Form 4 Physics Practice Paper: Force And Motion, Energy And Momentum, Waves And Light

Form 4 Physics Practice Paper: Force and Motion, Energy and Momentum, Waves and Light

2026-11-05

Form 4 Physics – 12 questions – 100 marks – 75 minutes

Form 4 Physics practice paper: 12 questions, 100 marks, 75 minutes on motion, Newton laws, work and power, energy, momentum, moments, waves, refraction and lenses, with full answers.

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Paper details　Form 4 Physics practice paper: 12 questions, 100 marks in 75 minutes on force, motion, energy, momentum, waves, refraction and convex lenses, with answers.

Section A: Force and Motion

Answer all parts and show your working and units in calculations.

#	Question	Answer and marking notes
1	A car travels at 8 m/s and then accelerates uniformly at 2.5 m/s² for 6 seconds. (a) Find its velocity at the end of the 6 seconds. (b) Find the distance travelled in these 6 seconds. (c) Find its average velocity over this interval.	(a) v = u + at = 8 + 2.5 × 6 = 23 m/s (3 marks). (b) s = ut + ½at² = 8 × 6 + ½ × 2.5 × 36 = 48 + 45 = 93 m (3 marks). (c) Average velocity = 93 ÷ 6 = 15.5 m/s, which equals (8 + 23) ÷ 2 (3 marks).
2	A 4 kg block on a horizontal table is pushed with a horizontal force of 30 N. The friction factor (friction force divided by normal force) between the block and the table is 0.25, and g = 10 m/s². (a) Find the friction force. (b) Find the net force on the block. (c) Find the acceleration of the block. (d) What is the acceleration if the friction factor rises to 0.5?	(a) Friction = 0.25 × 4 × 10 = 10 N (2 marks). (b) Net force = 30 − 10 = 20 N (2 marks). (c) a = 20 ÷ 4 = 5 m/s² (3 marks). (d) Friction becomes 0.5 × 4 × 10 = 20 N, so the net force is 10 N and a = 2.5 m/s² (1 mark).
3	A small ball is released from rest from the top of a tower 45 m above the ground. Air resistance is negligible and g = 10 m/s². (a) Find the time taken to reach the ground. (b) Find the speed on landing. (c) Find the time to land if the same experiment is done on the Moon, where g = 1.6 m/s².	(a) t = √(2 × 45 ÷ 10) = 3 s (3 marks). (b) v = 10 × 3 = 30 m/s (3 marks). (c) t = √(2 × 45 ÷ 1.6) = 7.5 s (2 marks).
4	(a) Explain, using Newton first law, why a standing passenger leans forward when a bus brakes suddenly. (b) State the state of motion implied by zero net force in Newton first law. (c) A 5 kg object experiences a net force of 20 N; find its acceleration.	(a) The passenger keeps the original state of motion while the bus slows down, so the body leans forward (3 marks). (b) With zero net force, a body at rest stays at rest and a moving body moves uniformly in a straight line (2 marks). (c) a = 20 ÷ 5 = 4 m/s² (3 marks).

Section B: Energy, Momentum and Moments

Answer all parts and give answers to appropriate significant figures.

