OCR A Physics · Module 3: Forces and motion

Motion: practice questions with answers

6 exam-style questions on motion. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.

Using the right equation2 marks
A car travelling at 30 m s⁻¹ brakes with a constant deceleration of 6.5 m s⁻². Calculate the distance it travels before it stops.
s = m
Show the answer and working
Answer s = 69.2 m
Use v² = u² + 2as with v = 0, u = 30 and a = -6.5: s = 900 ÷ 13 = 69.2 m.
- Got 139 m? You left out the factor of 2 in v² = u² + 2as. With v = 0: s = 30² ÷ (2 × 6.5) = 69.2 m.
- Got 34.6 m? You have divided by 4 × 6.5. The equation gives s = u² ÷ 2a = 900 ÷ 13 = 69.2 m.
Using the right equation3 marks
A stone is thrown horizontally at 14 m s⁻¹ from the top of a cliff 45 m above the sea. Ignore air resistance. Calculate the horizontal distance from the cliff base to where it lands.
x = m
Show the answer and working
Answer x = 42.4 m
Vertical: 45 = ½ × 9.81 × t² gives t = 3.03 s. Horizontal speed is constant: x = 14 × 3.03 = 42.4 m.
- Got 30 m? You left out the factor of 2 in the fall time. From s = ½gt², t = √(2 × 45 ÷ 9.81) = 3.03 s, so x = 14 × 3.03 = 42.4 m.
- Got 64.2 m? You used t = h ÷ g, which is not an equation of motion. Find t from s = ½gt²: t = 3.03 s, then x = 14 × 3.03 = 42.4 m.
Resolving vectors3 marks
A football is kicked at 22 m s⁻¹ at 35° above the horizontal. Ignore air resistance. Calculate the maximum height it reaches.
H = m
Show the answer and working
Answer H = 8.12 m
Vertical component uy = 22 sin 35° = 12.6 m s⁻¹. At the top vy = 0, so H = uy² ÷ 2g = 8.12 m.
- Got 16.6 m? You used the horizontal component. The vertical component is 22 sin 35° = 12.6 m s⁻¹, so H = 12.6² ÷ (2 × 9.81) = 8.12 m.
- Got 24.7 m? You used the full speed instead of the vertical component. Use uy = 22 sin 35° = 12.6 m s⁻¹, then H = uy² ÷ 2g = 8.12 m.
Reading graphs2 marks
A train's velocity increases uniformly from 0 to 8.0 m s⁻¹ in 4.0 s. It then travels at 8.0 m s⁻¹ for 6.0 s and finally decelerates uniformly to rest in 2.0 s. Calculate the total distance travelled using a velocity-time graph.
s = m
Show the answer and working
Answer s = 72 m
Area under the graph: ½ × 4 × 8 = 16 m, 8 × 6 = 48 m, ½ × 2 × 8 = 8 m. Total = 72 m.
- Got 96 m? You treated the whole journey as a rectangle of 8.0 × 12. The graph is a trapezium: 16 + 48 + 8 = 72 m.
- Got 64 m? You have left out the final deceleration. Its area is ½ × 2.0 × 8.0 = 8 m, so the total is 16 + 48 + 8 = 72 m.
Using the right equation1 mark
A ball is thrown vertically upwards. Which statement is correct at its highest point?
- A Its velocity is zero and its acceleration is 9.81 m s⁻² downwards.
- B Its velocity is zero and its acceleration is zero.
- C Its velocity is 9.81 m s⁻¹ downwards.
Show the answer and working
Answer A: Its velocity is zero and its acceleration is 9.81 m s⁻² downwards.
Gravity still acts at the top, so a = 9.81 m s⁻² downwards while v is momentarily zero.
- Why not B? Acceleration due to gravity acts throughout the flight, so it is not zero at the top.
- Why not C? The velocity changes from upwards to downwards, so it is zero at the top, not 9.81 m s⁻¹.
Reading graphs1 mark
On a displacement-time graph, what does the gradient at a point represent?
- A Acceleration at that instant
- B Velocity at that instant
- C Distance travelled so far
Show the answer and working
Answer B: Velocity at that instant
Gradient of displacement against time is velocity (rate of change of displacement).
- Why not A? Acceleration is the gradient of a velocity-time graph.
- Why not C? Distance travelled is not the gradient; the gradient is a rate of change of displacement.