Mechanics M2 Exam
Questions
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Moments
Centre of Mass
Work Energy Power
Kinematics (Vectors)
Work Energy Power 2
Collisions
Projectiles
Question 1
• A uniform rod AB, of length 8a and weight W, is free to
rotate in a vertical plane about a smooth pivot at A. One end
of a light inextensible string is attached to B. The other end
is attached to point C which is vertically above A, with AC =
6a. The rod is in equilibrium with AB horizontal, as shown
below.
C
6a
B
A
8a
(a) By taking moments about A, or otherwise, show that the
C
tension in the string is 5/6W.
(4)
Add the forces to diagram.
By Pythag CB = 10a
Taking moments about A
6a
X
Y
T
B
A
8a
4aW=8aTSinB
S in B 
6a
3

10a 5
4aW  8aT 
5
T  W
6
3
5
W
(b) Calculate the magnitude of the horizontal component of the
force exerted by the pivot on the rod.
C
(3)
Add the forces to diagram.
Resolving forces horizontally.
X = TCosB
8a
4
CosB 

10a 5
Therefore
6a
X
5
T  W
6
2
X  W
3
Y
T
B
A
8a
W
Question 2
Figure 2 shows a metal plate that is made by removing a circle
of centre O and radius 3 cm from a uniform rectangular lamina
ABCD, where AB = 20 cm and BC = 10 cm. The point O is 5 cm
from both AB and CD and is 6 cm from AD.
20 cm
A
B
5 cm
3 cm
10 cm
O
6 cm
D
C
(a)
Calculate, to 3 significant figures, the distance of the
centre of mass of the plate from AD.
(5)
Centre of mass will lie on
mirror line.
20 cm
A
B
Circle
Rectangle Plate
Masses
9∏
200
2009∏
3
cm
Centre
of mass
6
10
x
O
9  6  (200  9 )x  200  10
X = 10.7cm
5
cm
10
cm
6
cm
D
C
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.
B
A
G
Therefore angle GAB is given by:
D
 5 

Tan 1 

25

 10.7 
C
Question 3
A small package P is modelled as a particle of mass 0.6 kg. The
package 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 the
horizontal and ST is a line of greatest slope of the plane, as
shown in Figure 3. The speed of P at S is 10 m s–1 and the speed
of P at T is 9 m s–1.
S
P
12 m
T
30
Calculate
(a)
the total loss of energy of P in moving from S to T, (4)
KE at S = ½ × 0.6 × 100 = 30Js
S
KE at T = ½ × 0.6 × 81 = 24.3Js
12
m
KE lost = 5.7Js
PE Lost = mgh = 0.6 × 9.8 × 12Sin30º
= 35.28Js
Total loss of energy = 41.0Js
P
T
30

Calculate
(b)
the coefficient of friction between P and the plane. (5)
Add forces to diagram, and resolve
perpendicular 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
R
F
S
P
12
m
T
30

Question 4
A particle P of mass 0.4 kg is moving under the action of a single
force F newtons. At time t seconds, the velocity of P, v m s–1, is
given by
v = (6t + 4)i + (t2 + 3t)j.
When t = 0, P is at the point with position vector (–3i + 4j) m.
(a)
Calculate the magnitude of F when t = 4.
(4)
Using F = ma and the fact that the acceleration is differential
of the velocity vector.
v = (6t + 4)i + (t2 + 3t)j.
a = 6i + (2t + 3)j
When t = 4, a = 6i + 11j therefore F = 0.4 × (6i + 11j)
Magnitude of F =√(2.42 + 4.42) = 5.0
When t = 4, P is at the point S.
(b)
Calculate the distance OS.
(5)
The position vector is found by integrating the velocity vector
r = i (6t + 4)dt + j (t2 + 3t)dt
t3 3 2
r=(3t + 4t)i+(
+ t )j+C1i  C2 j
3
2
2
When t = 0, P has position vector (-3i + 4j).
t3 3 2
 r=(3t + 4t-3)i+(
+ t +4)j
3
2
2
When t = 4, r = 61i +49⅓j
2
 2 
1 
OS=  61   49    78m

