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Form 4 Physics Practice Paper: Kinematics, Forces, Work, Energy and Heat (12 questions, 100 marks, 75 minutes)
Form 4 Physics – 12 questions – 100 marks – 75 minutes
Form 4 Physics practice paper: 12 questions, 100 marks, 75 minutes on acceleration, free fall, friction, momentum, energy, heat and pressure, with full answers and marking points.
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Paper details Form 4 Physics practice paper with answers: 12 questions, 100 marks, 75 minutes on velocity, acceleration, free fall, friction, momentum, kinetic energy, work done, power, specific heat capacity, pressure, conduction and average speed.
Section A: Mechanics (Q1-Q4, 34 marks)
Answer all questions; show all workings and state the units in every answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. A car starts from rest and accelerates uniformly at 2 m/s² for 5 s. It then travels at that constant velocity for 10 s. (a) Find the velocity after 5 s. (3 marks) (b) Find the distance travelled during the acceleration. (3 marks) (c) Find the total displacement in the first 15 s. (3 marks) | Answer: (a) v = u + at = 0 + 2 × 5 = 10 m/s. (3 marks) (b) s = ut + ½at² = ½ × 2 × 5² = 25 m. (3 marks) (c) Uniform stage: s = 10 × 10 = 100 m, so total displacement = 25 + 100 = 125 m. (3 marks) |
| 2 | 2. A ball is released from rest at a height of 45 m above the ground. Take g = 10 m/s² and ignore air resistance. (a) Find the time taken to reach the ground. (3 marks) (b) Find the speed just before it hits the ground. (3 marks) (c) Give one reason why the real time of free fall is longer. (3 marks) | Answer: (a) s = ½gt², 45 = ½ × 10 × t², so t = 3 s. (3 marks) (b) v = gt = 10 × 3 = 30 m/s. (3 marks) (c) Air resistance acts upwards, so the net downward acceleration is less than g and the fall takes longer. (3 marks) |
| 3 | 3. A 5 kg block rests on a horizontal table and is pushed by a horizontal force of 30 N. The coefficient of friction is 0.2 and g = 10 m/s². (a) Find the friction acting on the block. (3 marks) (b) Find the acceleration of the block. (3 marks) (c) If the pushing force is increased to 40 N, find the new acceleration. (2 marks) | Answer: (a) N = mg = 5 × 10 = 50 N, so friction f = μN = 0.2 × 50 = 10 N. (3 marks) (b) Net force = 30 − 10 = 20 N, a = F/m = 20 ÷ 5 = 4 m/s². (3 marks) (c) Net force = 40 − 10 = 30 N, a = 30 ÷ 5 = 6 m/s². (2 marks) |
| 4 | 4. A 2 kg object moving at 3 m/s on a smooth horizontal surface collides with a stationary 1 kg object and they stick together. (a) Find their common velocity after the collision. (3 marks) (b) Find the total kinetic energy before and after the collision, and the energy lost. (3 marks) (c) Explain why momentum is conserved but kinetic energy is not. (2 marks) | Answer: (a) By conservation of momentum, 2 × 3 = (2 + 1)v, so v = 2 m/s in the original direction. (3 marks) (b) Before: Ek = ½ × 2 × 3² = 9 J. After: Ek = ½ × 3 × 2² = 6 J. Energy lost = 3 J. (3 marks) (c) No external horizontal force acts on the system, so momentum is conserved; part of the kinetic energy becomes heat and sound, so kinetic energy is not conserved. (2 marks) |
Section B: Work, Energy and Heat (Q5-Q8, 33 marks)
Take g = 10 m/s²; ignore air resistance unless stated; give answers to a suitable number of significant figures with units.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. A person pushes a box 10 m along a straight line with a horizontal force of 400 N in 20 s. (a) Find the work done by the pushing force. (3 marks) (b) Find the average power of the pushing force. (3 marks) (c) Give one reason why the actual work done is larger than the calculated value. (2 marks) | Answer: (a) Work done W = Fs = 400 × 10 = 4000 J. (3 marks) (b) Power P = W/t = 4000 ÷ 20 = 200 W. (3 marks) (c) Friction converts part of the work into heat, or the pushing force is not exactly parallel to the displacement. (2 marks) |
| 6 | 6. A 2 kg object slides from rest down a smooth slope from a height of 5 m. Take g = 10 m/s². (a) Find its gravitational potential energy at the top, taking the ground as zero. (3 marks) (b) Find its speed when it reaches the ground. (3 marks) (c) How would the speed change if the slope were rough? Explain. (2 marks) | Answer: (a) Ep = mgh = 2 × 10 × 5 = 100 J. (3 marks) (b) 100 = ½ × 2 × v², so v² = 100 and v = 10 m/s. (3 marks) (c) The speed would be smaller because part of the energy is used to overcome friction and becomes heat. (2 marks) |
