with the same linear speed and v

1 = 3 m/s,

then v2 in m/s is (g = 10 m/s2)

[JEE Advance 2015]

A v1 = 3m/s

B 30 m

C v2

D 27 m

Q.25 A ring of mass M and radius R is rotating with angularspeed about a fixed vertical axis passing through its

centre O with two point masses each of mass M

8 at

rest at O. These masses can move readially outwardsalong two massless rods fixed on the ring as shown in

the figure. At 8

9 and one of the masses is at a distance

of 3

5 R from O. At this instant the distance of the other

mass from O is [JEE Advance 2015]

O

(A) 2R

3 (B)

R

3 (C)

3R

5 (D)

4R

5

Q.26 The densities of two solid spheres A and B of the same

radii R vary with radial distance r as A (r) = k

r

R

and

B (r) = k

5r

R

, respectively, where k, is a constant.

The moments of inertia of the individual spheres aboutaxes passing through their centres are I

A and I

B,

respectively. If 10B

A

I n

I , the value of n is:

[JEE Advance 2015]

Q.27 A uniform wooden stick of mass 1.6 kg and length rests in an inclined manner on a smooth, vertical wall ofheight h (<) such that a small portion of the stickextends beyond the wall. The reactionforce of the wallon the stick is perpendicular to the stick. The stickmakes an angle of 30° with the wall and the bottom ofthe stick is on a rough floor. The reaction of the wall onthe stick isequal in magnitude to the reaction of thefloor on the stick. The ratio h/ and the frictional forcef at the bottom of the stick are: (g = 10 ms–2)

[JEE Advanced 2016]

(A)h 3 16 3

, f N16 3

(B) h 3 16 3

, f N16 3

(C) h 3 3 8 3

, f N16 3

(D) h 3 3 16 3

, f N16 3

Q.28 The position vector r of a particle of mass m is given

by the following equation 3 2ˆ ˆr t t i t j

,

where 310

ms3

, = 5 ms–2 and m = 0.1 kg. At t = 1 s,

which of the following statement(s) is(are) true about

the particle? [JEE Advanced 2016]

(A) The velocity v

is given by 1ˆ ˆv 10i 10j ms

(B) The angular momentum L

with respect to the origin

is given by 5 ˆL kNms3

(C) The force F

is given by ˆ ˆF i 2j N

(D) The torque

with respect to the origin is given by

20kNm

3

Q.29 Two thin circular discs of mass m and 4m, having radii

of a and 2a, respectively, are rigidly fixed by a massless,

right rod of length 24a through their center. This

assembly is laid on a firm and flat surface, and set rolling

without slipping on the surface so that the angular

speed about the axis of the rod is . The angular

momentum of the entire assembly about the point ‘O’

is L

(see the figure). Which of the following

statement(s) is(are) true ? [JEE Advanced 2016]

(A) The magnitude of angular momentum of theassembly about its center of mass is 17ma2/2

(B) The magnitude of the z-component of L is 55

ma2(C) The magnitude of angular momentum of center

of mass of the assembly about the point O is 81ma2

(D) The center of mass of the assembly rotates aboutthe z-axis with an angular speed of /5

Q.30 A wheel of radius R and mass M is placed at the bottom

of a fixed step of height R as shown in the figure. Aconstant force is continuously applied on the surfaceof the wheel so that it just climbs the step withoutslipping. Consider the torque about an axis normal tothe plane of the paper passing through the point Q.Which of the following options is/are correct?

[JEE Advanced-2017]

QR

X

P

S

(A) If the force is applied normal to the circumferenceat point P then is zero

(B) If the force is applied tangentially at point S then 0 but the wheel never climbs the step

(C) If the force is applied at point P tangentially then decreases continuously as the wheel climbs

(D) If the force is applied normal to the circumferenceat point X then is constant

Q.31 A rigid uniform bar AB of length L is slipping from itsvertical position on a frictionlesss floor (as shown inthe figure). At some instant of time, the angle made by

the bar with the vertical is . Which of the followingstatements about its motion is/are correct?

[JEE Advanced-2017]

B

A

O

L

(A) The trajectory of the point A is a parabola(B) Instantaneous torque about the point in contact

with the floor is proportional to sin .(C) When the bar makes an angle with the vertical,

the displacement of its midpoint from the initialposition is proportional to (1 – cos)

(D) The midpoint of the bar will fall vertically downward.

Comprehension # (Q. No. 32 and 33)One twirls a circular ring ( of mass M and radius R)near the tip of one’s finger as shown in Figure-1. In theprocess the finger never loses contact with the innerrim of the ring . The finger traces out the surface of acone, shown by the dotted line. The radius of the pathtraced out by the point where the ring and the finger isin contact is r. The finger rotates with an angularvelocity

0. The rotating ring rolls without slipping on

the outside of a smaller circle described by the pointwhere the ring and the finger is in contact (Figure-2).The coefficient of friction between the ring and thefinger is µ and the acceleration due to gravity is g.

[JEE Advanced-2017]

R

Figure-1

– r–

R

Figure-2

Q.32 The total kinetic energy of the ring is :

(A) 220

1M R r

2 (B) 22

0

3M R r

2

(C) 2 20M R (D) 22

0M R r

Q.33 The minimum value of 0 below which the ring will

drop down is :

(A) g

R r (B) g

2 R r

(C) 3g

2 R r (D) 2g

R r

Q.34 The potential energy of a particle of mass m at adistance r from a fixed point O is given by V(r) = kr2/2, where k is a positive constant of appropriatedimensions. This particle is moving in a circular orbitof radius R about the point O. If v is the speed of theparticle and L is the magnitude of its angularmomentum about O, what of the following statementsis(are) true? [JEE Advanced - 2018]

(A) k

v R2m

(B) k

v Rm

(C) 2L mkR (D) 2mkL R

2

Q.35 Consider a body of mass 1.0 kg at rest at the origin at

time t = 0. A force ˆ ˆF ( ti j)

is applied on the body,,

where = 1.0 Ns–1 and = 1.0 N. The torque acting on

the body about the origin at time t = 1.0 s is . Which

of the following statements is(are) true?[JEE Advanced - 2018]

(A) 1

| | Nm3

(B) The torque is in the direction of the unit vector

k(C) The velocity of the body at t = 1 s is

11v (i 2ms )

2

(D) The magnitude of displacement of the body at t = 1

s is 1

6m.

Q.36 A ring and a disc are initially at rest, side by side, at thetop of an inclined plane which makes an angle 60° withthe horizontal. They start to roll without slipping at thesame instant of time along the shortest path. If the timedifference between their reaching the ground is

(2 3) / 10 s, then the height of the top of the

inclined plane, in meters, is ____. Take g = 10 ms–2.[JEE Advanced - 2018]

Q.37 In the List-I below, four different paths of a particle aregiven as functions of time. In these functions and are positive constants of appropriate dimensions and . In each case, the force acting on the particle iseither zero or conservative. In List-II, five physical

quantities of the particle are mentioned; p

is the linear

momentum L

is the angular momentum about theorigin, K is the kinetic energy, U is the potentialenergy and E is the total energy. Match each path inList-I with those quantities in List-II, which areconserved for that path [JEE Advanced - 2018]List - I List - II

P. ˆ ˆr t t i t j

1. p

Q. ˆ ˆr t cos t i sin t j

2. L

R. ˆ ˆr t cos t i sin t j

3. K

S. 2ˆ ˆr t t i t j2

4. U

5. E(A) P Q 2, 5; R 2, 3, 4,5; S 5(B) P 1, 2, 3, 4, 5; Q 3, 5; R 2, 3, 4, 5; S 2, 5

(C) P 2, 3, 4 ; Q 5; R 1, 2, 4; S 2, 5

(D) P 1, 2, 3, 5; Q 2, 5 ; R 2, 3, 4, 5; S 2, 5

Q.38 A thin and uniform rod of mass M and length L is heldvertical on a floor with large friction. The rod is releasedfrom rest so that it falls by rotating about its contact-point with the floor without slipping. Which of thefollowing statement(s) is/are correct, when the rodmakes an angle 60º with vertical ?[g is the acceleration due to gravity]

[JEE Advanced - 2019](A) The radial acceleration of the rod's center of mass

will be 3g

4

(B) The angular acceleration of the rod will be 2g

L

(C) The angular speed of the rod will be 3g

2L(D) The normal reaction force from the floor on the will

be Mg

16

Q.39 A football of radius R is kept on a hole of radius r (r < R)made on a plank kept horizontally. One end of the plankis now lifted so that it gets tilted making an angle from the horizontal as shown in the figure below. Themaximum value of so that the football does not startrolling down the plank satisfies (figure is schematicand not drawn to scale) - [JEE Advanced - 2020]

RPlank

2r

(A) r

sinR

(B) r

tanR

(C) r

sin2R

(D) r

cos2R

Q.40 A small roller of diameter 20 cm has an axle of diameter10 cm (see figure below on the left). It is on a horizontalfloor and a meter scale is positioned horizontally on itsaxle with one edge of the scale on top of the axle (seefigure on the right). The scale is now pushed slowly onthe axle so that it moves without slipping on the axle,and the roller starts rolling without slipping. After theroller has moved 50 cm, the position of the scale willlook like (figures are schematic and not drawn to scale)-


(A)

x=0 x=50 cm

(B)

x=75 cmx=0

(C)

x=25 cmx=0

(D)

x=100 cmx=0

Q.41 Put a uniform meter scale horizontally on your extendedindex fingers with the left one at 0.00 cm and the rightone at 90.00 cm. When you attempt to move both thefingers slowly towards the center, initially only the leftfinger slips with respect to the scale and the right fingerdoes not. After some distance, the left finger stops andthe right one starts slipping. Then the right finger stopsat a distance x

R from the center (50.00 cm) of the scale

and the left one starts slipping again. This happensbecause of the difference in the frictional forces on thetwo fingers. If the coefficients of static and dynamicfriction between the fingers and the scale are 0.40 and0.32, respectively, the value of x

R (in cm) is ........