 3  

Question 5
A car of mass 1000 kg is towing a trailer of mass 1500 kg along a
straight horizontal road. The tow-bar joining the car to the
trailer is modelled as a light rod parallel to the road. The total
resistance to motion of the car is modelled as having constant
magnitude 750 N. The total resistance to motion of the trailer is
modelled as of magnitude R newtons, where R is a constant.
When the engine of the car is working at a rate of 50 kW, the
car and the trailer travel at a constant speed of 25 m s–1.
(a)
Show that R = 1250.
(3)
RN
750N
F
1500kg
T
T
1000kg
Power = Force × Velocity
Force = (50000/25) = 2000N
Constant velocity implies that F = 750 + R
Therefore R = 1250N
When travelling at 25 m s–1 the driver of the car disengages the
engine and applies the brakes. The brakes provide a constant
braking force of magnitude 1500 N to the car. The resisting
forces of magnitude 750 N and 1250 N are assumed to remain
unchanged.
Calculate
(b) the deceleration of the car while braking,
1250N
(3)
750N
Why have the arrows changed?
1500kg
T
T
1000kg
1500N
Using F = ma
2000N + 1500N = 2500a
Therefore a = 1.4ms-2
When travelling at 25 m s–1 the driver of the car disengages the
engine and applies the brakes. The brakes provide a constant
braking force of magnitude 1500 N to the car. The resisting
forces of magnitude 750 N and 1250 N are assumed to remain
unchanged.
Calculate
(c) the thrust in the tow-bar while braking,
1250N
1500kg
T
750N
T
1000kg
1500N
Equation of motion of car is
Thrust = 850N
T – 750 – 1500N = 1000 × -1.4
(2)
When travelling at 25 m s–1 the driver of the car disengages
the engine and applies the brakes. The brakes provide a
constant braking force of magnitude 1500 N to the car. The
resisting 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 car
and the trailer to rest.
(4)
Using
v2  u2  2as
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
When travelling at 25 m s–1 the driver of the car disengages the
engine and applies the brakes. The brakes provide a constant
braking force of magnitude 1500 N to the car. The resisting
forces of magnitude 750 N and 1250 N are assumed to remain
unchanged.
Calculate
(e) Suggest how the modelling assumption that the resistances to
motion are constant could be refined to be more realistic.
(1)
Resistance varies with respect to speed.
Question 6
A particle P of mass 3m is moving with speed 2u in a straight line on a
smooth horizontal table. The particle P collides with a particle Q of
mass 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).
Remember to always draw a diagram
2u
Before
P
3m
v1
u
Q 2m
After
P
By conservation of momentum
3m
v2
Q 2m
6mu – 2mu = 3mv1 + 2mv2
4u = 3v1 + 2v2
(1)
Coefficient of restitution = speed of separation/speed of approach
Therefore
Sub (2) into (1)
3eu = v2 - v1
and
4u = 3v2 - 9eu +2v2
v1 = v2 - 3eu
(2)
Hence v2 = 0.2u(9e+ 4)
As a result of the collision, the direction of motion of P is
reversed.
(b)
Find the range of possible values of e.
From a)
v2 = 0.2u(9e+ 4)
Therefore
v1 = 0.2u(9e+4) - 3eu
and
Hence v1 = 0.4u(2 – 3e)
But v1<0
Therefore (2 – 3e) < 0
So e>(2/3) and e<1
v1 = v2 - 3eu
(5)
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
e=7/9
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 below
A and OA = 20 m. The particle P moves freely under gravity
and passes through a point B, which is 16 m above ground,
before reaching the ground at the point C, as shown above.
Calculate
(a) the time of the flight from A to C,
Vertical component of velocity is vsinα = 19.2ms-1
Using
1
s  ut  at2
2
Therefore t = 4.77sec
With s = -20, u = 19.2, a = -9.8
(5)
Calculate
(b) the distance OC,
Horizontal component of velocity = vcosα =25.6ms-1
Distance = speed ×time = 25.6 × 4.77
= 122m
(3)
Calculate
(c) the speed of P at B,
Using v2  u2  2as
(4)
Where s = -4, a = -9.8 and u = 19.2
Vertical component of velocity = 21.14ms-1
Horizontal component remains constant = 25.6ms-1
Therefore by Pythagoras speed = 33.2ms-1
Calculate
(d) the angle that the velocity of P at B makes with the horizontal.
(3)
21.14
Tan 
25.6
   39.6