Lectures for F.5 - Mechanics

7/30/2019 Lectures for F.5 - Mechanics

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HKPhO Pre-Training 2005

1

Cover Page

The materials in this note serve as a guideline for the mechanics topics of HKPhO in addition

to the S4-5 syllabus before 2003. Videos of the lecture and the tutorial will be available inhttp://hkpho.phys.ust.hk/ . While the key issues are addressed explicitly, details of secondary

importance are mostly left out. Students are encouraged to spend additional time to furtherdigest the contents. Small gaps are left for the students to work out the details, and learnphysics through the process. Additional references should be sought to further understand the

topics like vectors, elementary calculus, etc., most of which can be found in relevant F6-7textbooks.

We hope that the pre-training will give a kick start for those who hope to do well in HKPhO

2006 and beyond. We also encourage the trainees to share the materials with their classmates

back in schools.

Z. YangHKPhO Committee Chairman

http://hkpho.phys.ust.hk/






2

Mechanics

1 Vector

1.1

Any vector000 z A y A x A A z y x

(1.1)

Addition of two vectors:

000 )()()( z B A y B A x B A B AC z z y y x x

(1.2)

Dot product of two vectors

z z y y x x B A B A B A AB B A cos

(1.3)

It is also referred to as the projection of A

on B

, or vise versa.

Amplitude of the vector

A A A A A A z y x

222|| (1.4)

1.2

Position vector of a particle: 000 z z y y x xr

(1.5).

If the particle is moving, then x, y, and z are function of time t .Velocity:

000000 zv yv xv zdt

dz y

dt

dy x

dt

dx

dt

r d v z y x

(1.6)

Acceleration

00002

2

02

2

02

2

000 za ya xa zdt

zd ydt

yd xdt

xd zdt

dv ydt

dv

xdt

dv

dt

vd a z y x

z y x

(1.7)

1.3 Uniform circular motion

Take the circle in the X-Y plane, so z = 0,

)cos( t R x , )sin( t R y (1.8)

is the angular speed. is the initial phase. Both are constants.

Using the above definition of velocity (1.6),

00 )cos()sin( yt R xt Rv

(1.9),

v

is always perpendicular to r

.

Its amplitude is Rv (1.10)

The acceleration is:

r yt R xt Ra 2

0

2

0

2 )cos()sin( (1.11).

Its amplitude is R

v Ra

22 (1.12)

A

B

A

B

x

y

t

x

y

t

( x,y,z)

x

y

z

r

o

( x,y,z)

x

y

z

r

( x,y,z)

x

y

z ( x,y,z)

x

y

z

r

o

C

B

A

C

B

A




3

2 Relative Motion

A reference frame is needed to describe any motion of an object.

Consider two such reference frames S and S’ with their origins at

O and O’, respectively. The X-Y-Z axes in S are parallel to the X’-Y’-Z’ axes in S’. One is moving relative the other.

Note: Rr r

' . (2.1)

So the velocity is: uvdt

Rd

dt

r d

dt

r d v

''

(2.2)

Similar for acceleration: Aaa

' (2.3)

This is the classic theory of relativity. If u

is constant, then 'aa

, i. e., Newton’s Laws work

in all inertia reference frames.Properly choosing a reference frame can sometimes greatly simplify the problems.

3 Forces

Pull through a rope, push, contact forces (elastic force and friction force), air resistance, fluid

viscosity, surface tension of liquid and elastic membrane, gravity, electric and magnetic,

strong interaction, weak interaction. Only the last four are fundamental. All the others are the

net effect of the electric and magnetic force.

3.1 TensionPulling force (tension) in a thin and light rope:

Two forces, one on each end, act along the rope direction. The

two forces are of equal amplitude and opposite signs becausethe rope is massless. It is also true for massless sticks.

3.2 ElasticsElastic contact forces are due to the deformation of solids.

Usually the deformation is so small that it is not noticed. The

contact force is always perpendicular to the contact surface. In

the example both 1F

and 2F

are pointing at the center of the

sphere.

