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Linear motion Questions
1. a) Distinguish between the terms ‘uniform velocity’ and ‘uniform acceleration’
b) The figure below shows a section of a ticker tape. The dots were made at a frequency of 50 Hz.
Determine the acceleration of the trolley pulling the tape
c) The graph below shows a part of the motion of a basket ball which is projected vertically
upwards from the ground and is allowed to bounce on the ground
i) Explain the motion of the ball relating it to its different positions along the following
I.AB II.BC III.CE
ii) From the graph calculate the acceleration due to gravity
c) State Newton’s second law of motion
2. One end of a metal rod is heated in a flame. After some time the other end becomes hot.
Explain this observation
3. A bullet of mass 150g moving at an initial velocity of 80m/s strikes a suspended block
of mass 2.5kg
3. (a)The block swings from point A to B. Determine the vertical displacement between A and B
. (b) What observations are you likely to observe on the block after collision
4. The diagram below shows a velocity – time graph of a certain motion.
From the graph, determine the average speed of the body.
5. The diagram below shows a ball being whirled in a vertical plane.
(a) Sketch on the same diagram, the path followed by the ball if the string cuts when the ball
is at position shown in the diagram.
6. The figure below shows a circuit diagram for controlling temperature of a room.
(i) Explain the purpose of the strip.
(ii) Describe how the circuit controls the temperature when the switch S is closed.
7. The figure 5 below shows a uniform bar of length 1.0m pivoted near one end. The bar is kept
in equilibrium by a spring balance as shown:
Given that the reading of the spring balance is 0.6N, determine the reaction force at the pivot
8. The
figure 8 shows the motion of a train over a section of track which includes a sharp bend
(a) The section of the track with the sharp bend has a maximum speed restriction. The train
decelerates approaching the bend so that at the start of the bend, it has just reached the
maximum speed allowed. The train is driven around the bend at the maximum speed
allowed and accelerates immediately on leaving the bend. Calculate the length of the bend
(b) The train has to slow down to go round the bend. Calculate the deceleration
(c) As the train is driven round the bend, there is an extra force acting, called the centripetal force.
(i) On the figure 9 below, draw an arrow to show the direction of this force
(ii) State the effect that this force has on the motion
(iii) State how this force is provided
(d) Figure 10 below shows a car with a dummy driver before and after a collision test:
The mass of the dummy driver is 90kg. The impact time to reduce the dummy’s speed from
45ms-1 to zero is 1.2 seconds:
(i) Calculate the average force on the dummy during impact
(ii) State the main energy transformation during the collision
(iii) Calculate how much of the dummy’s energy is transformed during the collision
9. (a) The velocity-time graph in the figure below illustrates the motion of a ball which has
been projected vertically upwards from the surface of the moon. The weight of the object on
earth’s surface is 20N, when the acceleration due to gravity is 10ms-2.
(i) State why the velocity becomes negative after 3seconds.
(ii) Determine the acceleration of free fall on the moon showing clearly your work
(iii) Determine the total distance travelled by the ball in 5.0sec
(iv) Find the weight of the ball on the moon
(v) If the ball was projected vertically upwards on the earth with the same velocity.
What difference would you expect to observe in the velocity-time graph above. Illustrate with a
sketch on the same axis
(b) The figure below represents part of a tape pulled through a ticker-timer of frequency 50Hz
moving down an inclined plane.
If the trolley was allowed to move down the inclined plane for 4 seconds, calculate the distance
it covers
10. (a) State Boyle’s law
(b) The volume of a bubble at the base of a container of water is 3cm3. The depth of water is
30cm. The bubble rises up the column until the surface
(i) Explain what happens to the bubble as it rises up the water column
(ii) Determine the volume of the bubble at a point 5cm below the water surface
(c) A faulty thermometer records 11oC instead of 0oC and 98oC instead of 100oC. Determine
the reading on the thermometer when dipped in liquid at a temperature of 56oC
11. Figure 9 is a velocity- time graph describing the motion of a particle
What does the shaded area represent?
