CAIE A-Level Physics AS 2 Kinematics Questions

Practise analysing straight-line and projectile motion with definitions, graphs, constant-acceleration equations, experiments, uncertainty and physical constraints.

Exam points

Define and calculate distance, displacement, speed, velocity and acceleration from motion descriptions, equations and graphs.
Use graph gradients and areas to determine velocities, accelerations and displacements with correct sign and units.
Derive and apply constant-acceleration equations to straight-line and free-fall problems.
Model perpendicular component motion and interpret trajectories, ranges, heights and physical constraints.
Plan or evaluate motion experiments and check measurement uncertainty, assumptions and feasibility.

Covered syllabus points

2.1 Equations of motion

[Maximum number: 1]

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 .

[Maximum number: 5]

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 $18 \mathrm{~ms}^{-1}$ 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 $t=24 \mathrm{~s}$ is shown in Fig. 3.2.

Fig. 3.2

(a)

State the time taken for the sledge to travel from A to B .
time =

[ 1 ]

time $=12 \mathrm{~s}$

(b)

Determine the displacement of the sledge up the slope from point A at time $t=24 \mathrm{~s}$ .

displacement =

[ 2 ]

distance (up slope) $=1 / 2 \times 12 \times 18$ (= 108)
distance ( down slope $)=1 / 2 \times 12 \times 6(=36)$ $\begin{aligned} \text { displacement from } \mathrm{A} =108-36 =72 \mathrm{~m} \end{aligned}$

(c)

Show that the acceleration of the sledge as it moves from B back towards A is $0.50 \mathrm{~m} \mathrm{~s}^{-2}$ .

[ 2 ]

v=u+ at or a= gradient or $a=\Delta v /(\Delta) t$ $a=6 / 12=0.50\left(\mathrm{~m} \mathrm{~s}^{-2}\right)$ (other points from the line may be used)
or

$v^{2}=u^{2}+2 a s \text { and } u=0$

or
$v^{2}=2 a s$ $a=6.0^{2} /(2 \times 36)=0.50\left(\mathrm{~ms}^{-2}\right)$
or

$s=u t+1 / 2 a t^{2} \text { and } u=0$

or
$s=1 / 2 a t^{2}$ $a=2 \times 36 / 12^{2}=0.50\left(\mathrm{~ms}^{-2}\right)$
or

$s=v t-\frac{1}{2} a t^{2}$ $a=2 \times(6 \times 12-36) / 12^{2}=0.50\left(\mathrm{~m} \mathrm{~s}^{-2}\right)$

[Maximum number: 4]

(a)

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

[ 4 ]

(i)

Use Fig. 2.2 to determine the acceleration of the cylinder at time $t=2.0 \mathrm{~s}$ .
acceleration = $\mathrm{ms}^{-2}$

[ 2 ]

a=(v-u) / t or $(\Delta) v /(\Delta) t$ or gradient
$\begin{aligned} =\text { e.g. } 8.0 \times 10^{-2} / 2.0 =4.0 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

(ii)

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 $t=4.0 \mathrm{~s}$ .
depth = m

[ 2 ]

$\begin{aligned} \text { distance } =(1 / 2 \times 2.5 \times 0.10)+(1 / 2 \times 1.5 \times 0.10) \text { or }(1 / 2 \times 4.0 \times 0.10) =0.20(\mathrm{~m})) \end{aligned}$ $\begin{aligned} \text { depth } =0.32-0.20 =0.12 \mathrm{~m} \end{aligned}$

[Maximum number: 4]

(a)

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

[ 3 ]

(i)

Describe qualitatively the motion of the train between time t=0 and time $t=1.0 \mathrm{~s}$ .

[ 1 ]

(ii)

Determine the speed of the train at time $t=2.0 \mathrm{~s}$ .

\text{speed = ................................................ } \mathrm{m\,s}^{-1}\quad \text{[2]}

[ 2 ]

speed $=(\Delta) d /(\Delta) t$ or gradient
$\begin{aligned} =\text { e.g. }(0.56-0.20) / 1.5 =0.24 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}$

(b)

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.

[ 1 ]

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Covered syllabus points

Question 1

Question 2

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Question 3

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Question 4

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