REVISION TEST 5 (Page 184)
This Revision Test covers the material contained in chapters 12 to 15. The marks available are
shown at the end of each question.
Problem 1. A train is travelling at 90 km/h and has wheels of diameter 1600 mm. (a) Find the
angular velocity of the wheels in both rad/s and rev/min. (b) If the speed remains constant for 2 km,
determine the number of revolutions made by a wheel, assuming no slipping occurs.
Marks
(a) 90 km/h = 90
=
km
m
1 h
1000

h
km 3600 s
90
m/s = 25 m/s i.e. the linear velocity, v, is 25 m/s.
3.6
1
The radius of a wheel is (1600/2) mm = 0.8 m
Linear velocity, v = r, hence  = v/r
i.e. the angular velocity,  =
25
= 31.25 rad/s
0.8
2
Angular velocity,  = 2n, where n is in revolutions per second.
Hence n = /2 and angular speed of a wheel in revolutions per minute is 60/2;
but  = 31.25 rad/s, hence
angular speed =
60  31.25
= 298.4 revolutions per minute
2
(b) Time taken to travel 2 km at a constant speed of 25 m/s =
2
2000 m
= 80 s
25 m / s
Since a wheel is rotating at 298.4 revolutions per minute, then
in 80/60 minutes it makes
298.4  80
= 398 revolutions
60
2
Total:
7
Problem 2. The speed of a shaft increases uniformly from 200 revolutions per minute to 700
revolutions per minute in 12 s. Find the angular acceleration, correct to 3 significant figures.
32
© John Bird & Carl Ross Published by Taylor and Francis
Marks
  1
Angular acceleration,  = 2
t
Initial angular velocity,  1 = 200 rev/min = 200/60 rev/s =
final angular velocity,  2 =
200  2
rad/s,
60
1
700  2
rad/s, and time, t = 12 s.
60
1
700  2 200  2

60
60
Hence, angular acceleration,  =
rad/s 2
12
=
500  2
= 4.36 rad/s 2
60  12
3
5
Total:
Problem 3. The shaft of an electric motor, initially at rest, accelerates uniformly for 0.3 s at
20 rad/s 2 . Determine the angle (in radians) turned through by the shaft in this time.
Marks
Angle turned through,  = 1 t +
1 2
t
2
Since the shaft is initially at rest, 1 = 0 and  =
1 2
t ,
2
the angular acceleration,  = 20 rad/s 2 and time t = 0.3 s
Hence, angle turned through,  =
1
 20  0.3 2 = 0.9 rad
2
4
Total:
4
Problem 4. Determine the momentum of a lorry of mass 10 tonnes moving at a velocity of 81 km/h
33
© John Bird & Carl Ross Published by Taylor and Francis
Marks
Momentum = mass  velocity
Mass = 10 t = 10000 kg and velocity = 81 km/h =
81
m/s = 22.5 m/s
3.6
2
2
Hence, momentum = 10000 kg  22.5 m/s = 225000 kg m/s
Total:
4
Problem 5. A ball of mass 50 g is moving with a velocity of 4 m/s when it strikes a stationary ball
of mass 25 g. The velocity of the 50 g ball after impact is 2.5 m/s in the same direction as before
impact. Determine the velocity of the 25 g ball after impact.
Marks
Mass, m 1 = 50 g = 0.050 kg, m 2 = 25 g = 0.025 kg
Initial velocity, u 1 = 4 m/s, u 2 = 0
Final velocity, v 1 = 2.5 m/s, v 2 is unknown.
Total momentum before impact = m 1 u 1 + m 2 u 2
= (0.050  4) + (0.025  0)
= 0.20 kg m/s
2
Total momentum after impact = m 1 v 1 + m 2 v 2
= (0.050  2.5) + (0.025 v 2 )
2
= 0.125 + 0.025 v 2
Total momentum before impact = total momentum after impact
Hence
1
0.20 = 0.125 + 0.025 v 2
from which,
velocity of 25 g ball after impact, v 2 =
0.20  0.125
= 3 m/s
0.025
2
Total:
7
34
© John Bird & Carl Ross Published by Taylor and Francis
Problem 6. A force of 24 N acts on a body of mass 6 kg for 150 ms. Determine the change in
velocity.