12. a) State Newton’s first law of motion
b) A parcel is to be dropped from an aeroplane traveling horizontally at 120ms-1, at an altitude
of 720m, to fall into a certain village.
Determine:
i) The time taken for the parcel to reach the ground
ii) How far ahead of the plane, the village should be when the parcel is released
c) A small stone, M1of mass 20g is attached to a string which in turn is passed through a smooth
thin cylinder. The other end of the string is tied to mass M2. The mass M1 is whirled in
a horizontal circle of radius 1m and mass M2 remains stationary as shown in figure 10
i) State two forces acting on the system other than the tension in the thread on M2
ii) Explain the observation made on mass M2 if the speed of M1, is increased
iii) Calculate the velocity of M1, if the mass M2 is 50g and the radius of the circle
is 1m
13. (a) Define uniform velocity
(b) The graph figure 10 below shows displacement –time graph of a in motion
fig 10
(i) Determine the instanteous velocities at t = 1second and at t = 4 seconds
(ii) Use the results in (b)(i) above to determine the acceleration of the body
14. A ball of mass 100g is kicked horizontally from the top of a cliff. If the ball takes 4 seconds
to hit the ground, determine the height of the cliff
15. A ball is kicked vertically upward from the ground with a velocity of 60m/s and reaches a maximum
height (h), it then falls freely back to the ground and bounces upwards to a height of 5M
(a) Sketch a velocity-time graph to represent the motion of the ball from the time it is kicked
vertically upwards until it bounces to a height of 5M
(b) Determine:
(i) the time taken by the ball to reach the maximum height(h)
(ii) The maximum height (h) reached by the ball
(iii) The velocity with which it bounces after striking the ground for the first time
(c) State any assumption made in your calculations in (b) above
16. In an experiment on momentum, trolley P of mass 800g was attached to a ticker timer of frequency 50Hz. Trolley P, initially moving with a velocity of 0.5m/s, was made to collide
with a stationary trolley Q of mass 400g. A copy of the tape as it appeared after the collision is presented in the figure below:-
(a) Determine the velocity of the trolley P after collision
(b) Calculate the impulsive force experienced by trolley P
(c) State the type of collision
17. I. (a) State the three equations of linear motion.
(b) A car is traveling uniformly at 100km/hr when the driver observes a road block ahead.
He takes 0.5 s before applying the brakes which brings the car to rest with a uniform
deceleration of 4m/s2. Determine the distance traveled by the car from the time the driver
observed the road block until the car comes to rest.
(c) A car moves at a constant speed of 20ms-1 for 50s and then accelerates uniformly to a speed
of 25ms-1 over a period of 10s. This speed is maintained for 50 s before the car is brought to
rest with uniform deceleration in 15s.
(d) Draw a graph of velocity (Y – axis) against time (graph paper to be availed)
(II) Calculate:
(i) The average speed for the whole journey.
(ii) The acceleration when the velocity changes from 20 ms-1 to 25ms-1 .
m, show that v2=2as +u2
18. Sketch a velocity-time graph for a body moving with zero acceleration
19. The figure below shows a velocity –time graph of a ball bouncing vertically upward from
the 
ground. The velocity upward is taken positive.
Determine the maximum height when the ball rises.
20. (a) On the axes provide below, sketch a graph of velocity V versus time (t) for uniformly
accelerated motion given that when t = 0, V is greater than zero.
(b) A car is brought to Rest from a speed of 20 ms-1 in time of 2 seconds. Calculate the
deceleration.
21. (a). State the law of linear momentum
(b). A marble of mass 50g moving on a horizontal surface at a velocity of V collides with another
glass marble of mass 75g resting on same horizontal surface. After collision, the marble
bounces back a long the path at a speed of 3.5m/s while the other marble moving with a
speed of 3.0m/s .Forward .
Determine the speed V.
(c). The paper below was attached to a trolley and pulled through a ticker tape times of frequency
50Hz. Determine the acceleration of the trolley.