3.3 Friction

The friction force between two contact surfaces is caused by the relative motion or the

tendency of relative motion. Its amplitude is proportional to the elastic contact force, so afriction coefficient can be defined.

When there is relative motion, the friction force is given by f =

k N , where k is the kinetic friction coefficient. The directionof the friction is always opposite to the direction of the

relative motion.

O

O’

r

'r

R

O

O’

r

'r

R

T

T

T

T

1F

2F

1F

2F

v

N

f

v

N

f




4

When there is no relative motion but a tendency for such

motion, like a block is being pushed by a force F

, theamplitude of f is equal to F until it reaches the limiting value

of s N if F keeps increasing, where s is the static friction

coefficient.

Once the block starts to move, the friction becomes f = k N . Usually k < s.

3.4 Viscosity

Air resistance and fluid viscous forces are proportional to the relative motion speed and the

contact area. A coefficient called viscosity is used in these cases.

3.5 Inertial force

Inertial force is a ‘fake’ force which is present in a reference frame (say S’) which itself is

accelerating. Recall that Aaa

' . Assume frame-S is not accelerating, then according to

Newton’s Second Law, )'( AamamF

. So in the S’-frame, if one wants to correctly

apply Newton’s Law, she will get 'am AmF

, i. e., there seems to be an additional force

AmF

int (3.1)

acting upon the object.

Example 3.1

A block is attached by a spring to the wall and placed on the

smooth surface of a cart which is accelerating. According to

the ground frame, the force F

acting on the block by the spring

is keeping the block accelerating with the cart, so amF

. Inthe reference frame on the cart, one sees the block at rest but

there is a force on the block by the spring. This force is

‘balanced’ by the inertial force amF

int

3.6 Gravity

Between two point masses M and m, the force is

r r

GMmF

2 (3.2)

The gravitation field due to M is r r

GM

g

2 (3.3).

The field due to a sphere at any position outside the

sphere is equal to that as if all the mass is concentrated atthe sphere center. (Newton spent nearly 10 years trying to

proof it.)

Potential energyr

GM U

(3.4).

N

f

F

N

f

F

M

M

=

M

M

=

r

M m

r

M m

amF

Ground frame

a

Cart frame

amF

amF

int

amF

Ground frame

a

Cart frame

amF

amF

int




5

Application of superposition: Uniform densitylarger sphere with a smaller spherical hole.

Near Earth surface, because the large radius

of Earth, the gravitation field of Earth can betaken as constant, and its amplitude is

2

E

E

R

GM g = 9.8 m/s

2(3.5).

Its direction is pointing towards the center of Earth, which in practice can be regarded as

‘downwards’ in most cases.

Example 3.2

The ‘weight’, or the force of the ground on a person ondifferent places on Earth. Take the radius of Earth as R, and

the rotation speed being (= 2 /86400 s-1

).

On the Equator, we have mg - N = ma = m 2 R,

so N = mg - m 2 R = m(g -

2 R)

At the South Pole (or North Pole), we have N = mg

At latitude , by breaking down the forces along

the X-direction and Y-direction, we have

ma f N mg sincos)( , and

0cossin)( f N mg .

Solving the two equations, we get cos)( ma N mg , sinma f , where cos2 Ra .

The negative sign of f means that its direction is the opposite of what we have guessed.

One can also break down the forces along the tangential and radial directions to obtain the

same answers. One can also take Earth as reference frame and introduce the inertia force toaccount for the rotational acceleration.

3.7 Buoyancy

In a fluid (liquid or gas) of mass density at depth H ,consider a column of it with cross section area A, then the

total mass of the column is AH , and the gravity on it is

AHg. The gravity must be balanced by the supporting forcefrom below, so the force of the column on the rest of the

liquid is F = AHg and pointing downwards. The pressure

=

1r

2r

1r

2

r

O O O

-=

1r

2r

1r

2

r

O O O

-

y

x

gm

gm

N

N

N

gm f

a

y

x

gm

gm

N

N

N

gm f

a

gm H gm H




6

P = F/A = Hg (3.6).