Marks
Impulse = applied force  time = change in linear momentum
24 N  0.15 s = mass  change in velocity
i.e.
= 6 kg  change in velocity
2
24 N  0.15s
= 0.60 m/s
6 kg
2
from which, change in velocity =
Total:
4
Problem 7. The hammer of a pile-driver of mass 800 kg falls a distance of 1.0 m on to a pile. The
blow takes place in 20 ms and the hammer does not rebound. Determine (a) the velocity of impact
(b) the momentum lost by the hammer (c) the average applied force exerted on the pile by the
hammer. Assume that g = 9.81 m/s 2 .
Marks
2
Initial velocity, u = 0, acceleration due to gravity, g = 9.81 m/s and distance, s = 1.0 m
(a) Using the equation of motion: v 2 = u 2 + 2gs
then
v 2 = 0 2 + 2(9.81)(1.0)
from which, impact velocity, v =
(2)(9.81)(1.0) = 4.429 m/s
3
(b) Neglecting the small distance moved by the pile and hammer after impact,
momentum lost by hammer = the change of momentum
= mv = 800 kg  4.429 m/s
= 3543 kg m/s
(c) Rate of change of momentum =
3
3543
change of momentum
=
= 177150 N
20 103
change of time
35
© John Bird & Carl Ross Published by Taylor and Francis
Since the impulsive force is the rate of change of momentum,
2
the average force exerted on the pile is 177.15 kN
Total:
8
Problem 8. Determine the mass of the moving head of a machine tool if it requires a force of 1.5 N
to bring it to rest in 0.75 s from a cutting speed of 25 m/min.
Marks
F
From Newton's second law, F = ma, thus m = , where force, F is 1.5 N
a
The law of motion v = u + at can be used to find acceleration a, where v = 0,
u = 25 m/min =
25
m/s, and t = 0.75 s
60
Thus,
from which, a = -
0=
25
+ a  0.75
60
3
25
= - 0.55 m/s 2 or a retardation of 0.55 m/s 2
60  0.75
Thus, the mass, m =
F
1.5
=
= 2.70 kg
.
a
0.55
2
Total:
5
Problem 9. Find the weight of an object of mass 2.5 kg at a point on the earth's surface where the
gravitational field is 9.8 N/kg.
36
© John Bird & Carl Ross Published by Taylor and Francis
Marks
weight = force acting vertically downwards
= mass  gravitational field
= 2.5  9.8 = 24.5 N
4
Total:
4
Problem 10. A van of mass 1200 kg travels round a bend of radius 120 m, at 54 km/h. Determine
the centripetal force acting on the vehicle.
Marks
2
The centripetal force is given by
mv
and its direction is towards the centre of the
r
circle. m = 1200 kg, v = 54 km/h =
Thus, centripetal force =
54
m/s = 15 m/s, and r = 120 m.
3.6
1200 152
= 2250 N or 2.25 kN
120
4
Total:
4
Problem 11. A spring, initially in a relaxed state, is extended by 80 mm. Determine the work done
by using a work diagram if the spring requires a force of 0.7 N per mm of stretch.
Force required for a 80 mm extension = 80 mm  0.7 N/mm = 56 N
Marks
1
The force/extension graph or work diagram representing the increase in extension in
proportion to the force, as the force is increased from 0 to 56 N, is shown below.
The work done is the area under the graph,
hence
work done =
1
1
 base  height =  80 mm  56 N
2
2
=
1
 80  10 3 m  56 N = 2.24 J
2
3
37
© John Bird & Carl Ross Published by Taylor and Francis
Total:
4
Problem 12. Water is pumped vertically upwards through a distance of 40.0 m and the work done
is 176.58 kJ. Determine the number of litres of water pumped. Assume that g = 9.81 m/s 2 and that
1 litre of water has a mass of 1 kg.