(d). Study the figure below
Calculate the pressure in the steam in the cylinder which would just raise the piston if area of
of the piston in contact with steam is 2cm2 and Atmospheric pressure is1.0 x 105 Nm-2.
(e) State a reason why the earth is colder at night than daytime during a sunny
21. A block of mass 20kg slides downward a plane inclined of 6o0 with the horizontal. The coefficient of friction between the plane and the block is 0.4.
Calculate the acceleration of the block.
22. A body accelerates uniformly from initial velocity of U m/s to a final velocity of V m/s
in time t seconds. If acceleration during the motion is a m/s2 and the distance covered is S
Linear motion Answers
1. a) Uniform velocity :- The change in displacement for equal time intervals is the same.
Uniform acceleration:- Change in velocity for equal time intervals is the same.
b) Determine the acceleration of the trolley pulling the tape
Va = 2 = 100cm/s Vb = 3 = 150 cm/s a = V-U
0.02 0.02 t
= (150 -100)/ (7X0.02 – 0.02)
a = 416.67 cm/s2
c) i) Determine the motion of the ball relating it to its different positions along the following
I AB the body is projected upwards and at B V = O
II BC the body falls back to the starting point (moving in the opposite direction)
III CE the body be rebounds on the ground (at starting point) and starts moving up again
ii) From the graph calculate the acceleration due to gravity
a = v-u a = -10m/s2
t = 10m/s2
= 0-20
2
2. Conduction
Free electrons at the heated and gain more kinetic energy and spread the heat energy to other parts of the rod
3. (a) Momentum before collision = momentum after collision
150 x 80 = 2.65 x V
1000
16 = 2.65V
V = 16
2.65
= 6.0377
But ½ mV2 = mgh
h = V2
= (6.0377)
2g 2 x10
h = 36.4538
20
= 1.82269m
(b)The block will be deformed
4. Total distance = Area under graph;
= ½ (12 + 5) x 20; OR
= 170m;
Average speed = 170m
12s
= 14.17m/s;
5.
6. (i) Acts as a thermostat
(ii) On closing, the switch, current becomes complete, the current flows causing heating effect,
the bimetallic strip bends downwards and contents separates.
when the room becomes cool the strip bends upward completing the current and the process
repeats itself on and off regulating the temperature
– Weight of the fluid in which it floats
7. Clockwise moments = anticlockwise moments at equilibrium
0.6 x 0.7 = W x 0.3;
W = 0.6 x 0.7
0.3
= 1.4N;
0.6 + R = 1.4
R = 0.8N;
8. (a) Length = area under curve
= 10 x (32-18);
= 10 x 14 = 140m;
(b) 10-25 = -15; = -1.875ms-1
18-10 8 Decal = 1.875ms-1;
(c)(i)
(ii) Keep the train in circular motion;
(iii) Friction force between the wheels and rails;
(d) (i) F = m(v-u)
t
= 90(0-45);
1.2
= -3375N;
(ii) Kinetic energy – Heat + sound + P.E(deformation;
(iii) E = ½ Mv2
= ½ x 90 x 45s;
= 91,125J;
9. (a) (i) When a body is projected vertically upwards, it under goes a uniform retardation due to
the gravitational pull. The body thus slows down, comes to rest them starts falling with an
increasing velocity (in opposite direction)
(ii) Acc of free fall = gradient / slope of the graph
= 5 – 0 = 5 = 1.66ms-2
3 – 0 3
(iii) Total distance = Area under the curve
(½ x 5 x 3) +(½ x 2 x 3.3.)
15 + 10
= 30 + 20 = 50 = 25 = 8 1/3m
2 3 6 6 3
(iv) – Wt in the moon = mg = 2kg x 5/3 = 10/3 = 31/3N
(v) – It will accelerate faster at 10ms-1 from the graph
– It will attain a maximum height after ½ second
(b) V1 = 1.5cm = 75cms-1 = V2 = 3.0cm = 150cms-1
0.025 0.02s
a = V2 – V1 = 150 – 75 = 75 = 937.5cm-2 or 9.375ms-2
t 0.02 x 4 0.08
S = ut + ½ ast2
= (0.75 x 4) + ½ x 9.375 x 42 = 3 + 75 = 78m
10. a) The volume of a fixed mass of gas is inversely proportional to its pressure provided
temperature is kept constant.
