S4 Physics Paper: Waves, Refraction, Heat, Pressure And Momentum (100 Marks)

S4 Physics Paper: Waves, Refraction, Heat, Pressure and Momentum (100 marks)

2026-11-12

S4 Physics Practice Paper

S4 Physics paper: 6 questions, 24 parts, 100 marks, 60 minutes on wave speed, refraction and lenses, heat capacity, pressure, momentum and experiment errors.

Download　Question paper PDF　|　Answer key PDF

Paper details　S4 physics practice paper with 6 questions and 24 parts on waves, refraction, heat, pressure and momentum.

Section A: Waves (15 marks)

A student uses a signal generator to produce a sound wave of frequency 250 Hz, and measures a wavelength of 1.36 m in air. The same source is then placed under water, where the speed of sound is 1500 m/s.

#	Question	Answer and marking notes
1	Find the speed of the sound wave in air.	Answer: v = f x wavelength = 250 x 1.36 = 340 m/s. (Marking: 2 marks for the formula, 2 marks for the value with unit.)
2	Find the period of the sound wave.	Answer: T = 1 divided by f = 1 divided by 250 = 0.004 s, that is 4 ms. (Marking: 1 mark for the formula, 1 mark for the value, 1 mark for the unit.)
3	How long does the sound take to reach a wall 850 m away?	Answer: t = d divided by v = 850 divided by 340 = 2.5 s. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
4	The frequency stays the same when the wave enters water. Find the wavelength in water.	Answer: wavelength = v divided by f = 1500 divided by 250 = 6.0 m. The frequency is fixed by the source, so it stays the same. (Marking: 1 mark for stating the frequency is unchanged, 2 marks for the calculation, 1 mark for the answer with unit.)

Section B: Refraction of Light and Lenses (18 marks)

A ray of light enters a glass block of refractive index 1.5 from air at an angle of incidence of 45 degrees. Take the speed of light in air as 3.0 x 10 to the power 8 m/s.

#	Question	Answer and marking notes
5	Find the angle of refraction in the glass.	Answer: sin r = sin 45 degrees divided by 1.5 = 0.471, so r = 28.1 degrees. (Marking: 2 marks for Snell law, 1 mark for the value, 1 mark for the answer.)
6	Find the speed of light in the glass.	Answer: v = c divided by n = 3.0 x 10 to the power 8 divided by 1.5 = 2.0 x 10 to the power 8 m/s. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
7	Find the critical angle at the glass to air boundary.	Answer: sin C = 1 divided by 1.5 = 0.667, so C = 41.8 degrees. (Marking: 2 marks for the formula, 1 mark for the value, 1 mark for the answer.)
8	A convex lens of focal length 10 cm is used with an object distance of 30 cm. Find the image distance and magnification, and state the nature of the image.	Answer: 1/f = 1/u + 1/v, so 1/10 = 1/30 + 1/v and v = 15 cm; m = v divided by u = 15 divided by 30 = 0.5; the image is real, inverted and diminished. (Marking: 2 marks for the lens formula, 1 mark for the image distance, 2 marks for the magnification, 1 mark for the nature.)

Section C: Heat (17 marks)

A heater of power 2000 W heats 0.5 kg of water from 20 degrees Celsius to 80 degrees Celsius. The specific heat capacity of water is 4200 J per kg per K.

#	Question	Answer and marking notes
9	Find the energy gained by the water.	Answer: E = m x c x temperature rise = 0.5 x 4200 x 60 = 126000 J. (Marking: 2 marks for the formula, 1 mark for the temperature rise, 1 mark for the value, 1 mark for the unit.)
10	Ignoring heat loss, how long does the heating take?	Answer: t = E divided by P = 126000 divided by 2000 = 63 s. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
11	0.05 kg of ice needs 1.7 x 10 to the power 4 J to melt completely. Find the specific latent heat of fusion of ice.	Answer: L = E divided by m = 1.7 x 10 to the power 4 divided by 0.05 = 3.4 x 10 to the power 5 J per kg. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
12	0.2 kg of water at 90 degrees Celsius is mixed with 0.3 kg of water at 20 degrees Celsius. Find the final temperature.	Answer: heat lost equals heat gained, so 0.2 x 90 + 0.3 x 20 = 0.5 x final temperature, giving 24 divided by 0.5 = 48 degrees Celsius. (Marking: 2 marks for conservation of energy, 1 mark for the calculation, 1 mark for the answer.)