Q.42 A rod of mass m and length L, pivoted at one of itsends, is hanging vertically. A bullet of the same massmoving at speed v strikes the rod horizontally at adistance x from its pivoted end and gets embedded in

it. The combined system now rotates with angularspeed about the pivot. The maximum angular speed

m is achieved for x=x

M. Then


x

L

v

(A) 2 2

3vx

L 3x

(B) 2 2

12vx

L 12x

(C) M

Lx

3 (D) M

v3

2L

ANSWER KEY

EXERCISE-IQ.1(4) Q.2 (4) Q.3 (2) Q.4 (2) Q.5 (4) Q.6 (3) Q.7 (3) Q.8 (2) Q.9 (3) Q.10 (2)

Q.11 (3) Q.12 (4) Q.13 (1) Q.14 (2) Q.15 (4) Q.16 (1) Q.17 (3) Q.18 (4) Q.19 (2) Q.20 (1)

Q.21 (1) Q.22 (1) Q.23 (1) Q.24 (3) Q.25 (2) Q.26 (1) Q.27 (2) Q.28 (4) Q.29 (3) Q.30 (1)

Q.31 (1) Q.32 (1) Q.33 (4) Q.34 (1) Q.35 (1) Q.36 (3) Q.37 (1) Q.38 (2) Q.39 (3) Q.40 (4)

Q.41 (4) Q.42 (2) Q.43 (4) Q.44 (2) Q.45 (1) Q.46 (4) Q.47 (3) Q.48 (1) Q.49 (2) Q.50 (4)

Q.51 (2) Q.52 (4) Q.53 (2) Q.54 (2) Q.55 (2) Q.56 (3) Q.57 (1) Q.58 (4) Q.59 (3) Q.60 (4)

Q.61 (1) Q.62 (4) Q.63 (2) Q.64 (2) Q.65 (3) Q.66 (2) Q.67 (4) Q.68 (2) Q.69 (1) Q.70 (1)

Q.71 (2) Q.72 (4) Q.73 (3) Q.74 (3) Q.75 (3) Q.76 (1) Q.77 (3) Q.78 (2) Q.79 (1) Q.80 (4)

Q.81 (3) Q.82 (4) Q.83 (3) Q.84 (4) Q.85 (2) Q.86 (2) Q.87 (4) Q.88 (3) Q.89 (1) Q.90 (1)

Q.91 (4) Q.92 (2) Q.93 (3) Q.94 (3) Q.95 (1) Q.96 (2) Q.97 (3) Q.98 (2) Q.99 (1) Q.100 (2)

Q.101 (4) Q.102 (1) Q.103 (2) Q.104 (1) Q.105 (3) Q.106 (2) Q.107 (2) Q.108 (3) Q.109 (1) Q.110 (2)

Q.111 (1)

EXERCISE-IIQ.1 (1) Q.2 (3) Q.3 (2) Q.4 (2) Q.5 (3) Q.6 (1) Q.7 (3) Q.8 (2) Q.9 (1) Q.10 (3)

Q.11 (2) Q.12 (4) Q.13 (2) Q.14 (1) Q.15 (3) Q.16 (1) Q.17 (1) Q.18 (2) Q.19 (2) Q.20 (2)

Q.21 (3) Q.22 (1) Q.23 (1) Q.24 (3) Q.25 (3) Q.26 (3) Q.27 (1) Q.28 (1) Q.29 (1) Q.30 (2)

Q.31 (3) Q.32 (1) Q.33 (2) Q.34 (1) Q.35 (1) Q.36 (3) Q.37 (2) Q.38 (4) Q.39 (2) Q.40 (4)

Q.41 (2) Q.42 (4) Q.43 (4) Q.44 (1) Q.45 (3) Q.46 (2) Q.47 (1) Q.48 (3) Q.49 (4) Q.50 (1)

Q.51 (4) Q.52 (1) Q.53 (4) Q.54 (3) Q.55 (2) Q.56 (3) Q.57 (3) Q.58 (2) Q.59 (2) Q.60 (2)

Q.61 (1) Q.62 (1) Q.63 (1) Q.64 (1) Q.65 (4) Q.66 (3) Q.67 (4)

EXERCISE-IIIQ.1 (A,B,C,D) Q.2 (B, D) Q.3 (B, C) Q.4 (B, D) Q.5 (B,D)

Q.6 (A,D) Q.7 (A,D) Q.8 (A,C) Q.9 (A,C,D) Q.10 (B,D)

Q.11 (A,C) Q.12 (A,B,D) Q.13 (A,B,C,D) Q.14 (A, B, C) Q.15 (A, D)

Q.16 (B, C) Q.17 (B, C, D) Q.18 (A, B, C) Q.19 (B, C) Q.20 (A,C,D)

Q.21 (A,C,D) Q.22 (A, C) Q.23 (A,C,D) Q.24 (A,B,D) Q.25 (C)

Q.26 (A) Q.27 (D) Q.28 (B) Q.29 (B,D) Q .30 (C)

Q.31 (C) Q.32 (B,C) Q.33 (B) Q.34 (A,B,C,D) Q.35 (A)

Q.36 (D) Q.37 (A) Q.38 (B) Q.39 (B) Q.40 (C)

Q.41 (D) Q.42 (D) Q.43 (C) Q.44 (A) Q.45 (C)

Q.46 (A) Q.47 (C) Q.48 (C)

Q.49 (A) p (B) q (C) p,r (D) q,s

Q.50 (A) p,q,r (B) p,q,r (C) p,q (D) p,q,r

Q.51 (A) p (B) q,s (C) p (D) q,s

Q.52 [0010] Q.53 [0025] Q.54 [2000] Q.55 [125] Q.56 [0019]Q.57 [5 rad/s] Q.58 [0006] Q.59 [0100] Q.60 [0004] Q.61 [0010]Q.62 [0002] Q.63 [28] Q.64 [0055] Q.65 [0005]

EXERCISE-IVJEE-MAINPREVIOUS YEAR’SQ.1 (4) Q.2 (1) Q.3 (4) Q.4 (2) Q.5 (3) Q.6 (3) Q.7 (2) Q.8 (3) Q.9 (1) Q.10 (3)

Q.11 (4) Q.12 (3) Q.13 (2) Q.14 (1) Q.15 (3) Q.16 (1) Q.17 (3) Q.18 (3) Q.19 (3) Q.20 [1]

Q.21 [8.00] Q.22 (2) Q.23 (2) Q.24 (4) Q.25 (4) Q.26 (1) Q.27 (2) Q.28 [10] Q.29 [23] Q.30 (4)

Q.31 (4) Q.32 (4) Q.33 (4) Q.34 (3) Q.35 [03] Q.36 (4) Q.37 (4) Q.38 [120.00]Q.39 (2,4) Q.40 (2)

Q.41 (3) Q.42 (3) Q.43 (3) Q.44 (2) Q.45 (1) Q.46 (2) Q.47 (2) Q.48 (2) Q.49 (3) Q.50 (1)

Q.51 (3) Q.52 (1) Q.53 (4) Q.54 (4) Q.55 (4) Q.56 (1) Q.57 (4) Q.58 (4) Q.59 (2) Q.60 (4)

Q.61 (4) Q.62 (1) Q.63 (3) Q.64 (2) Q.65 (4) Q.66 (4) Q.67 (3) Q.68 (4) Q.69 (2) Q.70 (4)

Q.71 (1) Q.72 [50.00] Q.73 (2) Q.74 (2) Q.75 (1) Q.76 (3) Q.77 [15] Q.78 (2) Q.79 (1) Q.80 (4)

Q.81 (1) Q.82 [25] Q.83 (4) Q.84 [09.00] Q.85 (3) Q.86 (2) Q.87 (1) Q.88 [20.00] Q.89 [11] Q.90 [20.00]

Q.91 [195.00] Q.92 (4) Q.93 (1) Q.94 (2)

JEE-ADVANCEDPREVIOUS YEAR’S

Q.1 (A,C) Q.2 (D) Q.3 [4] Q.4 [5] Q.5 (A) Q.6 (B) Q.7 (A,B) Q.8 (ABC) Q.9 (D) Q.10 [6.30]

Q.11 (B,C) Q.12 [0004] Q.13 (B) Q.14 (C) Q.15 [3] Q.16 (C) Q.17 (A) Q.18 (D) Q.19 (A,B) Q.20 (D)

Q.21 [8] Q.22 [4] Q.23 [2] Q.24 [7] Q.25 (C, D) Q.26 [6] Q.27 (D) Q.28 (A, B, D)

Q.29 (B) or (A, B) Q.30 (A, or AB) Q.31 (BCD) Q.32 (Bonus)Q.33 (A) Q.34 (B,C) Q.35 (A,C)

Q.36 [0.75 m]Q.37 (A) Q.38 (A,C,D) Q.39 (A) Q.40 (B) Q.41 [25.60] Q.42 (A, C, D)

EXERCISE-I

Q.1 (4) Centre of mass is a point which can lie within oroutside the body.

Q.2 (4) self explainatoryQ.3 (2)Q.4 (2) Centre of mass is nearer to heavier mass

Q.5 (4) Centre of mass are rcm

= h

4 =

40

4 = 10 cm

Q.6 (3) r = 1.27 Å

1 1 2 2cm

1 2

m r m rr

m m

Since centre of mass cannot go beyond bond length

cm

0 35.5 1.27 35.5 1.27r

35.5 10 36.5

= 1.24 Å

Q.7 (3)Q.8 (2)Q.9 (3)Q.10 (2)Q.11 (3)Q.12 (4)Q.13 (1) Body at rest may possess potential energy.Q.14 (2) Internal forces cannot change momentum but can

do work.

Q.15 (4) acm

= 1 2

1 2

m g m g

m m

= g

Q.16 (1) vector sum of internal forces on system is zero.

Q.17 (3) Vcm

= 1 1 2 2

1 2

M V M V

M M

= ˆ ˆ ˆ200 10i 500(3i 5j) 25ˆ ˆ5i j

700 7

Q.18 (4) VCM

= 0, because internal force cannot change the ve-locity of centre of mass

Q.19 (2)

Q.20 (1)

Q.21 (1)

Q.22 (1)

Q.23 (1)

Q.24 (3) P = 2mE

P m (if E = const.)

1 1

2 2

P m

P m

Q.25 (2)Q.26 (1)

m m m

v

C A B

Initial momentum of the system (block C) = mvAfter striking with A, the block C comes to rest andnow both block A and B moves with velocity V, whencompression in spring is maximum.By the law of conservation of linear momentum

mv = (m + m) V ; vV

2

By the law of conservation of energyK.E. of block C = K.E. of system + P.E. of system

2 2 21 1 1mv (2m)V kx

2 2 2

2

2 21 1 v 1mv (2m) kx

2 2 2 2

2 21

kx mv2

;

m

x v2k

Q.27 (2)Q.28 (4)Q.29 (3)Q.30 (1) Area of F-t curve = A = Impulse.

Impulse = dP = A = mv – 0

v = A

M.

Q.31 (1)Q.32 (1)Q.33 (4)Q.34 (1) If mass = m

first ball will stop v = 0

EXERCISE (Solution)

so K.E. = 0 (min)In other cases there will be some kinetic energy(K.E. can't be negative)

Q.35 (1)Q.36 (3) Change in the momentum

= Final momentum – initial momentum

m

v m

Lead ball m

v m

Tennis ball

v

For lead ball leadP 0 mv mv

For tennis ball tennisP mv mv 2mv

i.e. tennis ball suffers a greater change in momentum.

Q.37 (1)v1 =

1 2 1 2 2

1 2 1 2

(m – em )u m (1 e)u

m m m m

= 1(m – e2m)u 2m(1 e) 0

m 2m m 2m

= 0

0 = m – e2m e = 1/2

Q.38 (2)Q.39 (3)

Q.40 (4)

m2

v2=–5m/s v1=+3m/s

m1

As m1 = m

2 therefore after elastic collision velocities of

masses get interchangedi.e. velocity of mass m

1 = –5 m/s

and velocity of mass m2 = +3m/s

Q.41 (4) hn = he2n = 1 × e2 × 1 = 1 ×(0.6)2 = 0.36 m

Q.42 (2) When ball falls vertically downward from height h1 its

velocity 1 1v 2gh

and its velocity after collision 2 2v 2gh

Change in momentum

2 1 1 2P m(v v ) m 2gh 2gh

(because 1v

and 2v

are opposite in direction)

Q.43 (4)

Q.44 (2)by conservation of linear momentumPi = Pf

mv = (100 m) u u = v/100

Pi = Pf

mv = (100 m) u u = v/100

Q.45(1) After explosion m mass comes at rest and let Rest (M– m) mass moves with velocity v.By the law of conservation of momentum

MV = (M – m)v MV

v(M m)

Q.46 (4) Due to the same mass of A and B as well as due toelastic collision velocities of spheres get interchangedafter the collision.

Q.47 (3)Q.48 (1) Let mass A moves with velocity v and collides

inelastically with mass B, which is at rest.

v A

At rest

B

A

V B

m

m

m m

According to problem mass A moves in a perpendiculardirection and let the mass B moves at angle with thehorizontal with velocity v.Initial horizontal momentum of system(before collision) = mv ....(i)Final horizontal momentum of system(after collision) = mV cos ....(ii)From the conservation of horizontal linear momentummv = mV cos v = V cos ..(iii)Initial vertical momentum of system (before collision)is zero.