Now consider a very small cubic of fluid with all six side area of A at depth H . The force on

its upper surface is AHg and pointing down, the force on its lower surface is AHg but

pointing upwards so the cubic is at rest. However, for the cubic not to be deformed by the two

forces on its upper and lower surfaces, the forces on its side surfaces must be of the samemagnitude. This leads to the conclusion that the pressure on any surface at depth H is AHg,and its direction is perpendicular to the surface. One can then easily prove that the net force of

the fluid (buoyancy) on a submerged body of volume V is equal Vg. (See the HKPhO 2003

paper.) The buoyancy force is acting on the center of mass of the submerged portion of theobject.

3.8 Torque

When two forces of equal amplitude and opposite directionsacting upon the two ends of a rod, the center of the rod remains

stationary but the rod will spin around the center. The torque (of a force) is introduced to describe its effect on the rotationalmotion of the object upon which the force is acting. First, an

origin (pivot) point O should be chosen. The amplitude of the

torque of force F

is

= rF (3.7),

where r is the distance between F

and the origin O. Thedirection of the torque (a vector as well) is point out of the paper

surface using the right hand rule. One can choose any point as

origin, so the torque of a force depends on the choice of origin.However, for two forces of equal amplitude and opposite

directions, the total torque is independent of the origin.

The general form of torque is defined as

F r

(3.8),

which involves the cross-product of two vectors.

4 Oscillations

4.1 Simple Harmonic Motion

F

r

OF

r

F

r

O

F

O

r

F

O

r




7

Frequency f and Period T: f

T 1

)cos()( t xt xm

(4.1)

(a) Effects of different amplitudes

(b) Effects of different periods

(c) Effects of different phases

Since the motion returns to its initial value afterone period T ,

],)(cos[)cos( T t m xt m x

,)(2 T t t

.2 T

Thus .22

f T

(4.2)




8

Velocity

)],cos([)( t xdt

d

dt

dxt v

m(4.3a)

)].sin()( t xt vm

(4.3b)

Velocity amplitude:mm

xv (4.4).

Acceleration

)],sin([)( t xdt

d

dt

dvt a

m

.)]cos()( 2 t xt am

(4.5)

Acceleration amplitudemm

xa2 (4.6).

This equation of motion will be very useful in identifying simple harmonic motion and its

frequency.

4.2 The Force Law for Simple Harmonic Motion

Consider the simple harmonic motion of a block of mass m subject to the elastic force of a

spring

kxF (Hook’s Law) (4.7).

Newton’s law:

.makxF

.02

2

kxdt

xd m

.02

2

x

m

k

dt

xd (4.8)

Comparing with the equation of motion for simple harmonic motion,

2 k

m. (4.9)

Simple harmonic motion is the motion executed by an object of mass m subject to a force that

is proportional to the displacement of the object but opposite in sign.




9

Angular frequency:

m

k (4.10)

Period:

k

mT 2 (4.11)

Examples 4.1

A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 Nm-1

.

The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionlesssurface and released from rest at t = 0.

(a) What force does the spring exert on the block just before the block is released?(b) What are the angular frequency, the frequency, and the period of the resulting oscillation?

(c) What is the amplitude of the oscillation?

(d) What is the maximum speed of the oscillating block?(e) What is the magnitude of the maximum acceleration of the block?

(f) What is the phase constant for the motion?

Answers:

(a) N2.711.065 kxF

(b) 1-srad78.9

68.0

65

m

k

Hz56.12

f

s643.0

1

f T

(c) cm11m x

(d)-1ms08.1 mm xv

(e) 2-22 ms10.50.1178.9 mm xa

(f) At t = 0, 11.0cos)0( m x x (1)

0sin)0( m xv (2)

(2): 0sin 0

Example 4.2

At t = 0, the displacement of x(0) of the block in a linear oscillator is – 8.50 cm. Its velocity

v(0) then is – 0.920 ms-1

, and its acceleration a(0) is 47.0 ms-2

.