Section D: Pressure and Upthrust (16 marks)

A student of weight 600 N stands on both feet with a total contact area of 0.02 square metres. The density of water is 1000 kg per cubic metre; take g as 10 N per kg.

#	Question	Answer and marking notes
13	Find the pressure the student exerts on the ground.	Answer: p = F divided by A = 600 divided by 0.02 = 30000 Pa. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
14	Find the pressure due to the water at a depth of 1.2 m.	Answer: p = density x g x depth = 1000 x 10 x 1.2 = 12000 Pa. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
15	An object of volume 0.004 cubic metres is fully immersed in water. Find the upthrust on it.	Answer: upthrust = weight of water displaced = 1000 x 10 x 0.004 = 40 N. (Marking: 2 marks for the idea of displaced water, 1 mark for the calculation, 1 mark for the answer with unit.)
16	A wooden block of density 800 kg per cubic metre floats on water. Find the fraction of its volume that is under water.	Answer: upthrust equals the weight, so the submerged fraction is 800 divided by 1000 = 0.8, that is 80 percent. (Marking: 2 marks for upthrust equals weight, 1 mark for the ratio, 1 mark for the percentage.)

Section E: Momentum (17 marks)

A trolley of mass 0.4 kg moving at 5 m/s hits a stationary trolley of mass 0.6 kg, and the two trolleys stick together after the collision.

#	Question	Answer and marking notes
17	Find the momentum of the moving trolley before the collision.	Answer: p = m x v = 0.4 x 5 = 2.0 kg m per s. (Marking: 2 marks for the formula, 2 marks for the answer with unit.)
18	Find the common velocity of the trolleys after the collision.	Answer: momentum is conserved, so 0.4 x 5 = (0.4 + 0.6) x v, giving v = 2.0 m/s in the original direction. (Marking: 2 marks for the conservation equation, 1 mark for adding the masses, 1 mark for the answer, 1 mark for the direction.)
19	If the collision lasts 0.05 s, find the average force on the 0.6 kg trolley.	Answer: impulse = change in momentum = 0.6 x 2.0 = 1.2 N s; average force = 1.2 divided by 0.05 = 24 N. (Marking: 2 marks for the impulse, 1 mark for the force, 1 mark for the unit.)
20	Find the kinetic energy lost in the collision.	Answer: kinetic energy before = 0.5 x 0.4 x 25 = 5 J; after = 0.5 x 1.0 x 4 = 2 J; so 3 J is lost. (Marking: 1 mark for the value before, 2 marks for the value after, 1 mark for the difference.)

Section F: Experimental Errors and Graphs (17 marks)

A student repeats a length measurement three times and also measures a metal block with a balance and a measuring cylinder.

#	Question	Answer and marking notes
21	The three readings are 2.4 cm, 2.5 cm and 2.6 cm. Find the mean and the range.	Answer: mean = (2.4 + 2.5 + 2.6) divided by 3 = 2.5 cm; range = 2.6 minus 2.4 = 0.2 cm. (Marking: 2 marks for the mean, 2 marks for the range.)
22	A measured value is 9.6 while the accepted value is 10.0. Find the percentage error.	Answer: percentage error = \|9.6 minus 10.0\| divided by 10.0 x 100 percent = 4 percent. (Marking: 2 marks for taking the absolute value, 2 marks for the answer.)
23	The metal block has a mass of 78.0 g and a volume of 10.0 cm cubed. Find its density.	Answer: density = m divided by V = 78.0 divided by 10.0 = 7.80 g per cm cubed. (Marking: 2 marks for the formula, 2 marks for the value, 1 mark for significant figures and unit.)
24	A straight line passes through (0.10, 0.22) and (0.60, 0.82). Find its slope.	Answer: slope = (0.82 minus 0.22) divided by (0.60 minus 0.10) = 0.60 divided by 0.50 = 1.2. (Marking: 2 marks for the formula, 1 mark for the answer, 1 mark for leaving it unitless.)

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.