Final vertical momentum of system mv

mVsin3

From the conservation of vertical linear momentum

mvmVsin 0

3

vVsin

3 ...(iv)

By solving (iii) and (iv)2

2 2 2 2vv V (sin cos )

3

2

24vV

3

2V v

3 .

Q.49 (2)

Q.50 (4) Initial momentum = ˆ ˆP mvi mvj

P 2mv

Final momentum = 2m × VBy the law of conservation of momentum

2m × V = 2mv

vV

2

In the problem v = 10 m/s (given)

10

V 5 2 m /s2

Q.51 (2) By momentum conservation before and after collision.m

1V + m

2 × 0 = (m

1 + m

2) v

1

1 2

mv V

m m

i.e. Velocity of system is less than V.Q.52 (4)Q.53 (2)Q.54 (2)Q.55 (2) = t

27 30001

60

Q.56 (3)2

rad / s60 30

Q.57 (1)2

T

if time period same then will be same

aC = 2r 1 1

2 2

a r

a r

Q.58 (4)d

; tdt

= 60 rad/ sec. × 5 sec. = 300 rad.360 – 300 = 60 rad

Q.59 (3) v = r since is same for bothv

1 = r

v2 = (2r)

2

1

v

v = )r2(

r

= 2

1; (here x and y are displacement of

particle A & B is same time)So correct option (3).

Q.60 (4) Relation between linear acceleration (at) and angular

acceleration () is :-

at = × R

so, ta 10R

5

= 2m

Q.61 (1) v = rr is perpendicular distance of particle from rotationalaxis so correct option (1).

r v

v = r r is perpendicular distance of particle fromrotational axis so correct option (1).

Q.62 (4) v r

from above we get v

r but

d

dt

is not depend

on distance (r) from axis of rotation.

Q.63 (2) I = 2[5(0.2)2 + 2(0.4)4]

Q.64 (2)

Q.65 (3) As disc is lying in the x-z plane, so applying

perpendicular axis theorem :-

Ix + I

z = I

y

30 + Iz = 40

Iz = 40 – 30 = 10 kg m2

Q.66 (2)90°

I =2m2

12

I =1m2

12

So I = I1 + I

2 =

2 2 2

1 2m m m

I I I12 12 6

Q.67 (4) Disk 2ma

2Ring ma2

Square lamina 2ma

6

a

Four forming a square of side 2a 2 2ma a

m 412 4

2ma m a44

12 4 2

2 2 2ma ma ma

12 4 3

So that moment of inertia least about square lamina.

Q.68 (2) Iz = Ix + Iy Iz = 2I

Q.69 (1) 2

22MrI Mr

2

Q.70 (1)

Density increases linearly IA > IB becouse maximummass is away axis of rotation.

Q.71 (2) M.I. = mr2 = 4 × 12 = 4 kg m2.Q.72 (4) For particle I = I

x + I

y

= mrx2 + mr

y2

= m(rx2 + r

y2)

= 2(32 + 22) = 26 unit

For particle 2, m [12 + (–1)2] = 2 × (2) = 4 unitFor particle 3, m[12 + 12] = 2 × (2) = 4 unitFor particle 4, m(0) = 0Total inertia = 26 + 4 + 4 = 34 unit.

Q.73 (3) Given, Isolid sphere

= Ihallow

sphere

2 21 2

2 2Mr Mr

5 3

2

122

r 5

3r

1

2

r5 : 3

r

Q.74 (3) I = IA + I

B + I

C

B C

A

x'

x

2A

2I MR

5

2 2 2B C

2 7I I MR MR MR

5 5

2 2 22 7 7I MR MR MR

5 5 3

216I MR

5

Q.75 (3)

I=20 kg m2

I' = ?

as,2ML

I 206

ML2 = 120

2 2 2

C

L ML MLI ' I M

2 12 4

2ML 120

I '3 3

= 40 kg m2

Q.76 (1) M inertia about yy’ axis areI = I

1 + I

2 + I

3 = 2I

1 + I

3 ( I

1 = I

2)

2 22MR MR

2 MR2 2

22MR

I MR2

applying parallel axis theorem

= 7/2 MR2 = 7/2 PQ2.

Q.77 (3)

Q.78 (2) x

y

z M, L

M, L

M, L

I = I1 + I

2 + I

3

2 22ML ML 2

0 ML3 3 3

Q.79 (1) so,

2

I m 42

m m

mm

/

also, I = MK2

{Where M = 4m}

so,2

2m 4 4m K2

K2

Q.80 (4)I II III

IV

L/2M,L

I = ML2

12

for one rad, 22 2

1

ML L 4MLI M

12 2 12

Itotal

= 4I1 =

2 24ML 16ML4

12 12

so, Itotal

= 16(I)

Q.81 (3)R

P

Q

ICRing

Applying parallel axis theorem

IPQ

= IC

+ MD2

2

2 2PQ

MR 3I M(R) MR

2 2

Q.82 (4)

Q.83 (3) I = 3

m 2 × 2 +

12

m 2 + m ×

2

2

3

60° 3/2x

x

= 12

m9mm8 222 =

2

3m2 = 3mk2

cos 30° = x

k = 2

Q.84 (4) Torque can be taken about any point in sapce.

Q.85 (2)

For toppling

h af N

2 2

a

h

Q.86 (2) ˆ ˆ ˆ ˆ ˆ ˆF r (2i j 2k) (3i 2j 4k)

ˆ ˆ ˆ12i 14j 16k

2 2 212 14 16 596 = 24.4 N.M.

Q.87 (4)

Q.88 (3)

Q.89 (1) given, 0 = 20 rad/sec

= 0

I = 50 kg-m2

t = 10 sec

0 0 20

t 10

= –2rad/sec2

and = I = 50 × 2 = 100 kg-m2/s2

= 100 N-mQ.90 (1) = I;

2 2r mr II ' (2m)

2 2 2

’ = I’.

' I ' I ' 1; '

I I 2 2

Q.91 (4)2 2 21 1

I mr2 2

Q.92 (2)

By work energy theorem

22L 1 mL

my2 2 3

3g

L

Q.93 (3) as,2L

KE2I

if L = constant,

then L

KEI

as IA > I

B

so (KE)A < (KE)

B

Q.94 (3) I = constantMr2 = (Mr2 + 2mr2)’

M'

M 2m

Q.95 (1)

Initial Final

Applying conservation of angular momentum :-

I1

1 = I

2

2

1 2 21

2I I

T

22

2100 [100 50 (2) ]

10

on solving, 2

2

30

rad/sec

Q.96 (2)R

mv I2

2 2 22 mR mR 3mR

I mR4 4 4

2R 3mRmv

2 2

v

3R

Q.97 (3) From L = constantL = I

Because ext

= 0Due to drop the wax on disc moment of inertia of itswill be increase so will be decrease.

Q.98 (2) ext

= 0, So L = I = constant when girl moves fromedge towards centre I will decrease, and ‘’ willincrease.

Q.99 (1) ext

= 0;from conservation of angular momentumL = I = constantI

1

1 = (I

1 + I

2)

2

(1 =

60

2600 = 20,

2 = 400

60

2400 =

3

40 )

IP × 20 = (I

P + I

Q)

3

40

6 × 20 = (IP + I

Q)

3

40(Given I

P = 6)

After solving that we get I2 = 3 kg m2

Q.100 (2)

2

2 2

1I

2f1 1

I mv2 2

where v = r and 22

I I mR5

Q.101 (4)2 21 1

mgh mv I2 2

where w = v/r, I = 2/5 mR2

Q.102 (1)

Q.103 (2)

Q.104 (1) Sold cylinder has less moment of inclined so it masspasses less rotational energy as compare totranslational energy so it will reach first.

Q.105 (3) Because sphare has maximum translational lurchienergy first decrease in Potential energy.

Q.106 (2) Final speed only depend’s on initial height but timewill depend on length and inclination of plane.

Q.107 (2) Angular moment about lowest point is conservedbecause torque due to friction about point of constantwill be zero.

Q.108 (3)

F – f = MaFR + fR = Ia = RPutting I = MR2 and solving we get fR = 0.

Q.109 (1)

2

2Rotation

2Total

2

KK RFractionK K

1R

For disc, 2

2

K 1

2R ,

so, fraction

1

2 1: 31

12

Q.110 (2)

Q.111 (1)

EXERCISE-II

Q.1 (1) 5 5

6C B

A

ycm

= 0 (2 ) 2 5 2

2( )6 2 5

=

10

11

4 – 10

11 =

34

11

Q.2 (3)

Q.3 (2) ycm = 0

1

8 × 0.14 +

7

8×h = 0

7h

8 = –

0.14

8 h = –0.02 below x-axis.

Q.4 (2) Let x be the displacement of man. Then displacementof plank is L – x.For centre of mass to remain stationary

M

3 (L – x) = M . x

x = L

4

M

L–x x

M/3

Q.5 (3) Centre of mass hits the ground at the position whereoriginal projectile would have landed.

m COM 2mR/2 x1

m.R

2 = 2mx1 x1 =

R

4

Distance = R + R

4 =

5R

4

Q.6 (1) vcm =

11 2 6 102 m / sec

1 1/ 2 3

.

Q.7 (3) a = (nm – m)

nm mg

= (n –1)

(n 1) g

a1 = a2 = a

acm = 1 2nma – ma

(nm m) = (n –1)

a(n 1)

nm

m

a a

acm = 2

2

(n –1)g

(n 1) .

Q.8 (2)

Q.9 (1) vcm = 0mvB + m (vB + vrel) = 0

vB = relmv–

m M– sign means baloon moves downward

Q.10 (3) C1 will move but C2 will be stationary with respect tothe ground.

Q.11 (2) Vcom = V cos

V cos = 2–m0 mv

2m

v2 = 2V cos

vcos vcos

1 2

Q.12 (4) Speed is constant so K.E. ConstantGravitational potential energy change.

Momentum = vm

Direction of v changes

Momentum changes

Q.13 (2) Here net force = 0means momentum is conserved.pi = pf

0 = 1 2p p

1 2p p

K.E. = 2p

2m

1

2

K

K = 2

1

m

m

Q.14 (1)

m v

wall

Initial momentum of body = mv& final momentum of body = – mvChange in momentum = 2mv

Q.15 (3)netF

= 0

then p

= conserved

1 2 3p p p

= 0

3p

= – 1 2p p

3vm

= 1 2m v v

3v

= – ˆ ˆ ˆ ˆ3i 2 j i 4 j

3v

= ˆ ˆ2i 2j

Q.16 (1)netF

= 0

then p

= conserved

pi = pfm1v = m2(0) + (m1 – m2) v1

v1 = 1

1 2

m v

m m

Q.17 (1) As fnet = 0 from momentum conservation

1A 4 v 4v 1

4vv

A 4

Q.18 (2) Total travelled distance = 2d

then

Time between two collisions = 0

2d

v

So no. of collision/sec = 0v

2dImpulse in one collision = mv0 – (–mv0) = 2mv0

F = 2mv0 × 0v

2d =

20mv

d

Q.19 (2) Impulse = change in momentum–I = –m2u – muI = 3muW.D. = change in K.E.

2u I

u

W.D. = 1

2m(2u)2 –

1

2mu2

= 3

2mu2 W.D. =

Iu

2

Q.20 (2) mvi + mvj + 2mv3 = 0

3v

= ˆ(vi vj)

–2

= –

v

2(i + j ) = –

v

2.

kf = 1

2mv2 +

1

2mv2 +

1

22m

2v

2.

kf = 23mv

2.

