M2 Exam Questions

8/2/2019 M2 Exam Questions

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Mechanics M2 Exam

Questions

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Moments

Centre of Mass

Collisions

Work Energy Power

Kinematics (Vectors)

Work Energy Power 2

Projectiles

Question 1

• A uniform rod AB , of length 8a and weight W , is free torotate in a vertical plane about a smooth pivot at A . One endof a light inextensible string is attached to B . The other endis attached to point C which is vertically above A , with AC =6a . The rod is in equilibrium with AB horizontal, as shownbelow.

(a ) By taking moments about A , or otherwise, show that thetension in the string is 5/6W . (4)

5aW aT

Add the forces to diagram.

By Pythag CB = 10a

Taking moments about A

4aW=8aTSinB

(b ) Calculate the magnitude of the horizontal component of theforce exerted by the pivot on the rod.

a CosB

Add the forces to diagram.

Resolving forces horizontally.

X = TCosB

Therefore2

Question 2Figure 2 shows a metal plate that is made by removing a circleof centre O and radius 3 cm from a uniform rectangular laminaABCD , where AB = 20 cm and BC = 10 cm. The point O is 5 cmfrom both AB and CD and is 6 cm from AD .

(a ) Calculate, to 3 significant figures, the distance of thecentre of mass of the plate from AD . (5)

Circle Rectangle Plate

Masses 9∏ 200 200-9∏

Centreof mass

6 10 x

Centre of mass will lie onmirror line.

X = 10.7cm

9 6 (200 9 ) 200 10x

The plate is freely suspended from A and hangs in equilibrium.(b ) Calculate, to the nearest degree, the angle between AB

and the vertical. (3)

G will be directly below A. A B

Therefore angle GAB is given by:

10.7Tan

Question 3A small package P is modelled as a particle of mass 0.6 kg. Thepackage slides down a rough plane from a point S to a point T ,where ST = 12 m. The plane is inclined at an angle of 30 to thehorizontal and ST is a line of greatest slope of the plane, asshown in Figure 3. The speed of P at S is 10 m s–1 and the speedof P at T is 9 m s–1.

Calculate(a ) the total loss of energy of P in moving from S to T , (4)

KE at S = ½ × 0.6 × 100 = 30Js

KE at T = ½ × 0.6 × 81 = 24.3Js

KE lost = 5.7Js

PE Lost = mgh = 0.6 × 9.8 × 12Sin30º

= 35.28Js

Total loss of energy = 41.0Js

Calculate(b ) the coefficient of friction between P and the plane. (5)

Add forces to diagram, and resolveperpendicular to the plane.

R = 0.6 × 9.8 × Sin30º = 5.09

Using F = μR F = μ × 5.09

Work done against F = loss of energy

μ × 5.09 × 12 = 40.98

μ = 0.67

Question 4A particle P of mass 0.4 kg is moving under the action of a singleforce F newtons. At time t seconds, the velocity of P , v m s–1, isgiven by

v = (6t + 4)i + (t 2 + 3t ) j.When t = 0, P is at the point with position vector (–3i + 4 j) m.

(a ) Calculate the magnitude of F when t = 4. (4)

Using F = ma and the fact that the acceleration is differentialof the velocity vector.

v = (6t + 4)i + (t 2 + 3t ) j.

a = 6i + (2t + 3) j

When t = 4, a = 6i + 11 j therefore F = 0.4 × (6i + 11 j)

Magnitude of F =√(2.42 + 4.42) = 5.0

When t = 4, P is at the point S .(b ) Calculate the distance OS . (5)

t 3r=(3t + 4t)i+( + t )j+C i C j

2r = i (6t + 4)dt + j (t + 3t)dt

The position vector is found by integrating the velocity vector

When t = 0, P has position vector (-3i + 4 j).

When t = 4, r = 61i +49⅓j

32 2t 3

r=(3t + 4t-3)i+( + t +4)j

OS= 61 49 78m3

Question 5A car of mass 1000 kg is towing a trailer of mass 1500 kg along astraight horizontal road. The tow-bar joining the car to the

trailer is modelled as a light rod parallel to the road. The totalresistance to motion of the car is modelled as having constantmagnitude 750 N. The total resistance to motion of the trailer ismodelled as of magnitude R newtons, where R is a constant.When the engine of the car is working at a rate of 50 kW, thecar and the trailer travel at a constant speed of 25 m s–1.