(a) What are the angular frequency and the frequency f of this system?

(b) What is the phase constant ?

(c) What is the amplitude xm of the motion?




10

(a) At t = 0,

.085.0cos)cos()( mm xt xt x (1)

.920.0sin)sin()( mm xt xt v (2)

.0.47cos)cos()(22 mm xt xt a (3)

(3) (1): .)0()0( 2 x

a

.srad5.230850.0

0.47

)0(

)0( 1-

x

a

(answer)

(b) (2) (1):.tan

cos

sin

)0(

)0(

x

v

.4603.0)085.0)(51.23(

920.0

)0(

)0(tan

x

v

= – 24.7o

or 180o – 24.7

o= 155

o.

One of these 2 answers will be chosen in (c).

(c) (1): .cos

)0(

x xm

For = – 24.7o,

.cm49m)7.24cos(

085.0 . x

om

For = 155o, .cm49m

155cos

085.0 . x

om

Since xm is positive, = 155o

and xm = 9.4 cm.(answer)

Example 4.3

A uniform bar with mass m lies symmetrically across two rapidly rotating, fixed rollers, A and B, with distance L = 2.0 cm between the bar’s centre of mass and each roller. The rollers slip

against the bar with coefficient of kinetic friction k = 0.40. Suppose the bar is displaced

horizontally by a distance x, and then released. What is the angular frequency of the

resulting horizontal simple harmonic (back and forth) motion of the bar?

Newton’s law:

.0 mgF F F B A y (1)

.makB f kA f xF (2)

or .ma B

F k A

F k

Considering torques about A,.000)(20

kB f

kA f x Lmg L

BF

AF z (3)

or ).(2 x Lmg LF B

(3): .2

)(

L

x LmgF B

(1): .2

)(

L

x LmgF mgF B A




11

(2): .)]()([2

ma x Lmg x Lmg L

k

.0 x L

k g

a

Comparing with 02

2

2

xdt

xd for simple harmonic motion, L

gk

2 ,

.srad1402.0

)8.9)(40.0( 1- L

gk (answer)

4.3 Energy in Simple Harmonic Motion

Potential energy:

Since ),cos()( t xt xm

U t kx kx t m

( ) ). 1

2

2 1

2

2 cos (2 (4.12)

Kinetic energy:

Since ),sin()( t xt vm

).(sin)( 222

2

12

2

1 t xmmvt K

m

(4.13)

Since 2 k m / ,

K t kx t m

( ) ). 1

2

2sin (2

Mechanical energy:

E U K kx t kx t

kx t t

m m

m

1

2

2 1

2

2

1

2

2

cos ( sin (

[cos ( sin (

2 2

2 2

) )

) )].

Since cos ( sin (2 2 t t ) ) ,1

E U K kxm

1

2

2 .(4.14)




12

(a) The potential energy U (t ), kinetic energy K (t ), and mechanical energy E as functions of

time, for a linear harmonic oscillator. Note that all energies are positive and that the potentialenergy and kinetic energy peak twice during every period. (b) The potential energy U (t ),

kinetic energy K (t ), and mechanical energy E as functions of position, for a linear harmonic

oscillator with amplitude X m. For x = 0 the energy is all kinetic, and for x = X m it is all

potential.

The mechanical energy is conserved.

4.4 The Simple Pendulum

Consider the tangential motion acting on the mass.

Using Newton’s law of motion,

,sin2

2

dt

d mLmamg

d

dt

g

L

2

20

sin . (4.14)

When the pendulum swings through a small angle,

sin . Therefore

d

dt

g

L

2

20

.

(4.15)

Comparing with the equation of motion for simple harmonic motion,

L

g2 and

T L

g 2 .

5 The Centre of Mass

5.1 Definition

The centre of mass of a body or a system of

bodies is the point that moves as though all of

the mass were concentrated there and all

external forces were applied there.