Q.21 (3) From momentum conservationmu = 2mv

v = u

2from energy conservation

1

2 × 2m ×

2u

2

= 2 mgh

h = 2u

8g

Q.22 (1) In inelastic collision, due to collision some fraction ofmechanical energy is retained in form of deformationpotential energy. thus K.E. of particle is not conserved.In absence of external forces momentum is conserved.

Q.23 (1)1m 2gh + 0 = (m1 + m2) v

v = 1

1 2

m 2gh

(m m )

v2 – u2 + 2g × h

9= 6 + 2g ×

h

4 =

gh

2

v = gh

2

Also, gh

2 =

1 2

m 2gh

m m 2m1 + m1 + m2 ;

1

2

m1

m .

Q.24 (3) e = 1As collision is elastic therefore vi = vf

So K = 0 kf = ki = 1

2m 2 2

1 2u u

Q.25 (3) In absence of external force. Momentum of the systemis conserved.

Q.26 (3) If e = 1 and m1 = m2 then after collision velocityinterchange

Q.27 (1)V0

V = 5

V2 = Z

0

Vel. of Sep = Vel of approach ( elastic) 20 + 5 = V – 5 V = 30 m/s Ans.vb = –(v0 + 2v) m1 > > m2vb = –(20 + 10) = –30 m/sec.

Q.28 (1) mu = mv1 + mv2 .......(i)u = v1 + v2 .......(i)

2 1v – v

u = e ......(ii)

as solving have1

2

v

v = 1– e

1 e

.

Q.29 (1) 1st Collision m m 4mC

v0

A B

2nd Collision

Velocity of B v = mv 4m(0 v)

5m

=

3m

5

m m

3v/5

A BAfter collision of A and B.

m m3v/5

A B

Q.30 (2) Let mass of ball 2 is m and mass of ball 1 is 2 m.

1 1 2 2 2 2 11

1 2

m u m u m e u uv

m m

2m m

v

1 2

v

3 =

2mv em(0 v)

3m

e = 1

So elastic collision.

Q.31 (3) 1

5 55 10 0 v

2 2 v1 = 20 m/sec

2 21 5 1KE 20 5 10

2 2 2

= 500 – 250 = 250 J.

Q.32 (1) mgh = 1

2mv2

h

v = 2gh

By momentum conservation

m 2gh + 0 = 2mv'

v' = 2gh

2By energy conservation

1

2 (2m)v'2 = 2mgh', m

(2gh)

4 = 2mgh'

h' = h

4

Q.33 (2) 0 = 3000 rad/min

0 = 3000

60 rad/sec = (50 rad/sec)

t = 10 secf = 0f = 0 + t = 50 – (10) = 5 rad/sec2

= o t + 2

1 t2

= (50) (10) + 1

2 (–10) (10)2

= 500 – 250 = 250 rad

Q.34 (1)

= 2dmr = r2 dm = r2 m = mr2

Q.35 (1) B > A

B > Aso, If the axes are parallel A < B

Q.36 (3)

0 = 1 + 2

0 =

2

m / 22

3

+

2

m / 22

3

=

2m

12

Q.37 (2)

I II1

Moment of inertia about

diameter of sphere I = 22

mr5

Moment of inertia about tangent at their common point

2 2 21

2 14I mr mr 2 mr

5 5

I1 = 7I

Q.38 (4) Moment of inertia of disc

about diameter I = 2mr

4= 2,

mr2 = 8

I I 1

Moment of inertia about theaxis through a point on rim.

22

1mr

I mr4

= 10

Q.39 (2)

O

R

I = 2MR

2(pasing through 0)

Q.40 (4) M.O.I. about C.O.M. is MinimumI = I

C.M. + Mx

02

I = 2x2 – 12x + 27

d

dx

= 4x – 12 = 0

x = 3

Q.41 (2) = I = constant = increases

Q.42 (4) = 0 + t

100 = 10+(15) = 6 rad/sec2

= I60 Nm

Q.43 (4) = I2 = I × 2 I = 1 kgm2

I = MR2

1 = M(2)2

1M kg

4

Q.44 (1) = A = B

A A = B B

A < BA > B

A > B

Q.45 (3) F = 4 i – 10 j

r

= (–5 i – 3 j )

= r

× F

= (– 5 i – 3 j ) × (4 i – 10 j )

= 50 k + 12 k = 62 k

Q.46 (2) Torque about OF × 40 + F × 80 – (F × 20 + F × 60)In clockwise direction= F × 40

Q.47 (1) Initial velocity of each point onthe rod is zero soangular velocity of rod is zero.Torque about O =

20g (0.8) = 2m

3

20g (0.8) =

220(1.6)

3

3g

3.2 = = angular acceleration

= 15 g

16

Q.48 (3) Beam is not at rotational equilibrium, so force exertedby the rod (beam) decrcase

Q.49 (4)

xf

r

sin

Mg

N

fr = Mg sin = Mg cos f

r .

a

2 = N.x =

N

N

aMg sin

2

Q.50 (1)

F

3a/4mg

xx

For topling about edge xx _________

Fmin.

3a

4 = mg

a

2

Fmin.

= 2mg

3.

Q.51 (4)21

I 10002

10 rad / sec

2f = 10 5 300

f rad / sec rad / min

Q.52 (1)

Q.53 (4)

Q.54 (3) =

dL

dt

= 0 04A – A

4 =

03A

4

Q.55 (2)

Y

v

Xo

d

L = (mvd) = constant becouse v = const. and d =

const.

Q.56 (3) external torque ext

= 0

11 = 22

when he stretches his arms so 1 < 2

then (1 > 2)

so, (L = constant)

Q.57 (3)21

I 102

2510 2 rad / sec

2

Angular Momentum

L = I = 5 × 2 = 10 joule-sec.

Q.58 (2) I1

1 = I

2

2

2 21 1 2 2M R MR

1 2

2 1

R 3

R 1

Q.59 (2)

2 22

2

1 1 R vKE I

2 2 2 R

21Mv

4

Total KE = 2 21 1

I mv2 2

2 21 1Mv m

4 2

23mv

4

Ratio = 1

3

Q.60 (2)

When A point travels distance then B point 2 so, 2length of string passes through the hand of the boy .

Q.61 (1)

mg sin – f = ma

a = mgsin – f

m

.......(i)a is same for each body.f.R =

2

f .R

mk

For solid sphere k2 = 2

5 R2 is minimum there fore is

maximum hence, k.E. for solid sphere will be max atbottom.

Q.62 (1) Due to linear velocity body will move forward beforepure rolling.

Q.63 (1) Disk = 3 2

1

1 1I mv

2 2

222112

v1 MR 1mv

2 2 2R

21

3mv

4

Ring = 2

2 2 2 222 22

v1 1 1 1I mv mr mv

2 2 2 2R = 2

2mv

2 21 2

1mv mv

4

1/ 2

1

2

v 4

v 3

Q.64 (1) F

= Ma

a = F

M

For pure rolling

a = R

F

aR

SmoothSurface

M

R

F × R = I

= FR

F

m =

FR.R

I = MR2

MR2 is the moment of inertia of chin pipe.

Q.65 (4)

Q.66 (3) fsin

mg

sinmg

f

Q.67 (4) As the inclined plane is smooth, the sphere can never

roll rather it will just slip down.

Hence, the angular momentum remains conserved

about any point on a line parallel to the inclined plane

and passing through the centre of the ball.

EXERCISE-III

JEE-ADVANCEDCOMPREHENSION/STATEMENT/MATCHING/MCQ

Q.1 (A,B,C,D)

Q.2 (B, D) Center of mass of ring is at centre and centre of massof chord AB is at its mid point so centre of mass of thiscombination lie at the line which makes 45° with x axis.

Y

XA

B

Ring COM line

Possible combination

R R R R, ; ,

3 3 4 4

Q.3 (B, C)Q.4 (B, D)

Q.5 (B,D)

Pi = mv (i)Pf = (m + m) vat maximum compressionPi = Pf v' = v/2By energy compression

1

2mv2 + 0 =

1

2 (2m) (v)2 +

1

2 kx2

kx2 = 2mv

2 x =

mv

2k .

at maximum compression k = 1

2(m+m)v2 k = mv2 =

mv2/4.

Q.6 (A,D)Q.7 (A,D)

Q.8 (A,C)

mv = nvm v = v

n

time for first collisen is t1 = L

V (2nd block)

2nd collisions t2 = 2

V

= 2t1

(3rd block)so t = t1 + 2t1 + 3t1 + at1 ...........(n–1) t1.t = t1 [1 + 2 + 3] .......................(n–1)]

= (n –1)(n –1 1)

2

=

n(n –1)

2

so t = L

2xn (n – 1).

Q.9 (A,C,D) a = f

m for elastic collission e = 1

v12 = 0 + 2ad

vb12 =

2F.d

m vb1 =

2Fd

m

after collisin vb2 = 0.

Q.10 (B,D)

Q.11 (A,C) V

V

V

V(L – Vt)

2 2(Vt) (L vt) L 2V2t2 + L2 – 2LVt L2

Vt – L 0

t L

V

Q.12 (A,B,D) M 2M

V V

For minimum kinetic energyMV0 = 3MV V = V0/3

K = – 2200

V1 13m mv

2 3 2

= 2 Joule

Q.13 (A,B,C,D)Inelastic collision0 < e < 1

Q.14 (A, B, C)

Using perpendicular theorem0 = 4 + 3 3 = 40 = 1 + 2 2 = 13 = 2so, (0 = 1 + 3)

Q.15 (A, D)(1) For no slippingµmg cos mg sin .........(1)For toppling

mg sin 2

h mg cos.

2

a .........(2)

for minimum µ (by dividing)

µ.a

2=

h

2

µmin

= h

a.

[ Ans.: a/h ]

(2)

If f > mg sin mg cos > mg sin ( > tan ) block will topple before slidingtorque about point A A =0

mg sin 2h = mg cos

2

a

tan = ha

> ha

If > tan (block will slide)

Q.16 (B, C)

d

x d – x

NA NB

BA w

NA + NB = WW(d – x) = NA . d

Q.17 (B, C, D) Body is in equilibriumSo net = 0orFnet = 0

Q.18 (A, B, C)

(A) KE = 2

1 2

I depends on m KE depends on m

M

am

b

y-axis

x-axis

M

(B) axisy = 2Ma2

K.E. = 2

1 × 2Ma22 = Ma22

axisz = 2(Ma2 + mb2)

K.E. = 2

1 I2 = )mbMa( 22 2

Q.19 (B, C) External force will act at hinge so linear momentumof system will not remain const. but torque of external

force is zero about hinge so L = const., collision is

elastic so K.E = const.

Q.20 (A,C,D) is constant(mgr)

A r

Q.21 (A,C,D)

Q.22 (A, C) If bicycle is accelerating on a horizontal plane thenfriction on front wheel will be backward and on rearwheel it will be in forward direction.But if bicycle is accelerating down an inclined planethen friction on rear wheel may be backward or forwardboth.

Q.23 (A,C,D)

Q.24 (A,B,D)(A) Change in Angular Mom.

V0

O

= Lf –Li= ( – mV0R) – ( + mV0R)= – 2mV0R(B) Impulse = Change in momentum= – 2mV0R

Q.25 (C)(a) The acceleration of the centre of mass is

aCOM

= F

2mThe displacement of the centre of mass at time t

will be

x = 1

2a

COM t2 =

2Ft

4m Ans.