(b) (i) The bubble expands as it comes up finally bursts when at the surface
(ii) p1V1 = P2V2
(76 + 30) x 3 = (76 + 5) V2
106 x3 = 81 X V2
V2 = 106 x 3
81
= 3.93cm3
(c) 100oC – 0oC = 98 -11
1 division = 87
100
Reading = 8 x 56
1000
= 48.72oC
11. Distance traveled1
12. a) A body continues with its state of rest or uniform motion unless acted upon by some
external forces1
b) i) s = ½ gt2
720 = ½ x 10 x t21
t2 = 144
t= √144 = 12 sec 1
ii) Range = ut
= 120 x 121
= 1440m 1
c) i) – Centripetal force acting on M1 1
– Weight (M2g) acting on M21
ii) M2 moves upwards; 1
When the speed of M1 increases centripetal force remains the same, the radius of the
circle described by M1 increases1
iii) Centripetal force = weight of M2
M1V2/r = M2g
0.020 V2/1 = 0.050 x 101
V2 = 0.5/0.02 = 251
V= √ 25 = 5m/s√1
13. (a) Constant rate of change of displacement with time
OR- A body is said to be moving with uniform velocity it its rate of change of displacement with
time is constant
(b) (i) For one correct tangent
Velocity t = 1s = 42 – 20
2 – 0.5 (correct reading form graph and expression
= 14.67m/s
Velocity at t = 4s = 67.5 – 30
5 – 0.5 (correct reading from graph ad expression)
= 8.33m/s (accuracy)
(ii) a = V – u = 8.33 – 14.67 = 6.34
t 4-1 3
= 2.11m/s2
14. S = ½ gt2 since u = 0
= ½ x 10 x 4 x 4
= 80m
15. b) i) t = v – u
g
= 0 – 60
-10
= 6 secs
ii) h = ut – ½ gt2
= 60 x 6 – ½ x 10 x 62
= 360 – 180
= 180m
iii) V2 = U2 + 2aS
O = U2 + 2 x -10 x 5
O = U2 – 100
U = 10m/s
c) Resistance/ friction with air is negligible
16. a) Length of nine dots = 6.9cm
Time taken = 1/50 x 9 = 0.02 x 9
= 0.185
Velocity = 6.9cm
0.18s
= 38cm/s or 0.38m/s
b) Ft = 0.8 x 0.5 – 0.8 x 0.38
Ft = 0.096
F = 0.096
0.18 = 0.533N
17. a) Equations of linear motion.
i) V = u + at. 1
ii) V2 = u2 + 2as 1
iii) S = ut + ½ at2. 1
b) 100 km/h x 10 = 27.78m/s
36
In 0.5 sec the driver covers 27.78m/s x 0.55 = 13.89 M 1
After applying brake
a = -4m/s2
u = 27.78m/s.
v= 0
∴ v2 = u2 + 2as 1
– 2as = u2. since v = 0
S = u2 = (27.78m/s)2 = 96.47M 1
-2a (-2) (-4m/s2)
Total distance covered = (13.89 + 96.476)M = 110.36 M1
c) (i) See the graph paper
(ii)
(i) Average speed of the whole journey = Total distance covered
Total time taken
Distance = Area under the graph
= (20m/s x 50s) + (½ (20 + 25) x 10) + ½ (50 + 65) x 25
= 1000m + 225m + 1437.5m = 2662.5m 1
Total time = 125s
Speed = 2662.5 = 21.3 m/s 1
125s
= 21.3 m/s 1
(ii) a = v-u
t
= (25 – 20) m/s 1
10 s
= 0.5m/s2
1
18.