For two particles,

.2211

21

2211

M

xm xm

mm

xm xm x

cm




13

For n particles,

.11

M

xm xm x nn

cm

In general and in vector form,

.1

1

n

icm i

r i

m M

r

(5.1)

If the particles are in a uniform gravity field, then the

total torque relative to the center of mass is zero. The

same applies for the inertia force when the particlesare in an accelerating reference frame.

Proof:

0)()()(111

gr M r M gmr r mgr r m cmcmcm

n

ii

n

iii

n

icmii

5.2 Rigid Bodies

dV z M

dm z M

z

dV y M

dm y M

y

dV x M

dm x M

x

cm

cm

cm

11

,11

,11

(5.2)

where is the mass density.

If the object has uniform density,

.V

M

dV

dm (5.3)

Rewriting dV dm and V m , we obtain




14

.1

,1

,1

dV zV z

dV yV

y

dV xV

x

cm

cm

cm

(5.4)

Similar to a system of particles, if the rigid body is in a uniform gravity field, then the total

torque relative to its center of mass is zero. This is true even when the density of the object isnon-uniform. The same applies to the inertia force. The proof is very much the same as in the

case for particles. One only needs to replace the summation by integration operations.

5.3 Newton’s Second Law for a System of Particles

In terms of X-Y-Z components,

.

,

,

,,

,,

,,

zcm zext

ycm yext

xcm xext

MaF

MaF

MaF

(5.5)

5.4 Linear Momentum

For a single particle, the linear momentum is

.vm p

(5.6)




15

Newton’s Law:

.)(dt

pd vm

dt

d

dt

vd mamF

(5.7)

This is the most general form of Newton’s Second Law. It accounts for the change of mass as

well.

.i f

p pdt F I

(5.8)

The change of momentum is equal to the time integration of the force, or impulse.

For a system of particles, the total linear momentum is

.111 nnn

vmvm p pP

(5.9)

Differentiating the position of the centre of mass,

.11 nncm

vmvmv M

.cm

v M P

(5.10)

The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the centre of mass.

Apply Newton’s Law’s to the particle system,

,)( i jij

ex t

iii F F t am

i i jij

ext

ii

ii

iii

ii

iii

F F M

am M

A

vm M

V

masstotalm M xm M

X

11

1

)(,1

According to the Third Law, jiij F F

. So

0i j

ijF

(5.11)

and

M

F F

M A

ex t

tot

i

ex t

i

)0(

1 (5.12)




16

Newton’s law:

.cm

a M dt

cmvd

M dt

Pd

(5.13)

Hence

.dt

Pd F ext

(5.14)

If 0 ext F

, then const vmV M i

ii

(5.15)

The total momentum of a system is conserved if the total external force is zero.

5.5 Rigid body at rest

The necessary and sufficient conditions for the balance of a rigid body is that net externalforce = 0, and the net torque due to these external forces = 0, relative to any origin (pivot).

Choosing an appropriate origin can sometimes greatly simplify the problems. A common trick is choosing the origin at the point where an unknown external force is acting upon.

Example 5.1A uniform rod of length 2l and mass m is fixed on one end by

a thin and horizontal rope, and on the wall on the other end.

Find the tension in the rope and the force of wall acting upon

the lower end of the rod.Answer:The force diagram is shown. Choose the lower end as the

origin, so the torque of the unknown force F

is zero. By

balance of the torque due to gravity and the tension, we get

mglsin - Tlcos = 0, or T = mg tan

Breaking F

along the X-Y (horizontal-vertical) directions, we get F x = T , and F y = mg.It is interesting to explore further. Let us choose another point of origin for the consideration

of torque balance. One can easily verify that with the above answers the total torque isbalanced relative to any point of origin, like the center of the rod, or the upper end of the rod.

Can you prove the following?If a rigid body is at rest, the total torque relative to any pivot point is zero.

