OxfordAQA International AS and A-level Physics: Projectiles and terminal speed (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.2.4 (Projectile motion) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4. It is shared International AS and A-level content. Use g = 9.81 m s⁻²; ignore air resistance unless told otherwise.

Study first with the study guide or the revision notes. For the equations, see motion along a straight line and Newton’s laws.

Questions

1. A projectile moves through the air with only its weight acting on it. (a) State its horizontal acceleration. [1] (b) State the magnitude and direction of its acceleration at the highest point of its path. [1] (c) Explain why its horizontal velocity stays constant. [1]

2. A marble rolls off the edge of a bench at 2.4 m s⁻¹ in a horizontal direction. The bench is 0.95 m above the floor. Calculate the time the marble takes to reach the floor and the horizontal distance from the bench edge at which it lands. [4]

3. A box slides down a rough slope. (a) State the direction of the friction force on the box. [1] (b) Explain why friction reduces the box’s acceleration. [1] (c) State what happens to the energy transferred against friction. [1]

4. A skydiver jumps from a hovering helicopter and falls before opening her parachute until her speed becomes constant. Describe and explain how her acceleration changes during this fall. [5]

5. A ball is thrown from the ground at 17.0 m s⁻¹ at 40° above the horizontal. It lands at the same height. Calculate (a) the horizontal and vertical components of the initial velocity [2], (b) the maximum height [2], (c) the time of flight [1], (d) the range [1].

6. A rescue buoy is thrown from a pier 4.2 m above the sea at 8.0 m s⁻¹ at 25° above the horizontal. Calculate the horizontal distance from the pier at which it lands in the sea. [6]

7. A car reaches a maximum speed on a flat road. (a) Explain why the car cannot accelerate further at its maximum speed. [2] (b) A roof box is fitted and the engine is unchanged. Explain the effect on the maximum speed. [2]

8. An aeroplane flies in level flight at constant velocity. (a) State the relationship between lift and weight, and between thrust and drag. [2] (b) State the direction of the lift force relative to the airflow. [1] (c) The pilot increases thrust and holds lift equal to weight. Describe what happens to the speed. [1]

9. A football is kicked from the ground. Compare its path with the path it would follow if air resistance were negligible. You should refer to range, maximum height, the shape of the path, horizontal velocity and landing speed. [5]

10. A ball of mass 0.045 kg is released from rest in a tall column of liquid. Its speed is recorded at intervals.

Time / s	0	0.05	0.10	0.15	0.20	0.25	0.30	0.35
Speed / m s⁻¹	0	0.42	0.71	0.89	1.00	1.06	1.09	1.09

(a) State the terminal speed of the ball. [1] (b) Calculate its acceleration over the first 0.05 s. [2] (c) Calculate the acceleration between 0.25 s and 0.30 s and explain why it is smaller. [2] (d) Calculate the average resultant force on the ball during the first 0.05 s. [1]

11. A motorcyclist leaves the end of a ramp at 21.0 m s⁻¹ at 25° above the horizontal. A gap of width 33.0 m separates the ramp from a landing platform, which is 8.0 m long and 1.5 m lower than the end of the ramp. (a) Calculate the horizontal and vertical components of the launch velocity. [2] (b) Calculate the time taken to land at the level of the platform. [3] (c) Show whether the motorcyclist lands on the platform. [2] (d) Explain how air resistance would change the landing position. [2]

12. A ball is struck from a point 1.2 m above level ground at 22.0 m s⁻¹ at 30° above the horizontal towards a wall 30.0 m away. The wall is 5.0 m high. (a) Calculate the time taken to travel 30.0 m horizontally. [2] (b) Calculate the height of the ball above the ground at that time. [2] (c) State whether the ball clears the wall. [1] (d) Calculate the speed of the ball at that time. [2] (e) State whether the ball is rising or falling then, with a reason. [1]

Answers

1. (a) The horizontal acceleration is zero [1]. (b) The acceleration is 9.81 m s⁻² downwards [1]; this is the same at every point of the flight, including the top. (c) The only force is the weight, which is vertical, so there is no horizontal resultant force [1] [3] Examiner insight: say which quantity is zero at the top (vertical velocity) and which is not (acceleration).

2. Vertical: u = 0, s = 0.95 m, a = 9.81 m s⁻², so s = ½at² applies [1]. t = √(2 × 0.95 / 9.81) = 0.44 s [1]. Horizontal distance = ut [1] = 2.4 × 0.440 = 1.06 m [1] [4] Examiner insight: stating that the time is shared by both directions makes the link between the two parts clear.

3. (a) Up the slope (parallel to the surface, opposite to the motion) [1]. (b) It opposes the component of weight down the slope, so the resultant force down the slope is smaller [1] and acceleration is smaller (a = F/m). (c) It is transferred to internal (thermal) energy of the box and the slope [1] [3] Examiner insight: linking the resultant force to the acceleration through F = ma is clearer than saying friction “slows it down”.

4. Just after release the speed is zero, so there is no air resistance and only weight acts: a = g [1]. As she speeds up, the air resistance grows [1]. The resultant force (weight minus air resistance) decreases, so the acceleration decreases [1]. Once the air resistance has grown to equal her weight, the resultant force is zero [1], so the acceleration is zero and her speed stays constant at the terminal speed [1] [5] Examiner insight: a clear chain of speed, drag, resultant force and acceleration is the easiest to follow.