Marks
Work done = force  distance
i.e.
from which ,
176580 = force  40.0
force =
176580
= 4414.5 N
40.0
1
The force to be overcome when lifting a mass m kg vertically upwards is mg, i.e.
(m  9.81) N
Thus, 4414.5 = m  9.81, from which, mass, m =
4414.5
= 450 kg.
9.81
2
Since 1 litre of water has a mass of 1 kg,
450 litres of water are pumped
1
Total:
4
Problem 13. 3 kJ of energy are supplied to a machine used for lifting a mass. The force required is
1 kN. If the machine has an efficiency of 60%, to what height will it lift the mass?
38
© John Bird & Carl Ross Published by Taylor and Francis
Marks
60
useful output energy
output energy
Efficiency,  =
i.e.
=
100
input energy
3000 J
from which,
output energy =
60
 3000 = 1800 J
100
2
1800 J
= 1.8 m
1000 N
2
Work done = force  distance moved, hence
1800 J = 1000 N  height, from which, height =
Total:
4
Problem 14. When exerting a steady pull of 450 N, a lorry travels at 80 km/h. Calculate (a) the
work done in 15 minutes and (b) the power required.
Marks
(a) Work done = force  distance moved.
1
The distance moved in 15 min, i.e. h, at 80 km/h = 20 km.
4
Hence,
2
work done = 450 N  20000 m = 9000 kJ or 9 MJ
9 106 J
work done
(b) Power required =
=
= 10000 W or 10 kW
time taken
15  60 s
2
Total:
4
Problem 15. An electric motor provides power to a winding machine. The input power to the
motor is 4.0 kW and the overall efficiency is 75%. Calculate (a) the output power of the machine,
(b) the rate at which it can raise a 509.7 kg load vertically upwards.
39
© John Bird & Carl Ross Published by Taylor and Francis
Marks
(a) Efficiency,  =
power output
power input
from which, power output =
i.e.
75
power output
=
100
4000
2
75
 4000 = 3000 W or 3 kW
100
(b) Power output = force  velocity, from which, velocity =
power output
force
Force acting on the 509.7 kg load due to gravity = 509.7 kg  9.81 m/s 2 = 5000 N
Hence, velocity =
3000
= 0.60 m/s or 600 mm/s
5000
2
Total:
4
Problem 16. A tank of mass 4800 kg is climbing an incline at 12 to the horizontal. Determine the
increase in potential energy of the tank as it moves a distance of 40 m up the incline.
Marks
With reference to the diagram below, sin 12 =
from which,
h
opposite
=
40
hypotenuse
1
h = 40 sin 12 = 8.316 m
Hence, increase in potential energy = mgh
= 4800 kg  9.81 m/s 2  8.316 m
3
= 391584 J or 391.6 kJ
Total:
4
40
© John Bird & Carl Ross Published by Taylor and Francis
Problem 17. A car of mass 500 kg reduces speed from 108 km/h to 36 km/h in 20 s. Determine the
braking power required to give this change of speed.
Marks
Change in kinetic energy of car =
1
1
m v1 2 - m v 2 2
2
2
where m = mass of car = 500 kg,
v 1 = initial velocity = 108 km/h =
v 2 = final velocity = 36 km/h =
108
m/s = 30 m/s and
3.6
36
m/s = 10 m/s
3.6
Hence, change in kinetic energy =
1
1
m( v12 - v 2 2 ) = (500)(30 2 - 10 2 )
2
2
3
= 200000 J
Braking power =
change in energy
200000 J
=
= 10000 W or 10 kW
time taken
20 s
1
Total:
4
TOTAL MARKS FOR REVISION TEST 5: 80
41
© John Bird & Carl Ross Published by Taylor and Francis