5.6 Conservation of Linear Momentum

If the system of particles is isolated (i.e. there are no external forces) and closed (i.e. noparticles leave or enter the system), then

gmF

T

gmF

gmF

T




17

.constantP

(5.16)

Law of conservation of linear momentum:

. f i

PP

(5.17)

Example 5.2

Imagine a spaceship and cargo module, of total mass M , traveling in deep space with velocityvi = 2100 km/h relative to the Sun. With a small explosion, the ship ejects the cargo module,

of mass 0.20 M . The ship then travels 500 km/h faster than the module; that is, the relativespeed vrel between the module and the ship is 500 km/h. What then is the velocity v f of the

ship relative to the Sun?

Using conservation of linear momentum,

f iPP

f rel f i Mvvv M MV 8.0)(2.0

rel f i vvv 2.0

reli f vvv 2.0

= 2100 + (0.2)(500)

= 2200 km/h (answer)

Example 5.3

Two blocks are connected by an ideal spring and are free to slide on a frictionless horizontal

surface. Block 1 has mass m1 and block 2 has mass m2. The blocks are pulled in oppositedirections (stretching the spring) and then released from rest.

(a) What is the ratio v1 / v2 of the velocity of block 1 to the velocity of block 2 as theseparation between the blocks decreases?

(b) What is the ratio K 1 / K 2 of the kinetic energies of the blocks as their separation decreases?

Answer

(a) Using conservation of linear momentum,

f i PP

22110 vmvm

1

2

2

1

m

m

v

v

(b)1

2

2

1

2

2

1

2

2

1

2

1

2

22

2

11

2

1

2

12

1

m

m

m

m

m

m

v

v

m

m

vm

vm

K

K




18

Example 5.4

A firecracker placed inside a coconut of mass M , initially at rest on a frictionless floor, blows

the fruit into three pieces and sends them sliding across the floor. An overhead view is shown

in the figure. Piece C , with mass 0.30 M , has final speed v fc=5.0ms

-1

.

(a) What is the speed of piece B, with mass 0.20 M ?

(b) What is the speed of piece A?

Answer:

(a) Using conservation of linear momentum,

(b) fxix PP

fyiy PP

050cos80cos oo fA A fB B fC C vmvmvm (1)

050sin80sinoo

fB B fC C vmvm (2)m A = 0.5 M , m B = 0.2 M , mC = 0.3 M .

(2): 050sin2.080sin3.0 oo fB fC Mv Mv

1-1-

o

o

ms6.9ms64.950sin2.0

80sin)5)(3.0( fBv (answer)

(b) (1): fA fB fC

Mv Mv Mv 5.050cos2.080cos3.0 oo

1-oo

ms0.35.0

50cos)64.9)(2.0(80cos)5)(3.0(

fAv

(answer)

5.7 Elastic Collisions in One Dimension

In an elastic collision, the kinetic energy of each colliding body can change, but the total

kinetic energy of the system does not change.

In a closed, isolated system, the linear momentum of each colliding body can change, but the

net linear momentum cannot change, regardless of whether the collision is elastic.

In the case of stationary target, conservation of

linear momentum:

.221111 f f i

vmvmvm

Conservation of kinetic energy:




19

.222

211

211 2

1

2

1

2

1 f f i

vmvmvm

Rewriting these equations as

,)( 22111 f f i vmvvm

.)(222

21

2

11 f f ivmvvm

Dividing,

.211 f f i

vvv

We have two linear equations for v1f and v2f . Solution:

.2

,

121

12

121

21

1

i f

i f

vmm

mv

vmm

mmv

Motion of the centre of mass: .1

21

1

21icm

vmm

m

mm

Pv

Example 5.5

In a nuclear reactor, newly produced fast neutrons must be slowed down before they canparticipate effectively in the chain-reaction process. This is done by allowing them to collidewith the nuclei of atoms in a moderator .(a) By what fraction is the kinetic energy of a neutron (of mass m1) reduced in a head-on

elastic collision with a nucleus of mass m2, initially at rest?(b) Evaluate the fraction for lead, carbon, and hydrogen. The ratios of the mass of a nucleus

to the mass of a neutron (= m2 / m1) for these nuclei are 206 for lead, 12 for carbon and

about 1 for hydrogen.