#	Question	Answer and marking notes
5	A 1500 kg car accelerates from rest to 20 m/s in 10 seconds; resistance is negligible. (a) Find the gain in kinetic energy. (b) Find the average power of the engine in kW. (c) Why is the power of a real engine higher than the answer in (b)?	(a) Kinetic energy = ½ × 1500 × 20² = 300 000 J = 300 kJ (3 marks). (b) Power = 300 000 ÷ 10 = 30 000 W = 30 kW (3 marks). (c) A real car must also overcome air resistance and road friction, so some energy is lost as heat (2 marks).
6	A small ball is dropped from rest from a height of 20 m. Air resistance is negligible and g = 10 m/s². (a) Find its speed on reaching the ground. (b) If air resistance removes 25% of the mechanical energy, find the speed on landing to three significant figures. (c) State two energy conversions during the fall.	(a) v = √(2 × 10 × 20) = 20 m/s (3 marks). (b) With 75% of the energy remaining, v = √(2 × 0.75 × 10 × 20) ≈ 17.3 m/s (3 marks). (c) Gravitational potential energy becomes kinetic energy, and some becomes heat and sound (3 marks).
7	A 2 kg trolley moving to the right at 5 m/s collides with a stationary 3 kg trolley and they move together after the collision. (a) Find their common velocity. (b) Find the total kinetic energy before and after the collision and the energy lost. (c) State the type of this collision.	(a) By conservation of momentum, 2 × 5 = (2 + 3) × v, so v = 2 m/s (3 marks). (b) Kinetic energy before = ½ × 2 × 25 = 25 J; after = ½ × 5 × 4 = 10 J; 15 J is lost (3 marks). (c) A completely inelastic collision (2 marks).
8	A force of 60 N is applied to a spanner at 1.5 m from the pivot, and a downward force F at 0.5 m from the pivot balances it. (a) Find F. (b) Find the upward reaction at the pivot. (c) Explain why a longer spanner needs less force.	(a) Taking moments about the pivot, 60 × 1.5 = F × 0.5, so F = 180 N (3 marks). (b) The reaction at the pivot is 60 + 180 = 240 N (3 marks). (c) A longer moment arm needs a smaller force for the same moment (2 marks).

Section C: Waves and Light

Answer all questions and quote the relevant physics laws in explanations.

#	Question	Answer and marking notes
9	A sound source emits a sound wave of frequency 250 Hz with a wavelength of 1.36 m in air. (a) Find the speed of sound. (b) Find the period of the wave. (c) If the same wave enters another medium where the wavelength halves but the frequency is unchanged, how does the speed change?	(a) v = fλ = 250 × 1.36 = 340 m/s (3 marks). (b) T = 1 ÷ 250 = 0.004 s (3 marks). (c) The frequency is unchanged and the wavelength halves, so the speed halves to 170 m/s (2 marks).
10	A beam of light passes from air into glass of refractive index 1.5 at an angle of incidence of 30°. (a) Find the angle of refraction to three significant figures. (b) Find the critical angle for light passing from this glass into air. (c) State the two conditions for total internal reflection.	(a) sin r = sin 30° ÷ 1.5 = 0.3333, so r ≈ 19.5° (3 marks). (b) sin c = 1 ÷ 1.5 = 0.6667, so c ≈ 41.8° (3 marks). (c) Light must travel from an optically denser to a less dense medium, and the angle of incidence must exceed the critical angle (2 marks).
11	A ship sends a sonar pulse to the seabed and receives the echo after 0.60 s; the speed of sound in sea water is 340 m/s. A medical ultrasound scanner uses 1500 m/s in soft tissue and the pulse takes 0.08 s for the round trip. (a) Find the depth of the water. (b) Find the thickness of the tissue. (c) Explain why medical imaging uses ultrasound rather than audible sound.	(a) Depth = 340 × 0.60 ÷ 2 = 102 m (3 marks). (b) Thickness = 1500 × 0.08 ÷ 2 = 60 m (3 marks). (c) Ultrasound has a short wavelength and hence high resolution, and it cannot be heard by the human ear (2 marks).
12	An object is placed 15 cm in front of a convex lens of focal length 10 cm. (a) Find the image distance. (b) Find the magnification and describe the image. (c) If the object is moved to 5 cm in front of the lens, find the image distance and magnification and describe the image.	(a) Using 1/f = 1/u + 1/v, v = 10 × 15 ÷ (15 − 10) = 30 cm (3 marks). (b) Magnification = 30 ÷ 15 = 2; the image is real, inverted and magnified (3 marks). (c) v = 10 × 5 ÷ (5 − 10) = −10 cm and magnification = 2; the image is virtual, upright and magnified (4 marks).

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