IGCSE Physics: Motion, Forces and Energy (Extended) — Practice Questions (Cambridge 0625)

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 — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Tier note: every question here is marked (Extended): each one is on Supplement content of the 0625 syllabus — the resultant of two vectors at right angles (1.1 Supplement 7), acceleration and deceleration (1.2 Supplement 9 and 12), F = ma (1.5.1 Supplement 11), momentum and impulse (1.6, all Supplement), kinetic and gravitational potential energy (1.7.1 Supplement 4–5) and pressure change with depth, Δp = ρgΔh (1.8 Supplement 4). Core candidates do not need them.

After each answer there is a common mistake to avoid.

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Questions

1. (Extended) A small boat is steered straight across a river, pointing at right angles to the bank, and moves through the water at 3.0 m/s. The river current carries it downstream at 1.6 m/s. Calculate the size of the boat’s resultant velocity and the angle between this resultant and the direction the boat is pointing. [3]

2. (Extended) A goalkeeper catches a ball of mass 0.15 kg that is travelling at 20 m/s and brings it to rest in 0.12 s. (a) Calculate the change in momentum of the ball. (b) Calculate the average force on the ball. (c) Explain why pulling the hands backwards while catching reduces this force. [3]

3. (Extended) Two ice skaters stand still, facing each other, and then push apart. The first skater has a mass of 60 kg and moves off at 1.5 m/s. The second skater has a mass of 45 kg. Calculate the velocity of the second skater, stating its direction. [3]

4. (Extended) A ball of mass 0.40 kg is thrown vertically upwards with a speed of 12 m/s. Air resistance can be ignored and g = 9.8 N/kg. Calculate (a) the kinetic energy of the ball as it is thrown and (b) the greatest height the ball rises above the point of release. [3]

5. (Extended) A diver works 25 m below the surface of the sea. The density of sea water is 1030 kg/m³, g = 9.8 N/kg and the pressure of the atmosphere at the surface is 1.0 × 10⁵ Pa. (a) Calculate the pressure due to the water alone at this depth. (b) Calculate the total pressure on the diver. [3]

6. (Extended) A car of mass 1200 kg slows down steadily from 18 m/s to 6.0 m/s in 4.0 s. Calculate the acceleration of the car, and the size and direction of the resultant force acting on it. [3]

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Answers

1. (Extended) The two velocities are at right angles, so resultant = √(3.0² + 1.6²) = √11.56 [1] = 3.4 m/s [1]. Angle: tan θ = 1.6 ÷ 3.0, so θ = 28° downstream of the direction the boat is pointing [1]. (A scale drawing giving 3.3–3.5 m/s and 27–29° also earns the marks.)

Common mistake: simply adding the two speeds (4.6 m/s). Velocities at right angles must be combined with Pythagoras or a scale drawing.

2. (Extended) (a) Change in momentum = mv = 0.15 × 20 = 3.0 kg m/s (3.0 N s) [1]. (b) F = Δp ÷ Δt = 3.0 ÷ 0.12 = 25 N [1]. (c) Moving the hands back increases the time taken to stop the ball; the change in momentum is the same, so the force is smaller [1].

Common mistake: saying the hands “absorb the momentum”. The momentum change is fixed; it is the longer time that lowers the force.

3. (Extended) Total momentum before = 0, so total momentum after = 0 [1]. 60 × 1.5 = 45 × v, so v = 90 ÷ 45 = 2.0 m/s [1], in the opposite direction to the first skater [1].

Common mistake: leaving out the direction. Momentum is a vector, so the two skaters’ momenta are equal in size but opposite in direction.

4. (Extended) (a) Eₖ = ½mv² = ½ × 0.40 × 12² = 28.8 J (29 J) [1]. (b) At the top all the kinetic energy has become gravitational potential energy, so mgΔh = 28.8 [1]; Δh = 28.8 ÷ (0.40 × 9.8) = 7.3 m [1].

Common mistake: forgetting to square the speed in ½mv², which gives 2.4 J and a height of about 0.6 m.

5. (Extended) (a) Δp = ρgΔh = 1030 × 9.8 × 25 [1] = 252 350 Pa ≈ 2.5 × 10⁵ Pa [1]. (b) Total pressure = 2.5 × 10⁵ + 1.0 × 10⁵ = 3.5 × 10⁵ Pa [1].

Common mistake: forgetting that the atmosphere also presses on the surface of the water, so the total pressure is the water pressure plus atmospheric pressure.

6. (Extended) a = Δv ÷ Δt = (6.0 − 18) ÷ 4.0 = −3.0 m/s² (a deceleration of 3.0 m/s²) [1]. F = ma = 1200 × 3.0 = 3600 N [1], acting opposite to the direction of motion (backwards) [1].

Common mistake: working out 18 − 6.0 and losing the minus sign. A deceleration is a negative acceleration, and the resultant force is in the same direction as the acceleration.

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Where marks are usually lost

• Adding perpendicular velocities or forces as if they were in a straight line.
• Leaving the direction out of momentum and velocity answers.
• Forgetting to square the speed in ½mv².
• Leaving out atmospheric pressure when a question asks for the total pressure at a depth.
• Dropping the minus sign for a deceleration, or giving a force without its direction.