Answer

(a) Conservation of momentum

f if i vmvmvm 22111

For elastic collisions,

2

22

2

11

2

112

1

2

1

2

1 f f i vmvmvm

f f i vmvvm 22111 )( (1)

2

22

2

1

2

11 )( f f ivmvvm (2)




20

Dividing (1) over (2), f f i vvv 211 (3)

(1): )()( 112111 f i f i vvmvvm

i f v

mm

mmv 1

21

211

Fraction of kinetic energy reduction

2

1

2

1

2

1

2

1

2

1

2

11

2

11

2

11

1

2

12

1

2

1

i

f

i

f i

i

f i

i

f i

v

v

v

vv

vm

vmvm

K

K K

2

21

21

2

21

21

)(

41

mm

mm

mm

mm

(answer)

(b) For lead, m2 = 206m1,

Fraction %9.1207

)206(4

)206(

)206(422

11

11

mm

mm(answer)

For carbon, m2 = 12m1,

Fraction %2813

)12(4

)12(

)12(422

11

11

mm

mm(answer)

For hydrogen, m2 = m1,

Fraction %100)(

)(42

11

11

mm

mm(answer)

In practice, water is preferred.

5.8 Inelastic Collisions in One Dimension

In an inelastic collision, the kinetic energy of thesystem of colliding bodies is not conserved.

In a completely inelastic collision, the collidingbodies stick together after the collision.

However, the conservation of linear momentum still holds.

,)(211

V mmvm or .21

1 vmm

mV

Examples 5.6




21

The ballistic pendulum was used to measure the speeds of bullets before electronic timing

devices were developed. Here it consists of a large block of wood of mass M = 5.4 kg,hanging from two long cords. A bullet of mass m = 9.5 g is fired into the block, coming

quickly to rest. The block + bullet then swing upward, their centre of mass rising a vertical

distance h = 6.3 cm before the pendulum comes momentarily to rest at the end of its arc.

(a)

What was the speed v of the bullet just prior to the collision?(b) What is the initial kinetic energy of the bullet? How much of this energy remains as

mechanical energy of the swinging pendulum?

Answer

(a) Using conservation of momentum during collision,

V m M mv )(

Using conservation of

energy after collision,

ghm M V m M )()(2

1 2

ghV 2

V m

m M v

1-ms630)063.0)(8.9(2

0095.0

0095.04.52

gh

m

m M

(b) Initial kinetic energy

J1900630)0095.0(2

1

2

1 22 mvK

Final mechanical energy

J3.3)063.0)(8.9)(0095.04.5()( ghm M E

(only 0.2%) (answer)

Example 5.7

(The Physics of Karate) A karate expert strikes downward with his fist (of mass m1 = 0.70 kg),

breaking a 0.14 kg wooden board. He then does the same to a 3.2 kg concrete block. The

spring constants k for bending are 4.1 104

Nm-1

for the board and 2.6 106

Nm-1

for theblock. Breaking occurs at a deflection d of 16 mm for the board and 1.1 mm for the block.

(a) Just before the board and block break, what is the energy stored in each?

(b) What fist speed v is required to break the board and the block? Assume that mechanicalenergy is conserved during the bending, that the fist and struck object stop just before the

break, and that the fist-object collision at the onset of bending is completely inelastic.

Answer

(a) For the board, J248.5016.0)101.4(2

1

2

1 242 kd U 5.2 J (answer)

For the block, J573.10011.0)106.2(2

1

2

1 262 kd U 1.6 J (answer)




22

(b) For the board, first the fist and the board undergoes

an inelastic collision. Conservation of momentum:

V mmvm )( 211 (1)

Then the kinetic energy of the fist and the board isconverted to the bending energy of the wooden board.