Q.26 (A)

Q.27 (D)(Solution of Q. 26 and Q. 27)Suppose the displacement of the first block is x

1 and

that of the second is x2. Then,

x = 1 2mx mx

2m

or,

21 2x xFt

4m 2

or, x1 + x

2 =

2Ft

2m ...(i)

Further, the extension of the spring is x1 – x

2. Therefore,

x1 – x

2 = x

0...(ii)

From Eqs. (i) and (ii), x1 =

1

2

2

0

Ftx

2m

and x2 =

1

2

2

0

Ftx

2m

Ans.

Q.28 (B)As net force in x direction is zero. So from momentumconservation.mV0 = (M + m)V2

M

h

m v0

V2 = 0MV

M m

Q.29 (B,D) Velocity of center of mass

VCOM = MV mV

M m

= V

So both are at rest with respect to centre of mass. Andkinetic energy is converted into potential energy.

Q .30 (C) By Energy conservation

1

2mv0

2 = 1

2(M + m)

2

0mv

M m

+ mgh

After solving

h = M

M m

20V

2g

Q.31 (C)V1 is the velocity of particel and V2 is the velocity ofwedge.

O

V2V1

(V1 + V2) = vel. of particle w.r.t. wedge

–0 0mV M( V )

M m

+ 0 0mV mV

M m

= V0

Vel. of particle V1 = V0 M m

M m

VCOM = 2 1MV ( mV )

M m

= 0mv

M m

Q.35 (A)As Fnet in x direction = 0mx1 = mx2 [ Fx = 0]x1 = x2Now x1 + x2 = L sin

CMf = Lsin

2

L sin

v2

x1

/2 x2

CMf

L

L

CMi

L/2 cos

Q.36 (D)VCMx = 0 and Fx = 0from momentum conservationmv1 = mv2 v1 = v2 = v(let)Now energy conservation

mg (1 – cos ) = 221

mv2

v2 = g (1 – cos )

Distance from centre of mass = R = 2

So T = 2mv

R =

mg (1 cos )

/ 2

T = 2mg (1 – cos )

Q.37 (A)from previous question

vmax = V = 1/2g (1 cos )

Q.38 (B)Only in vertical direction[ fx = 0 always]

So displacement = L

2 –

L

2 cos =

L

2[1 – cos ]

Q.39 (B)

2

1

2

1

2

m

2

m

30

Q.32 (B,C) As net force in x direction is zero.So by momentum conservationMv2 – mv1 = mV0

and V1 + V2 = V0

MV1

V2

m

Q.33 (B)As net force in x direction is zero.So by momentum conservationMV2 – mV1 = mV0 .......(1)V1 + V2 = V0 .......(2)By solving

V1 =V0

M m

M m

Q.34 (A,B,C,D)(a) V1 + V2 = V0

V2 = V0 – V0 M m

M m

= 0 0 0(M m)V V M V m

M m

= 02mV

M m

K.E. = 1

2 × M ×

2 20

2

4m V

(M m)

[ h = 20VM

(m M) 2g ]

K.E. = 24m

gh(m M)

(b) V2 = 02mv

M m(c) K.E. = kf – ki

= 1

2M

2 20

2

4m V

(M m)

– 0

= 2

4mM

(m M)20

1mV

2

(d) vel. of wedge V2 = 02mV

M m

Q.40

30

2

1

2mR2

1

2

1

2mR2

1

2

1

2

m

2

1

2

m

3

R2 0

NOTE : 2r

2 v)R(

Q.41

2

m

2

1

9

mR 20

2

Q.42 (D)Q.43 (C)Q.44 (A)

Solution of Question number 42 to 44The free body diagram of plank and disc isApplying Newton's second law

F – f = Ma1

.... (1)f = Ma

2.... (2)

FR = 2

1MR2 .... (3)

from equation 2 and 3

a2 =

2

R

From constraint a1 = a

2 + R

a1 = 3a

2.... (4)

Solving we get a1 =

M4

F3 and =

MR2

F

If sphere moves by x the plank moves by L + x. Thefrom equation (4)

L + x = 3x or x = 2

L

Q.45 (C)Q.46 (A)Q.47 (C)Q.48 (C)Q.49 (A) p (B) q (C) p,r (D) q,s

(A) If velocity of block A is zero, from conservation ofmomentum, speed of block B is 2u. Then K.E. of

block B = 2

1m(2u)2 = 2mu2 is greater than net

mechanical energy of system. Since this is notpossible, velocity of A can never be zero.

(B) Since initial velocity of B is zero, it shall be zero formany other instants of time.

(C) Since momentum of system is non-zero, K.E. ofsystem cannot be zero. Also KE of system isminimum at maximum extension of spring.

(D) The potential energy of spring shall be zerowhenever it comes to natural length. Also P.E. ofspring is maximum at maximum extension of spring.

Q.50 (A) p,q,r (B) p,q,r (C) p,q (D) p,q,rSince all forces on disc pass through point of contactwith horizontal surface, the angular momentum of discabout point on ground in contact with disc is conserved.Also the angular momentum of disc in all cases isconserved about any point on the line passing throughpoint of contact and parallel to velocity of centre ofmass.The K.E. of disc is decreased in all cases due to workdone by friction.From calculation of velocity of lowest point on disc,the direction of friction in case A, B and D is towardsleft and in case C is towards right.The direction of frictional force cannot change in anygiven case.

Q.51 (A) p (B) q,s (C) p (D) q,s(A) Speed of point P changes with time(B) Acceleration of point P is equal to 2x ( = angular

speed of disc and x = OP). The acceleration isdirected from P towards O.

(C) The angle between acceleration of P (constant inmagnitude) and velocity of P changes with time.Therefore, tangential acceleration of P changes withtime.

(D) The acceleration of lowest point is directedtowards centre of disc and remains constant with time

Q.52 [0010] 2v2 + 3v3 = 0

2

1kx2 =

2

1 × 2v2

2 + 2

1 × v3

2

2

1 × 750 × (0.4)2 = v2

2 + 2

3 ×

2

2v3

2

375 × 0.16 = 3

5v2

2

v2 = 16.0225 = 6

v3 = 3

2 + 6 = 4

v2 + v3 = 10

Q.53 [0025] M0 = r × 3

4R3

Mc = × 3

4

3

4

R

= 64

M0

xc =

c0

c0

MM4R

M0M

=

64

MM

R256

M

00

0

= 25 km

Q.54 [2000] mv = Mv1 + M v/2

v1 =

M2

mv = g5

v = m

M2g5

v = m

M2g5 =

01.0

12100 = 2000 m/s

Q.55 [125] 3mv = FT

4m (vx) = FT = 3 mv

vx = 4

v3 and vy = v

v' = 22 75100 = 25 22 34 = 125 m/s

Q.56 [0019] Area = mv – mv0–42 = 2 (v – 2)–21 = v – 2 v = –19 m/s

Q.57 [5 rad/s] Angular momentum conservation

Li = Lf

mJohn VJohn

2

d+ IMGR MGR = Isystem

(30) (5) (2) + (500) (5.6) = [500 + 30(2)2] = 5 rad/s.

Q.58 [0006] f = mg × 8

R3 cos 60°

30°60°

mg

3R/8

30°

N1

N2

60°

N2 = fN1 = mg

f = 16

mg3

f µN1

µ 16

3 32µ = 6

Q.59 [0100]

kx70g

40g

T2T1

T1

T2

T1 = 70g + kx ...........(i)T2 = 40 g ...........(ii)T1 (0.3) = T2 (0.6) ...........(iii) 70g + kx = 80g

kx = 10 g x = 1m

Q.60 [0004] P – T = ma .......... (i)TR = I

C .......... (ii)

a = R .......... (iii)

T

m g0

Fx

Fy

mg

N

T P

a

P = R

IC + mR = 4 r/s2

Q.61 [0010] = t ; 10 = × 10 ; = 1 rad/s2

net = 2

1 × 10 × 102 × 1 = 500 Nm = 100 × 10 –

= 500 Nm

' = I

= – 1 rad/s

0 = 0 + 't = 10 – 1 × t t = 10 sec.

Q.62 [0002] Fl = 3

m 2l

A

A =

lm

F3

F2

l =

12

m 2l

B

B =

lm

F6

= 2

1

A × 32 =

lm2

F27

lm2

F27 = =

2

1 × a

B × t2 =

2

1 ×

lm

F6t2

t = 2

3 sec. 2sec.

Q.63 [28] 152 × 0.5 – T × 0.2 = I

0.2

0.5

F = 152N

T

30030g 0.8

T – 300 = 0.8 × 30T = 32476 – 324 ×0.2 = I0.2 = 0.8 = 411.2 = I × 4I = 2.8 kg-m2.

Q.64 [0055] Isys = 2

mr2

+

2

2

)r2(m2

mr× 6

= 2

55mr2 = 55

Q.65 [0005] mg × 2R = 2

1mv2 +

2

1 × mR2 ×

2

R

v

v = gR2 = 5 m/s

EXERCISE-IVJEE-MAINPREVIOUS YEAR'S

Q.1 (4) Let initial speed of neutron is v0 and kinetic energy is

K.1st collision

by momentum conservationmv

0 = mv

1 + 2mv

2 v

1 + 2v

2 = v

0

by e = 1 v2 – v

1 = v

0

v2 = 02v

3; v

1 = – 0v

3

fractional loss =

22 00

20

v1 1mv m

2 2 31

mv2

Pd =

8

9 ·89

2nd collision

by momentum conservationmv

0 = mv

1 + 12mv

2

v1 + 12v

2 = v

0

by e = 1 v2 – v

1 = v

0

v2 = 02v

13 ; v

1 = 011v

13

Now fraction loss of energy

22 0

c 0

20

11v1 1P mv m

2 2 131

mv2

= 48

1690.28

Q.2 (1) Initial

2 2 21 2 0

1 1 3 1mv mv mv

2 2 2 2

2 2 21 2 0

3v v v

2 ......(1)

from momentum conservationmv

0 = m(v

1 + v

2) ......(2)

(v1 + v

2)2 = v

02

2 2 21 2 1 2 0v v 2v v v

2v1v

2 =

20v

2

(v1 – v

2)2 =

22 2 2 01 2 1 2 0

v3v v 2v v v

2 2

v1 – v

2 =

02v

Q.3 (4) F = dp

dt = 2n mv cos 45°

Pressure = F

A =

2nmv cos 45

Area

=

23 27 3

4

12 10 3.3 10 10

22 10

= 2.35 × 103 N/m2

Q.4 (2)dp

F ktdt

3P T

P 0dP kt dt

2KT p2p ; T 2

2 k

Q.5 (3) Apply LMC (Linear Momentum Conservation)mv = (2m + M)v’

v’ = mv

2m MInitial energy

21mv

2Final energy

21 mv

(2m M)2 2m M

Initial kinetic energy - Final kinetic energy = 5

6 of initial

kinetic energy.

