CAIE A-Level Physics 2.1 Equations of Motion

Practise selecting kinematics definitions, graphs and equations for motion, free fall, projectiles and experiments, checking uncertainty and feasibility.

Exam points

Represent motion with distance, displacement, speed, velocity and acceleration graphs, reading gradients, areas, signs and turning features.
Determine displacement from velocity–time area and velocity/acceleration from displacement–time or velocity–time gradients.
Derive and apply uniformly accelerated-motion equations, including sign conventions and vertical free fall.
Describe the free-fall experiment and control timing, distance, uncertainty and safety.
Resolve perpendicular uniform-velocity and uniform-acceleration components to analyse projectile-style paths.

Covered syllabus points

2.1.1—Distance, displacement, speed, velocity and acceleration 2.1.2—Distance, displacement, speed, velocity and acceleration 2.1.3—Displacement from the area under a velocity–time graph 2.1.4—Velocity using the gradient of a displacement–time graph 2.1.5—Acceleration using the gradient of a velocity–time graph 2.1.6—Derive, from the definitions of velocity and acceleration, equations that 2.1.7—Equations that represent uniformly accelerated motion in a straight line 2.1.8—The acceleration of free fall using a falling object 2.1.9—Motion due to a uniform velocity in one direction and a uniform acceleration

[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 $=12 \mathrm{~s}$

A1

(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)

C1

distance ( down slope $)=1 / 2 \times 12 \times 6(=36)$

C1

$\begin{aligned} \text { displacement from } \mathrm{A} =108-36 =72 \mathrm{~m} \end{aligned}$

A1

(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 $a=\Delta v /(\Delta) t$

C1

$a=6 / 12=0.50\left(\mathrm{~m} \mathrm{~s}^{-2}\right)$ (other points from the line may be used)

A1

or

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

or
$v^{2}=2 a s$

(C1)

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

(A1)

or

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

or
$s=1 / 2 a t^{2}$

(C1)

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

(A1)

or

$s=v t-\frac{1}{2} a t^{2}$

(C1)

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

(A1)

[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

C1

$\begin{aligned} =\text { e.g. } 8.0 \times 10^{-2} / 2.0 =4.0 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

A1

(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 ]

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

C1

$\begin{aligned} \text { depth } =0.32-0.20 =0.12 \mathrm{~m} \end{aligned}$

A1

All question bank results loaded

Covered syllabus points

2.1 Equations of motion question 1

2.1 Equations of motion question 2

Question (a)

Question (b)

Question (c)

2.1 Equations of motion question 3

Question (a)

Question (i)

Question (ii)