(a ) Show that R = 1250. (3)

Power = Force × Velocity

Force = (50000/25) = 2000N

Constant velocity implies that F = 750 + R

Therefore R = 1250N

1000kg1500kg

750NRNF

When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remainunchanged. Calculate

(b ) the deceleration of the car while braking, (3)

Using F = ma 2000N + 1500N = 2500a Therefore a = 1.4ms-2

1000kg1500kg

750N1250N

Why have the arrows changed?

When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remain

unchanged. Calculate

(c ) the thrust in the tow-bar while braking, (2)

Equation of motion of car is T – 750 – 1500N = 1000 × -1.4

1000kg1500kg

750N1250N

Thrust = 850N

When travelling at 25 m s–1 the driver of the car disengagesthe engine and applies the brakes. The brakes provide aconstant braking force of magnitude 1500 N to the car. Theresisting forces of magnitude 750 N and 1250 N are assumed

to remain unchanged. Calculate

(d ) the work done, in kJ, by the braking force in bringing the carand the trailer to rest. (4)

u = 25ms-1 a = -1.4ms-2 v = 0, therefore s =223.2m

Work done = force x distance

Work done = 1500 x 223.2 = 335KJ

Using 2 2v u 2as

When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remainunchanged. Calculate

(e ) Suggest how the modelling assumption that the resistances tomotion are constant could be refined to be more realistic. (1)

Resistance varies with respect to speed.

Question 6A particle P of mass 3m is moving with speed 2u in a straight line on asmooth horizontal table. The particle P collides with a particle Q ofmass 2m moving with speed u in the opposite direction to P . The

coefficient of restitution between P and Q is e .

(a ) Show that the speed of Q after the collision is 0.2u (9e + 4).

By conservation of momentum 6mu – 2mu = 3mv1 + 2mv2

4u = 3v1

Coefficient of restitution = speed of separation/speed of approach

Therefore 3eu = v2 - v1

Remember to always draw a diagram

2m3mBefore

After QP

and v1 = v2 - 3eu (2)

Sub (2) into (1) 4u = 3v2 - 9eu +2v2 Hence v2 = 0.2u(9e+ 4)

As a result of the collision, the direction of motion of P isreversed.

(b ) Find the range of possible values of e . (5)

From a) v2 = 0.2u(9e+ 4) and v1 = v2 - 3eu

Therefore v1 = 0.2u(9e+4) - 3eu

Hence v1 = 0.4u(2 – 3e)

But v1<0 Therefore (2 – 3e) < 0

So e>(2/3) and e<1

Given that the magnitude of the impulse of P on Q is 6.4mu ,(c ) find the value of e . (4)

Impulse = change in momentum

6.4mu = 2m(0.2u(9e+4) +u)

6.4u = 3.6eu + 1.6u + 2u

2.8 = 3.6e

Question 7

A particle P is projected from a point A with speed 32 m s–1 at an angle of elevation , where sin = 3/5

The point O is on horizontal ground, with O vertically belowA and OA = 20 m. The particle P moves freely under gravityand passes through a point B , which is 16 m above ground,before reaching the ground at the point C , as shown above.

(a) the time of the flight from A to C , (5)

Vertical component of velocity is vsinα = 19.2ms-1

Therefore t = 4.77sec

Calculate

21s ut at

Using With s = -20, u = 19.2, a = -9.8

(b ) the distance OC , (3)

Horizontal component of velocity = vcosα =25.6ms-1

Distance = speed ×time = 25.6 × 4.77

= 122m

Calculate

(c ) the speed of P at B , (4)

Vertical component of velocity = 21.14ms-1

Horizontal component remains constant = 25.6ms-1

Calculate

2 2v u 2as Where s = -4, a = -9.8 and u = 19.2

Therefore by Pythagoras speed = 33.2ms-1

(d ) the angle that the velocity of P at B makes with the horizontal. (3)

Calculate

21.14Tan

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