| 7 | 7. 0.5 kg of water at 20 °C is heated to 80 °C. The specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹. (a) Find the energy absorbed by the water. (3 marks) (b) If a 1500 W heater is used with no heat loss, find the time needed. (3 marks) (c) Give one reason why the actual heating time is longer. (3 marks) | Answer: (a) E = mcΔT = 0.5 × 4200 × 60 = 126000 J. (3 marks) (b) t = E/P = 126000 ÷ 1500 = 84 s. (3 marks) (c) Heat is lost by conduction, convection and radiation, the container absorbs energy, and the heater is not 100% efficient. (3 marks) |
| 8 | 8. 0.2 kg of water at 90 °C is mixed with 0.3 kg of water at 20 °C in a vacuum flask. (a) Find the final temperature. (4 marks) (b) State two assumptions made in the calculation. (2 marks) (c) If heat is lost to the flask, will the final temperature be higher or lower? (2 marks) | Answer: (a) 0.2 × 4200 × (90 − T) = 0.3 × 4200 × (T − 20), giving 18 − 0.2T = 0.3T − 6, so 0.5T = 24 and T = 48 °C. (4 marks) (b) No heat is lost to the surroundings, the specific heat capacity of water is constant, and the flask absorbs no energy. (2 marks) (c) Lower, because some energy escapes to the surroundings. (2 marks) |
Section C: Pressure, Heat Transfer and Practical Skills (Q9-Q12, 33 marks)
Marks are given for the correct formula, clear steps and correct units; workings and answers carry separate marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. A student weighing 600 N stands on both feet. The contact area of each foot with the ground is 0.015 m². (a) Find the pressure the student exerts on the ground. (3 marks) (b) What is the pressure if the student stands on one foot only? (3 marks) (c) Explain why snowshoes stop a person sinking into snow. (2 marks) | Answer: (a) Total contact area = 2 × 0.015 = 0.03 m², P = 600 ÷ 0.03 = 20000 Pa. (3 marks) (b) The area halves, so P = 600 ÷ 0.015 = 40000 Pa. (3 marks) (c) Snowshoes increase the contact area, so for the same weight the pressure is smaller and the snow deforms and sinks less. (2 marks) |
| 10 | 10. Heat can be transferred by conduction, convection and radiation. (a) Describe the mechanism of conduction, convection and radiation. (3 marks) (b) Explain why the handle of a metal spoon becomes hot in hot soup. (3 marks) (c) State one way to reduce heat loss from a home and the transfer method involved. (2 marks) | Answer: (a) Conduction: particles vibrate or free electrons pass energy along; convection: a heated fluid expands, becomes less dense and rises, setting up a circulation; radiation: energy travels as electromagnetic waves and needs no medium. (3 marks) (b) Metal is a good conductor with many free electrons, so energy is conducted quickly from the bowl of the spoon to the handle. (3 marks) (c) Double glazing, where the trapped air reduces conduction and convection; or aluminium foil, which reflects radiation. (2 marks) |
| 11 | 11. A student uses a stopwatch to measure the time for 20 oscillations of a simple pendulum as 40 s. (a) Find the period of the pendulum. (3 marks) (b) Find its frequency. (3 marks) (c) Explain why timing 20 oscillations is more accurate than timing one. (3 marks) | Answer: (a) Period T = 40 ÷ 20 = 2 s. (3 marks) (b) Frequency f = 1/T = 0.5 Hz. (3 marks) (c) Human reaction time is about 0.2 s, so a single timing has a large relative error; dividing the time for 20 oscillations by 20 spreads that error and gives a more accurate period. (3 marks) |
| 12 | 12. A runner completes 100 m in 20 s, covering 30 m in the first 10 s. (a) Find the average speed for the whole run. (3 marks) (b) Find the average speed in the first 10 s. (3 marks) (c) Distinguish between average speed and instantaneous speed. (2 marks) | Answer: (a) Average speed = 100 ÷ 20 = 5 m/s. (3 marks) (b) 30 ÷ 10 = 3 m/s. (3 marks) (c) Average speed is total distance divided by total time; instantaneous speed is the speed at one instant, given by the gradient of the tangent on a distance-time graph. (2 marks) |
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