After solving, we get M

4m

Q.6 (3)40 m

60 m

COM

100 m/s

u = 0

Ycm

from ground = 0.03 100

60 m0.05

Vcm

= 0.02 100

40 m / s0.05

H = 2cmV 40 40

80 m2g 20

Height above building = 80 – 40 = 40 m

Q.7 (2) Linear momentum conservation

m 2v + 2m v = m × 0 + v

m 22

v 2 2vQ.8 (3)

mv

rest

M = 4mm

vc

vch

Applying Linear momentum conservation mv = (m +M)v

c

vc =

v

5applying work energy theorem

– mgh = 1

2 (m + M)v

c2 –

1

2mv2

Solve, h = 22v

5g

Q.9 (1)

A

B C

D

Y

X

aa

a

a

Aâ ai

Bâ aj

Câ ai

Dâ aj

a a b b c c d dcm

a b c d

m a m a m a m aa

m m m m

cm

ˆ ˆ ˆ ˆmai 2maj 3mai 4maja

10m

ˆ ˆ2mai 2maj

10m

a aˆ î j5 5

a ˆ î j5

Q.10 (3) Note:Pressure is defined as normal force per unit area.Force is calculatedas change in momentum/timeBy this answer is 2N/m2

None of the option matches of this question must beBonusDetailed solution is as following.

m

v

v m

Magnitude of change in momentum per collision = 2mV

Pressure = N 2mvForce

Area 1

=22 26 410 2 10 10

1

= 2N/m2

Q.11 (4)

(0, 0) (a, 0)

(0, b) (a, b)3a4

3b4

,

a M 3aM

2 4 4xM

M4

a 3a 5a5a2 16 16

3 3 124 4

y =

b M 3bM 5b2 4 4

M 12M4

Q.12 (3) Applying linear momentum conservation

m1v

1 i + m2v

2 i = m1v

3 i + m2v

4 im

1v

1 + 0.5 m

1v

2 = m

1(0.5 v

1) + 0.5 m

1v

4

0.5 m1v

1 = 0.5 m

1(v

4 – v

2)

v1 = v

4 – v

2

Q.13 (2)2kg

Rest

m

v0

2kg m

v /40 v

By conservation of linear momentum:-

2v0 = 2

0v

4

+ mv 2v0 = 0v

2 + mv

03v

2 = mv .........(1)

Since collision is elastic V

separation = v

approach

v – 0v

4 = v

0 05v

4 = v ......(2)

equating (2) and (1)

0 03v 5v 6m m 1.2kg

4 4 5

Option (2)

Q.14 (1) Momentum p = mv ...(i)

and for motion under gravity 2 2u v

h2g

...(ii)

2 2 2u p / mh

2g

p

h

Q.15 (3)

150g

m = 100g2

1.0m0.5m

50g = m1

0

Y

60º

(1/2, 3/2)

The co-ordinates of the centre of mass

cm

1 3ˆ ˆ ˆ0 150 i j 100 i2 2

r300

cm

7 3ˆ ˆr i j12 4

= Co-ordinate 7 3

, m12 4

Q.16 (1)v1 v2

Man Son

50 20

0 = 50V1 – 20V

2 and V

1 + V

2 = 0.7

V1 = 0.2

Q.17 (3)Official Ans. by NTA (3)

M × 10 cos30º + 2M × 5 cos45º= 2M × v

1 cos 30º + M v

2 cos 45º

21

v35 3 5 2 2v

2 2

10 × M sin 30º – 2M × 5 sin 45ºM v

2 sin45º – 2M v

1 sin30º

5 – 5 2 = 2v

2– v

1

Solving v1 =

17.5

2.7= 6.5 m/s

v2 6.3 m/s

Q.18 (3)

Q.19 (3)

(0, 3) (2, 3)

(0, 0) (1, 0)

(0,5,1)

(1,2)

(1,2,5)

cm

ˆ î 5jˆ ˆ1 j 1 i2 2

r2

cm

3 7ˆ ˆr i j4 4

Q.20 [1] For elastic collision KEi = KEf

21 1 1 1m 25 m 9 m 32 mv

2 2 2 2

34 = 32 + v2

1 1KE 0.1 2 0.1J

2 10

x = 1

Q.21 [8.00]

h=100 m

u = 0

V(m/s)

(0,0)t (s)

tt 12

Area of shaded trapezium

1g t t

1219

2 2

..... (1)

21gt 100

2 ...... (2)

200t

g

1g 2t 76

2

2004 1

g76

g 2

g = 8 m/s2

Q.22 (2)3

14

M R3

32

4M 1

3

1 1 2 2

1 2

M X M XXcom M M

3 3

3 3

4 4R 0 (1) ( ) R 1

3 34 4

R (1)3 3

3

R 1(2 R)

R 1

(R 1)

2

(R 1)2 R

(R 1)(R R 1)

(R2 + R + 1) (2 – R) = 1

Alternative :M

remaining (2 – R) = M

cavity (1 – R)

(R3 – 13) (2 – R) = 13 [R – 1] (R2 + R + 1) (2 – R) = 1

Q.23 (2)B

h

V = 2gh

A

Time for collision 1h

t2gh

After t1

A 1gh

V 0 gt2

B 11

and V 2gh gt gh 22

at the time of collision

i fP P

A B fmV mV 2mV

fgh 1

gh 2 2V2 2

Vf = 0

and height from ground 2

11 h 3h

h gt h2 4 4

so time

3h3h4

2g 2g

Q.24 (4) Conserving mementum

1 2

v vˆ ˆ ˆ ˆ ˆmvi m i j 2m(v i v j)2 2

on solving

1

3vv

4 and 2

vv

4

Change in K.E.

221 1 v

mv m 22 2 2

–2 21 9v v

(2m)2 16 16

2 2 23mv 5mv mv

4 8 8

Q.25 (4) Conserving momentum

f

m v mmv m v

2 2 2

f

5mV 5Vv

3m 642

vf < v

orb (= v) thus the combined mas will go on to an

elliptical path

Q.26 (1)ucos

m um

= 60º

2m v

pi = pf

mu + mucos = 2mv

u(1 cos60º ) 3v u

2 4

so horizontal range after collision

max2Hv

g

2 2

2

3 2u sin (60º )u

4 2g

22

33 3 3u4u4 g 8g

Q.27 (2)x

cm0

1x x.dM

M

2x

dM .dx a b dx.

2

20

cm 2

20

bxx a dx

xdM x dxx

dM dx bxa dx

2 4

20 0

3

0 20

x b xa

2 4

b xa(x)

3

2 2

a b (2a b)2 4 3b 4(3a b)a3

3 2a b

4 3a b

Q.28 [10]

m ûi

m

10m

1

2

y

x

Since energy of m reduced by half u1 =

u

2

and 2

21

1 1 u10m.v m

2 2 2 v

1 =

u

20

Now, momentum in y direction will remain conserved. mu

1sin = 10mv

1sin

2

1 2u u

sin 10 sin2 20

sin1 = 10 sin

2

Q.29 [23]

Let the density is Then original mass m

0 = a2

Remaining mass m' =

22 a

a4

= a2 4 1

4

Removed mass m = a

.4

a2 (0) = 2a a a

. (4 1)r4 2 4

r = a a a

2(4 1) 23.13 23

Q.30 (4)

y

x

mu m

3m3m

v

v'BeforeCollision

AfterCollision

Conservation of linear momentum

ˆ ˆ3mv ' mui mvj

For elastic collision

2 2 21 1 13mv' mv mu

2 2 2

Elliminating v' gives

v = u

2

Q.31 (4) Velocity of block after collision = 0.1 20

1.9 0.1

= 1m/s

Kinetic energy just before striking the floor

212 1 2 10 1

2 = 21J

Q.32 (4) I = M

2 2R L

4 12

M = R2L

I = M

2M L

4 L 12

dI

0dL

L3 = 6 M

4

3

2

L

LR=

6

4

L 3

R 2

Q.33 (4) i fP P

mv = (16 m)v0 v

0 =

v

16

(KE)loss

= 2

21 1 vmv (16m)

2 2 16

% Loss = Loss

2

( KE) 1100 1 100

1 16mv2

= 94%

Q.34 (3) F = dp dM

vdt dt

M (t) = a – (bv2)v a = 2bv

M(t)

Q.35 [03] 3R

x 3cm8

Q.36 (4) M =

L L

0

0 0

xdx 1 dx

L = 0

3L

2

I =

L2 3

0

0

7( dx)x L

12

I = 27

ML18

Q.37 (4) 45º60º

1V

2V

m2 = m and m

1 = 2m

i fp p

2ˆ ˆ ˆ ˆ2m 3i j 2m i 3j mV

2ˆ ˆV 2 3 2 i j 2 3 2

ˆ î 3j

Angle = 105º

Q.38 [120.00] mv0 × cos × 2 = 2

n ×

0V

2

v0

v0

m

m

2m v0

2

cos = 1

2

= 60º 2 = 120º

Q.39 (2, 4) L mr v

Q.40 (2)

Q.41 (3) I = 2 2m mR

12 4

or I =

22m

R4 3

......(1)

Also m = R2

R2 = m

Put in equation (1)

I =

2m m

4 3

For maxima and minima

2

dI m 2 m0

d 4 3

2

2 m

3

2

2

2 R

3

or22 R

3

2

2

3

2R

or3

R 2

Q.42 (3)

mg

mgsin

Taking torque about pivot = I

mg sin 2

=

2m

3

3gsin

2

Q.43 (3)

I0 = I

cm + md2

= 27MR

2 + 6 (M × (2R)2) =

255MR

2I

p = I

0 + md2

= 255MR

2 + 7M(3R)2 =

81

2

MR2

Q.44 (2)

2 22MR MR

I 2 MR2 4

2 2

2 2MR MR2MR 3MR

2 2

Q.45 (1) 2.5 = 1 × 5 sin

Sin = 0.5 = 1

2 =

6

Q.46 (2) 40 + f = m(R) .......(i)40 × R – f × R = mR2 = 2

40 – f = mR .......(ii)

40

f

a

From (i) and (ii)

= 40

mR = 16

Q.47 (2) Lets consider mass of each rod is m for stableequilibrium the torque about point O should be zero.

mga

2 sin = mg

acos a sin

2

tan =1

3

O

a

mg

mg

a

tan–1 1

3

Q.48 (2)

221 1 m

mg2 2 2 3

3g

302

Q.49 (3) Just before collision speed of m

0v 2gL(1 cos )

Just after collision speed of M

1 1v 2gL(1 cos )

And v1 =

M m

M m

v ; 1v M m

v M m

1

0

1 cos M m

1 cos M m

1

0

Sin( / 2) M m

Sin ( / 2) M m

[ 1 – cos 2 = 2 sin2 ]

1

0

M m

M m

M1 + m

1 = M

0 – m

0

M = 0 1

0 1

m–

Q.50 (1) I = 2 2 22

MR M4R5

+ M24R

12

= MR2 1 4

83 5

= 137

15MR2

Q.51 (C) v = 2 25 2gh 5 2 10 10 = 225

= 15 m/s h = rmv = 20 × (20 × 10–3 kg) × (15)

= 6 kg m2/sec

Q.52 (1) I

o o2 2

o o

5M g 4M g

5M 2M g

o2

o

M g

13M

g

13

Q.53 (4) F – f = ma

fR = 1

2mR2

a = R

Q.54 (4) I = ICM

+ Mx2

ICM

= 2

5MR2

I = 2

5MR2 + Mx2

Q.55 (4) d dN x

= 2

F2 dx

R

x

xdx

R

0

d

Q.56 (1) Given moment of inertia ‘I’ = 1.5 kgm2

Angular Acc “” = 20 Rad/s2

21KE I

2

211200 1.5

2

2 1200 21600

1.5

= 40 rad/s2

0 + t

40 = 0 + 20 tt = 2 sec.