Conservation of energy:

(2):21

2

mm

U V

1-ms534.3

14.07.0

)248.5(2

(1): V m

mmv

1

21 534.3

7.0

14.07.0

4.2 ms

-1(answer)

For the concrete block,

21

2

mm

U V

1-

ms8981.02.37.0

)573.1(2

.

V m

mmv

1

21 8981.0

7.0

2.37.0

5.0 ms-1 (answer)

The energy to break the concrete block is 1/3 of that for the wooden board, but the fist speed

required to break the concrete block is 20% faster! This is because the larger mass of the

block makes the transfer of energy to the block more difficult.

5.9 Collisions in Two Dimensions

Conservation of linear momentum:

x component: ,22211111

coscos f f i

vmvmvm

y component: .222111

sinsin0 f f

vmvm

Conservation of kinetic energy:

.2

2

12

2

12

2

12

2

122112211 f f ii

vmvmvmvm

Typically, we know1

m ,2

m ,i

v1

and 1. Then we can solve for f v1 , f

v2

and 2.




23

Examples 5.8

Two particles of equal masses have an elastic collision, the target particle being initially at

rest. Show that (unless the collision is head-on) the two particles will always move off

perpendicular to each other after the collision.

Using conservation of momentum,

f f i vmvmvm 211

f f i vvv 211

The three vector form a triangle.

In this triangle, cosine law:

cos2 21

2

2

2

1

2

1 f f f f i vvvvv (1)

Using conservation of energy:

2

2

2

1

2

12

1

2

1

2

1 f f i mvmvmv

2

2

2

1

2

1 f f i vvv (2)

(1) – (2): 0cos2 21 f f vv

o90 (answer)

Example 5.9

Two skaters collide and embrace, in a completely inelastic collision. That is, they stick

together after impact. Alfred, whose mass mA is 83 kg, is originally moving east with speed vA

= 6.2 km/h. Barbara, whose mass mB is 55 kg, is originally moving north with speed vB = 7.8km/h.

(a) What is the velocity V

of the couple after impact?

(b) What is the velocity of the centre of mass of the two skaters before and after thecollision?

(c) What is the fractional change in the kinetic

energy of the skaters because of thecollision?

(a) Conservation of

momentum:

cos)( V mmvm B A A A (1)

sin)( V mmvm B A B B (2)

(2) (1):




24

A A

B B

vm

vm tan 834.0

)2.6)(83(

)8.7)(55( , so = 39.8o 40o (answer)

(1):o

8.39cos)5583(

)2.6)(83(

cos)(

B A

A A

mm

vmV = 4.86 km/h 4.9 km/h (answer)

(b) Velocity of the centre of mass is not changed by the collision. Therefore V = 4.9 km/h and

= 40o

both before and after the collision. (answer)

(c) Initial kinetic energy 22

2

1

2

1 B B A Ai vmvmK = 3270 kg km

2 /h

2.

Final kinetic energy 2)(2

1V mmK B A f = 1630 kg km

2 /h

2.

Fraction %503270

32701630

i

f i

K

K K (answer)

6 General equations of motion in the X-Y plane with constant forces

6.1 General formulae

t m

F vt v

t m

F vt v

y

yo y

x

o x x

)(

)(

(6.1)

2

2

2

1)(

2

1)(

t m

F t v yt y

t m

F t v xt x

y

yoo

x

o xo

(6.2)

These general formulae can be applied to any situations, once the initial conditions ( x0 , y0 , v x0 ,

v y0) are given.

A word of caution: Friction forces are not always ‘constant’. Pay attention to their directionsbecause they change with the direction of the velocity.

6.2 Projectile motion near Earth surface

The only force on the object is the gravity which is along the – Y direction. Accordingly, wehave

gt vt m

F vt v

vt m

F vt v

yo

y

yo y

o x

x

o x x

)(

)(

(6.3)



Lectures for F.5 - Mechanics

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