CAIE A-Level Physics 2 Kinematics
Practise analysing straight-line and projectile motion with definitions, graphs, constant-acceleration equations, experiments, uncertainty and physical constraints.
- Syllabus
- 2028–2030
- Course
- Physics 9702
- Level
- AS
Practise analysing straight-line and projectile motion with definitions, graphs, constant-acceleration equations, experiments, uncertainty and physical constraints.
A pendulum consists of a solid sphere suspended by a string from a fixed point P , as shown in Fig. 3.1.

Fig. 3.1 (not to scale)
The sphere swings from side to side. At one instant the sphere is at its lowest position X , where it has kinetic energy 0.86 J and momentum 0.72 Ns in a horizontal direction. A short time later the sphere is at position Y , where it is momentarily stationary at a maximum vertical height h above position X.
The string has a fixed length and negligible weight. Air resistance is also negligible.
On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X .
solid straight line drawn between centre of sphere at X and at Y
B1
A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1.

Fig. 3.1 (not to scale)
The sledge passes point A with speed 18 ms−1 at time t=0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A .
The variation with time t of the velocity v of the sledge from t=0 to t=24 s is shown in Fig. 3.2.

Fig. 3.2
State the time taken for the sledge to travel from A to B .
time =
time =12 s
A1
Determine the displacement of the sledge up the slope from point A at time t=24 s.
distance (up slope) =1/2×12×18 (= 108)
C1
distance ( down slope )=1/2×12×6(=36)
C1
displacement from A=108−36=72 m
A1
Show that the acceleration of the sledge as it moves from B back towards A is 0.50 m s−2.
v=u+ at or a= gradient or a=Δv/(Δ)t
C1
a=6/12=0.50( m s−2) (other points from the line may be used)
A1
or
v2=u2+2as and u=0
or
v2=2as
(C1)
a=6.02/(2×36)=0.50( ms−2)
(A1)
or
s=ut+1/2at2 and u=0
or
s=1/2at2
(C1)
a=2×36/122=0.50( ms−2)
(A1)
or
s=vt−21at2
(C1)
a=2×(6×12−36)/122=0.50( m s−2)
(A1)
The string is now used to move the cylinder in (a) vertically upwards through the water. The variation with time t of the velocity v of the cylinder is shown in Fig. 2.2.

Fig. 2.2
Use Fig. 2.2 to determine the acceleration of the cylinder at time t=2.0 s.
acceleration = ms−2
a=(v-u) / t or (Δ)v/(Δ)t or gradient
C1
= e.g. 8.0×10−2/2.0=4.0×10−2 m s−2
A1
The top face of the cylinder is at a depth of 0.32 m below the surface of the water at time t=0.
Use Fig. 2.2 to determine the depth of the top face below the surface of the water at time t=4.0 s.
depth = m
distance =(1/2×2.5×0.10)+(1/2×1.5×0.10) or (1/2×4.0×0.10)=0.20( m))
C1
depth =0.32−0.20=0.12 m
A1
A toy train moves along a straight section of track. Fig. 1.1 shows the variation with time t of the distance d moved by the train.

Fig. 1.1
Describe qualitatively the motion of the train between time t=0 and time t=1.0 s.
decelerates
or
speed/velocity decreases
B1
Determine the speed of the train at time t=2.0 s.
speed =(Δ)d/(Δ)t or gradient
C1
= e.g. (0.56−0.20)/1.5=0.24 m s−1
A1
The straight section of track in (b) is part of the loop of track shown in Fig. 1.2.

Fig. 1.2
The train completes exactly one lap of the loop.
State and explain the average velocity of the train over the one complete lap.
displacement is zero (so) average velocity is zero
B1