Q.57 (4) Applying angular momentum conservation, about axisof rotationL

i = L

f

22 2

0ML ML L

m 212 12 2

= 0M

M 6m

Q.58 (4) M =

R 30

00

2 Rr(2 rdr)

3

0(MOI about COM)

I =

R 52 0

00

2 Rr(2 rdr) r

5

by parallel axis theoremI = I

0 + MR2

= 5

0 2 R

5

+

30 2 R

3

× R2 =

02R5 ×

8

15

= MR2 × 8

5

Q.59 (2) Official Ans. by NTA (2)

=225 2

10 rad / sect 5

= 25

MR4

= 5

4× 5 × 10–3 × (10–2)2×10

= 1.9625 × 10–5 Nm

2.0 × 10–5 Nm

Q.60 (4)

L m[r v]

m = 2 kg

2ˆr 2t i – 3t j

= 4 ˆ î –12j (At t = 2 sec)

dr ˆ ˆ ˆ ˆv 2i – 6tj 2i –12 jdt

ˆ ˆ ˆ ˆr v (4i –12j) (2i –12j)

= –24 k

L

= m (r v)

= – 48 k

Q.61 (4)

I1 =

2

2

7M(2R)

7 784 MR

2 16 4

MR2

I2 = 2 2

21

2 M 2 M R MRR

5 8 5 8 4 80

3 31

4 4R 8 R

3 3

R3 = 8 31R

R = 2R1

2

12

2

I 7 / 4MR 780

I 4MR

80

= 140

Q.62 (1) for solid sphere

22 2

sph2

1 1 2 vmv . mR . mgh

2 2 5 R

for solid cylinder

22 2

cyl.2

1 1 1 vmv . mR . mgh

2 2 2 R

sph.

cyl.

h 7 / 5 14

h 3 / 2 15

Q.63 (3) Angular impulse = change in angular momentum t = L

mg2

× .01 =

2m

3

3g 0.01

2

3 10 0.1

2 0.3

= 1

2 = 0.5 rad/s

time taken by rod to hit the ground

t = 2h 2 5

1sec.g 10

in this time angle rotate by rod = t = 0.5 × 1 = 0.5 radian

Q.64 (2)2

2

1 1m v mgh

2 R

if radius of gyration is k, then

22

2ring

ring

k1 v

kRh , 1,

2g R

solid cylinder

solid cylinder

k 1

R 2

solidsphere

solidsphere

k 2

R 5

h1: h

2 : h

3 :: (1 + 1) :

1 21 : 1 : : 20 :15 :14

2 5

Therefore most appropriate option is (2)athough which in not in correct sequence

Q.65 (4) Kinetic energy KE 2 21

I k2

2

2 2k

I

2k

I

....(i)Differentiate (i) wrt time

d 2k d

dt I dt

= 2k 2k

I I {by (i)}

= 2k

I

Q.66 (4)

dm

drr

dI = (dm)r2

= (dA)r2

20 2 rdr rr

= (2)r2dr

b2

0

a

I dI 2 r dr

=3 3

0

b a2

3

m dm dA b

0

a

2 dr m =

0 2(b – a)

Radius of gyration

3 3b aIk

m 3 b a

3 3a b ab

3

Q.67 (3)

dr

r

IDisc

=

R2

0

(dm)r IDisc

=

R2

0

( 2 rdr)r

IDisc

=

R2 2

0

(kr 2 rdr)r Mass of disc

IDisc

=

R5

0

2 k r dr

R2

0

M 2 rdr kr

IDisc

=

R6

0

r2 k

6

M =

R3

0

2 k r dr

IDisc

= 2k 6R

6M = 2k

R4

0

r

4

IDisc

=

6 4 2kR kR R 2

3 2 3

4RM 2 k

4

IDisc

= 2M2R

3

IDisc

= 22

MR3

Q.68 (4) Ei =

22 1 1

1 1

I1 1I

2 2 2 4

=

221 1

1 1

I 9I

2 8 16

I1

1 + 1 1 1I 3I

4 2

11 1

3I5I

4 2

= 1

5

6

Ef =

1

2 ×

211

3I 25

2 36

= 2

1 1

25I

48

Ef – E

i = I

1

12

25 9

48 16 =

21 1

2I

48

= 2

1 1I

24

Q.69 (2) 2 21 1 2 2

ˆ ˆF m(a cos ti b sin tj)

0 1 0 2ˆ ˆr (x a cos t)i (y bsin t) j

20 1 2 2

ˆr F m(x a cos t)b sin t k

20 2 1 1

ˆm(y bsin t)a cos tk 21 0

ˆma y k

Q.70 (4) mgh 2 21 1

mv I2 2

v = R (no slipping)

22 2 21 1 mR

mgh m R2 2 2

mgh2 23

m R4

2

4gh 1 4gh

R 33R

Q.71 (1)/4

222M

M MK12 4

2 22K

12 16

7K

48

Q.72 [50.00] For no toppling

a aF b mg

2 2

a ab

2 2

0.2a + 0.4b 0.5a0.4b 0.3a

3ab

4

b 0.75a (in limiting case)

max

b100 50.00

a

Q.73 (2) = A + Br

dm (A Br)2 rdr 2I dm r

a3

0

(A Br)2 r dr 4 5

4a a A Ba2 A B 2 a

4 5 4 5

Q.74 (2)

V

2

4m

O

2

Loi = Lof

mv

22

= 2 24m m

12 4

×

= 6v 3 2 v

77 2

Q.75 (1) K.E. of the sphere = Translational K.E + Rotational K.E.

22

2

1 Kmv 1

2 R

K = Radius of gyration

21 1 5 2

12 2 100 5

–43510 J

4

Q.76 (3) M.I about P

2222 d d 13

3 M M Md5 2 103

M.I about B2 2

2 22 d 2 d 232 M M(d) M Md

5 2 5 2 10

Now ratio =13

23

Q.77 [15] mg2

sin 30° =

221 m

2 3

Solving2 = 15

15

Q.78 (2) ki + U

i = k

1 + k

2

2 2 22 1 1 2

1 1 10 0 m v m v l m gh m gh

2 2 2

2 2 21 2 2 1

1 1 1m m gh m ( R) m ( R) I

2 2 2

1 2

21 2 2

2(m m )ghI

m m RR

1 2

1 2 2

2(m m )gh1IR

m mR

Q.79 (1) (I1 + I

2) = I

1

1 + I

2

2

= 0.1 10 0.2 5

0.1 0.2

= 20

3rad/s

K.E. = 2

1 21

(I I )2

= 20

J3

Q.80 (4)

T

A B

50 cm

mg 2mg

25cm

Net torque about B = 0 T × 100 = mg × 50 + 2mg × 25 T = mg

Q.81 (1) R–a R

ax

F

To step up,F × R Mg × x

Fmin

= 2 2Mg

R (R a)R

= 2

R aMg 1

R

Q.82 [25] x1 = 2

a;x

2 = a and x

3 = 0

m mx2

x1 m

I = 2

2mama 0

4

= 25

ma4

I = 2 225 N

ma ma20 20

N = 25

Q.83 (4)

x

O

dmr2

dmg

From a rotating frame rod will appear in equilibrium. Nettorque about suspension point must be zero.

L 2

0

M Ldx (x sin ) x cos Mg sin

L 2

cos = 2

3g

2 l

Q.84 [09.00]

M0

m

M0 = 200 kg

m = 80 kg

I = (IM

+ Im) =

220M R

mR2

I1W

1 = I

2W

2

W2 =

220

20

M R 5mR

2 M R

2

= 2

2

5R (80 100)

100R

W2 = 9 RPM

Q.85 (3) mvl =

22MI

m3

l

221 MI I

m Mg mgI (1 cos )2 3 2

l

= 18

5rad/s

1 5 324(10 10)(1 cos )

2 3 25

20(1–cos) = 324

30

cos = 1–324 276

0.46600 600

Q.86 (2) 2 2O

MI (80) (60)

12

2O' OI I M(50)

O

O'

M(10000)I 112

MI 4(10000) (M)(2500)12

Q.87 (1) I = MR2

M R2

I R4

41 1

2 2

I R

I R

Q.88 [20.00] I1

1 = I

2

2

2 =

2

1 21 1

22 2

MR

2 8 4MR

2 5MR 5MR MR

2 8

2 22 21 1

221

1 MR 1 5MR 16. .E 2 2 2 8 25100 P 100

E 1 MR

2 2

=2

21 2

21

41

1 MR 5100

2 2 1 MR

2 2

= 20

Q.89 [11]

m1

Let I0 is about ‘G’ mass be m, and length be a

kma2 = I0 ...(i)

Then MI of (m1) =

20Ikm a

4 4 16

Let moment of inertia of one triangular part (side) beI

1 about ‘G’ then

3I1 +

00

II

16 3I

1 = I

0 –

0 0I 15I

16 16

I1 =

015I

16

So, moment of inertia of remaining portion

I'' = I0 –

015I

16

I'' = 011

I16

Q.90 [20.00]

O

Angular momentum will remain conserved about pivotal

point

(0.1× 80 × 1) = 1 9 (1)

110 10 3

= 8 10

4

= 20 rad/s

Q.91 [195.00] r f

ˆ ˆ ˆ ˆ ˆ ˆr (4i 3j k) (i 2j k)

ˆ ˆ ˆr (3i j 2k)

ˆ ˆ ˆ ˆ ˆ ˆ(3i j 2k) (i 2j 3k)

ˆ ˆ ˆ ˆ ˆ ˆ6k 9j k 3i 2 j 4i

ˆ ˆ ˆ7i 11j 5k

(t) = 195

Q.92 (4) I = 4I'

' = 4

(K)LOSS

=

221 1

I (4I)2 2 4

Fractional loss = 2LOSS

11

( K) 1K I

K 2

=

3

4

Q.93 (1) 2

/ 2

/ 2

I = m2

2 22 m ( 2 ) 3m2

L = I = 3 m2

Q.94 (2) : I = 2MR

2

Moment of inertia of this cone will same as circular diskof mass (M) and radius R.

R

M

M

R

JEE-ADVANCEDPREVIOUS YEAR'S

Q.1 (A,C)

Since collision is elastic, so e = 1Velocity of approach = velocity of separationSo, u = v + 2 .............(i)By momentum conservation :

1 × u = 5v – 1 × 2u = 5v – 2v + 2 = 5v – 2

So, v = 1 m/sand u = 3 m/sMomentum of system = 1 × 3 = 3 kgm/sMomentum of 5kg after collision = 5 × 1 = 5 kgm/s

So, kinetic energy of centre of mass = 1

2(m1 + m2)

21

1 2

m u

m m

= 1

2(1 + 5)

21 3

6

= 0.75 J

Total kinetic energy = 1

2 × 1 × 32 = 4.5 J.

Q.2 (D) R = 2h

ug

20 = 12 5

V10

and 100 = 2

2 5V

10

V1 = 20 m/s , V

2 = 100 m/sec.

Applying momentum conservation just before and justafter the collision(0.01) (V) = (0.2)(20) + (0.01)(100) V = 500 m/s

Q.3 [4] = 0.1

21mu

2 = mg × 0.06 +

1

2kx2

1

2 × 0.18 u2 = 0.1 × 0.18 × 10 × 0.06

0.4 = N

10

N = 4 Ans.

Q.4 [5]

To complete the vertical circle

1g = 25g

2

1

= 5

Q.5 (A) At the highest point

v1 = 0u cos

2

(by applying momentum

conservation in horizontal direction)

v2 = 0u cos

2

(by applying momentum

conservation in vertical direction)

(H = 2 20u sin

2g

)

= 45°

Q.6 (B)

K = 2 21

mg t2

2tK : parabolic graph

then during collision kinetic energy first decreases toelastic potential energy and then increases.Most appropriate graph is B.

Q.7 (AB) If speed of point mass is v, then using conservation of

linear momentum mv

VM

2

21 1 mvmgR mv M ;

2 2 M

21 mmgR mv 1

2 M

2gRv

m1

M

; M

mRX

M m

Q.8 (ABC) u’ = u, = constant

v

u' u'

A

w. r. t plane

A

(u' – v) (u' + v)

Before collisionBefore collision

(u' – v)(u' + v)

After collision

After collision

A

F

FleadingFtrailing

Ftrailing

= 2A(u’ – v)2

Fleading

– Ftrailling

= 2A(4u’v) = 8Au’vPressure difference

leading trailingF F8 u 'v 8 uv

Area

Net force on plate

net

mdvF F 8 A uv

dt

After long time v will be sufficient so F = 8AuvAfter that v = constant, i.e. plate will achieve terminalvelocity.