5. (a) Horizontal 17.0 cos 40° = 13.0 m s⁻¹ [1]; vertical 17.0 sin 40° = 10.9 m s⁻¹ [1]. (b) At the top v = 0, so v² = u² + 2as gives 0 = 10.93² − 2 × 9.81 × H [1], H = 6.09 m [1]. (c) T = 2 × 10.93 / 9.81 = 2.23 s [1]. (d) Range = 13.02 × 2.228 = 29.0 m [1] [6] Examiner insight: carrying unrounded components into later parts keeps the final answers consistent.

6. Components: horizontal 8.0 cos 25° = 7.25 m s⁻¹, vertical 8.0 sin 25° = 3.38 m s⁻¹ [1]. Take up as positive: s = −4.2 m, so −4.2 = 3.38t − 4.905t² [1]. Rearranged: 4.905t² − 3.38t − 4.2 = 0 [1]. Solving gives t = 1.33 s or t = −0.643 s [1]. The negative time is rejected [1]. Distance = 7.25 × 1.33 = 9.7 m [1] [6] Examiner insight: say why one root of a quadratic is rejected.

7. (a) Resistive forces (air resistance and rolling resistance) increase with speed [1]. At maximum speed they equal the driving force, so the resultant force is zero and the acceleration is zero [1]. (b) The roof box increases the resistive force at any given speed (greater cross-sectional area and a less streamlined shape) [1]. The resistive force reaches the driving force at a lower speed, so the maximum speed decreases [1] [4] Examiner insight: refer to both the driving force and the resistive force; mentioning only drag leaves the balance incomplete.

8. (a) Lift = weight [1] and thrust = drag [1]. (b) Perpendicular to the airflow [1]. (c) The speed increases until drag has risen to equal the new thrust [1], after which the velocity is constant again [4] Examiner insight: “perpendicular to the flow” is the precise wording, whatever the attitude of the aeroplane.

9. The range is smaller [1]. The maximum height is smaller [1]. The path is asymmetrical, with a steeper descent than ascent [1]. The horizontal velocity decreases throughout the flight, whereas without air resistance it stays constant [1]. The landing speed (at the launch height) is less than the launch speed [1] [5] Examiner insight: in a comparison, name both cases: “smaller than without air resistance”, not just “smaller”.

10. (a) The speed stops changing at 1.09 m s⁻¹, so the terminal speed is 1.09 m s⁻¹ [1]. (b) a = Δv / Δt = (0.42 − 0) / 0.05 [1] = 8.4 m s⁻² [1]. (c) a = (1.09 − 1.06) / 0.05 = 0.6 m s⁻² [1]. The drag is larger at higher speed, so the resultant force is smaller [1]. (d) F = ma = 0.045 × 8.4 = 0.38 N (average over the first 0.05 s) [1] [6] Examiner insight: writing the change in speed over the change in time keeps the method visible.

11. (a) Horizontal 21.0 cos 25° = 19.0 m s⁻¹ [1]; vertical 21.0 sin 25° = 8.87 m s⁻¹ [1]. (b) Take up as positive: s = −1.5 m, so −1.5 = 8.875t − 4.905t² [1]. Rearranged: 4.905t² − 8.875t − 1.5 = 0, which gives t = 1.97 s or −0.156 s [1]. The negative root is rejected, so t = 1.97 s [1]. (c) Horizontal distance = 19.03 × 1.965 = 37.4 m [1]. The platform spans 33.0 m to 41.0 m, so the motorcyclist lands on the platform, 4.4 m past the near edge [1]. (d) Air resistance reduces the horizontal velocity [1], so the horizontal distance is smaller and the landing point moves towards the near edge of the platform [1] [9] Examiner insight: “show whether” needs the calculated value compared with the platform’s edges.

12. (a) u_x = 22.0 cos 30° = 19.05 m s⁻¹ [1]; t = 30.0 / 19.05 = 1.57 s [1]. (b) u_y = 22.0 sin 30° = 11.0 m s⁻¹. Height = 1.2 + 11.0 × 1.575 − 4.905 × 1.575² [1] = 6.36 m [1]. (c) 6.36 m is greater than 5.0 m, so the ball clears the wall by 1.36 m [1]. (d) v_y = 11.0 − 9.81 × 1.575 = −4.45 m s⁻¹ [1]; speed = √(19.05² + 4.45²) = 19.6 m s⁻¹ [1]. (e) v_y is negative, so the ball is falling [1] [8] Examiner insight: speed combines both components; a vertical velocity alone does not answer it.

Where marks are usually lost

• Using the launch speed in a vertical equation without resolving it first.
• Stating that the acceleration is zero at the highest point.
• Giving one root of a quadratic without saying why the other is rejected.
• Quoting the vertical velocity as the speed when both components are needed.
• Saying that forces are zero at terminal speed rather than the resultant force.
• Rounding intermediate values early.

Next steps

• Read the revision notes and revisit the study guide.
• Return to the OxfordAQA A-level Physics course page and tick off outcomes on the checklist.
• Try all the free 10-minute diagnostics.
• Want help with this topic? Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams May/June 2020 onwards, published by OxfordAQA. Section 3.2.4, Projectile motion.