Q.9 (D) /n R

h

R = Maximumheight

hcos

n R

hR h h

cosn

1h

cos 1n

Q.10 [6.30]

t0v v e

0J

v 2.5m sm

t0v v e

t0

dxv e

dt

xt

00 0

dx v e dt

x

x ee dx

1

t

0

0

ex v

1

1 0x 2.5 4 e e

x 25 4 0.37 1

x 6.30

Q.11 (B,C)

L

v0 V

(1) average rate of collision = 2L

v(2) speed of particle after collision = 2V + v

0

change in speed = (2V + v0) – v

0

After each collision = 2V

no. of collision per unit time (frequency) = v

2Lchange in speed in dt time = 2V × number of collision indt time

dv = 2V v

2L

. dL

V

dv = vdL

L

Now, dv = – vdL

L {as L decrease}

0

0 0

L / 2v

v L

dv dL

v L

0

0

L / 2vv L[In v] [In L]

v = 2v0

0LKE =

1

220mv

0L / 2

0

KE4

KE

0L / 2KE = 1

2m(2v

0)2

or

(dt)v 2mv

F2x dt

x

F = 2mv

x

– m v2dv mv

v x

– dv dx

v x

ln 2

1

v

v = In 1

2

x

x

vx = constant on decreasing lenth to half K.E. be-comes 1/4vdx + xdv = 0

Q.12 [0004] 2–f2 =2a=0.6 f2=1.4

=I (f2–f1)R=MR2Ra

1.4 – f1 = Ma = 0.6

stick

f2

f1

2N

f1=0.8 = (2)= 10P

×2

P = 4

Q.13 (B) = dtdL

= )I(dtd

= dIdt

= )II(dtd

mrod

as Irod

= com = in sec t

wdI

dt

= )mr(dtd 2 = m

dr2r

dt

= 2m rv

= 2m(vt)v t

Q.14 (C)L0 remains cons. in magnitude and direction but L

Pchanges its direction continously hence L

P is variable

L0

v x

L (varies direction)P

Q.15 [3]

2R

2RO P

I0 =

2

)R2()m4( 2 –

2

3 mR2

= mR2 [8 – 2

3]

= 2

13mR2

IP =

2

3 (4m) (2R)2 –

]R)R2[(m

2

mR 222

= 24 mR2 – 2mR

2

11

= 2mR

2

37

=

2132

37

I

I

O

P = 3

1337

Q.16 (C)Q

R

P

y

xO

45°

At 45° P & Q both land in unshaded region.The general motion of a rigid body can be consideredto be a combination of (i) a motion of its centre of massabout an axis, and (ii) its motion about an instantaneousaxis passing through the centre of mass. These axesneed not be stationary. Consider, for example, a thinuniform disc welded (rigidly fixed) horizontally at itsrim to a massless stick, as shown in the figure. Whenthe disc-stick system is rotated about the origin on ahorizontal frictionless plane with angular speed , themotion at any instant can be taken as a combination of(i) a rotation of the centre of mass of the disc about thez-axis, and (ii) a rotation of the disc through aninstantaneous vertical axis passing through its centreof mass (as is seen from the changed orientation ofpoints P and Q). Both these motions have the sameangular speed in this case.

Now consider two similar systems as shown in thefigure: Case (a) the disc with its face vertical and parallelto x-z plane; case (b) the disc with its face making anangle of 45o with x-y plane and its horizontal diameterparallel to x-axis. In both the cases, the disc is welded

at point P, and the systems are rotated with constantangular speed about the z-axis.

Case (a) Case (b)

Q.17 (A)Consider case (a)

Q

P

A (out of paper)

(insidepaper)

B

at t = 0

AB

at t = T/4

at t = T/2 at t = 3T/4

Q

P

A(inside)(outside) B

A B

Hence axis is vertical.For case (b)

Q.18 (D) Angular Velocity of rigid body about any axes which

are parallel to each other is same . So angular velocity

is .

Q.19 (A,B) V0 = 3R i

VP (3R –

2

R cos 60º) i +

2

R sin 60 j

= i4

R3i

4

R11

Q.20 (D)IP > IQ

a = 2

gsin1 I / MR

Hence ap < a0

tp > tQVp < vQ

And as = v/RSo P < Q

Q.21 [8] Angular momentum conservation

I1

1 = I

2

2

22 2 2

1 2

MRMR 2(mr mr )

2

= 250 0.4

2× 10

=

2

2 250 0.42 6.25 0.2 0.2

2

2

40 = [4 + 1] 2

2 = 8 rad/s

Q.22 [4] Applying conservation of angular momentum.

2mvr – 02

MR2

2MR

mvr4

41

1045

41

)9()105()4(

2

2

= 4 rad/s

Q.23 [2]

I

dt=

I

t

0

dtR30sinF3

= 2

5.05.1

)1()5.0()5.0()5.0(.32

= 2 rad/s

Q.24 [7]

Q.25 (C or D)

Q.26 [6]

Q.27 (D)At equilibrium, reaction of the wall on the stick cannotbe equal in magnitude to the reaction of the floor onthe stick.

Q.28 (A, B, D) 3 2ˆ ˆr t t i t j

2ˆ ˆ3 2dr

v t i t jdt

1

10 ˆ ˆ53t

r i j

1

ˆ ˆ10 10t

v i j

1

ˆ ˆt

p i j

5 ˆ3

L r p k

dvF m

dt

1

ˆ ˆ2t

F i j

20 ˆ3

r F k

Hence, (a, b, d)

Q.29 (B) or (A, B)As the discs are rolling without slipping

' 5 '5

a a

Angular momentum of system about CM through anaxis along rod is

= 22 24 2 17

2 2 2

m ama ma

24a

24a

5a 5a

2a

a

'

Hence, (B) or (a, b)

Q.30 (A, or AB)

Q.31 (BCD)

90– x

y(x , y)

(0,0)

x sin2

y = cos2 2

2 2

y 4x1

Path of A is ellipse(B) torque about point of contact

mg sin I2

hence torque sin

(C) cm

Ly 1 cos

2

(D) midpoint will fall vertically downwards

Q.32 (Bonus)Q.33 (A)

Q.34 (B,C) v = 2kr

2

F = – kr (towards centre) dv

Fdr

v

rO

At r = R,

kR = 2mv

R[Centripetal force]

v = 2kR

m =

kR

m

L = m2k

Rm

Q.35 (A,C) ˆ ˆF ( ti j)

[At t = 0, v = 0, = 0

]

= 1, = 1

ˆ ˆF ti j

dv ˆ ˆm ti jdt

On integrating

2t ˆ ˆmv (i tj)2

[m = 1kg]

2dr t ˆ î tjdt 2

[r 0 at t 0]

On integrating

At t = 1 sec, 1 1ˆ ˆ ˆ ˆ(r F) i j (i j)6 2

1k

3

2t ˆ ˆv i tj2

At = t = 1 1 ˆ î j2

= 1 ˆ î 2 j m / sec2

At t = 1 1 0s r r

1 1ˆ î j6 2

= [0]

1 1ˆ ˆs i j6 2

2 21 1

| s |6 2

10

6 m

Q.36 [0.75 m]

=60º

h

cC

2

gsina

I1

MR

ringgsin

a2

disc2gsin

a3

21 1 2

h 1 gsin 4h 16ht t

sin 2 2 3ggsin

22 2 2

h 1 2g sin 3h 4ht t

sin 2 3 ggsin

16h 4h 2 3

3g g 10

4h 2 2 3

3

2 3 3 3 3

h h 0.75m2 44 2 3

Q.37 (A) (P) ˆ ˆr t ti tj

dr t ˆ ˆv i j constantdt

dva 0

dt

P mv

(remain constant)

21k mv

2 (remain constant)

U Uˆ ˆF i i 0x y

U constantE = K + U

dLr F 0

dt

L

constant

(Q) ˆ ˆr cos t i sin t j

dr ˆ ˆv sin t i cos t jdt

2 2dv ˆ â cos t i sin t jdt

2 ˆ ˆcos t i sin t j 2a r

r F 0 { r

and F

are parallel}

r2

0

U F.dr m .r.dr

22 r

U m2

2U r

2 2 2 2r cos t sin t

r is a function of time (t)U depends on r hence it will change with timeTotal energy remain constant because force is central.

(R) ˆ ˆr t cos ti sin t j

dr t ˆ ˆv t sin t i cos t jdt

v

(Speed remains constant)

2 2dv t ˆ â t cos t i sin t jdt

2 ˆ ˆcos t i sin t j

2t r

F r 0

r

(remain constant)

Force is central in nature and distance from fixed pointis constant.Potential energy remains constantKinetic energy is also constant (speed is constant)

(S) 2ˆ ˆr ti t j

2

dr ˆ ˆv t i t jdt

(speed of particle depends on ‘t’)

dv â jdt

{constant}

F ma = {constant}

t

0

ˆ ˆ Û F.dr m j. i tj dt

2 2m tU

2

2 2 2 21 1k mv m t

2 2

E = 21

k U m2

(remain constant)

Q.38 (A,C,D) We can treat contact point as hinged.Applying work energy theoremWg = K.E.

mg4

=

1

2

22m

3

=

3g

2

radial acceleration of C.M. of rod = 2 3g

2 4

60º 30º

C.M.

atar2

C.M.

Using = I about contact point

mg

2

sin60º =

2m

3

= 3 3

4g

Net vertical acceleration of C.M. of roda

v = a

r cos 60º + a

t cos 30º

= 3g 1

cos30º4 2 2

= 3g 3 3 3

4 4 2 2

= 3g 9 15

g8 16 16

Applying Fnet

= ma in vertical direction on rod as sys-tem

`mg – N = mav = m

15g

16

N = mg

16

Q.39 (A)

mg N1

N2

max

mg

max

r

For max

, the football is about to roll, then N2=0 and all

forces (Mg and N1) must pass through contact point

cos (90º–max

) max

r rsin

R R

Q.40 (B) For no slipping at the ground,V

centre =R (R is radius of roller)

Velocity of scale = (Vcenter

+r) [r is radius of axle]Given, V

center t= 50 cm

Distance moved by scale = (Vcenter

+r)t

center centercenter

V r 3VV t t 75cm

R 2

Therefore relative displacement (with respect to centreof roller) is (75 – 50) cm = 25 cm

Q.41 [25.60]N1

Mg10 cm

N2

f2

90 cmf1

xL

50<

40 cm

Initially

1 2N N Mg 1

4MgN

9

N 1aboutcentre

0 N 50 2

5MgN

9

5N1=4N

2

K

K

K

1 K 1

1 1

2 2

f N

f 0.32N

f 0.32N

L

L

L

1 S 1

1 1

2 2

f N

f 0.4N

f 0.4N

Suppose xL=distance of left finger from centre when

right finger starts moving(

n=0)

about centre N

1x

L=N

2(40)

1 2K L 1 2f f 0.32N 0.40N

4N1=5N

2

11 L

4NN x 40

5

xL=32

Now XR=distance when right finger stops and left finger

starts moving(

n=0)about centre N

1x

L=N

2(x

R)

1 2L K 1 2f f 0.4N 0.32N

5N1=4N

2

22 R

4N32 N x

5

R

128x 25.6cm

5

Q.42 (A, C, D)

vm

x

by the angular momentum conservation about thesuspension point.

22m

mvx mx3

2 22

mvx 2vx

m 3xmx

3

For maximum d

0dx

Mx3

So the V

32

Motion of System of Particles and Rotational Motion - SelfStudys

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