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Form 5 Physics Practice Paper: Mechanics, Waves and Electricity
Form 5 Physics – 12 questions – 100 marks – 75 minutes
Form 5 Physics paper: 12 questions, 100 marks, 75 minutes on kinematics, force, work, energy, momentum, power, waves, refraction, circuits, electrical power and heat.
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Paper details Form 5 Physics practice paper with 12 structured questions on forces, motion, waves and electricity; includes full worked answers and two printable PDFs.
Section A: Mechanics (Q1-Q5)
Answer all parts; show the formula, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A car starts from rest and accelerates uniformly at 2.5 m/s2 for 8 seconds. (a) Find its speed at the end of the 8 seconds. (b) Find the distance travelled in that time. |
(a) v = u + at = 0 + 2.5 x 8 = 20.0 m/s. (b) s = ut + 0.5at^2 = 0 + 0.5 x 2.5 x 8^2 = 80.0 m. Marking notes: correct kinematics equation (2 marks), speed with unit (3 marks), distance with unit (3 marks). |
| 2 | A stone is released from rest on a platform 45 m above the ground. Take g = 10 m/s2 and neglect air resistance. (a) Find the time the stone takes to reach the ground. (b) Find its speed on landing. |
(a) s = 0.5gt^2, so 45 = 0.5 x 10 x t^2, giving t = 3.0 s. (b) v = gt = 10 x 3.0 = 30.0 m/s downwards. Marking notes: free fall equation (2 marks), time (2 marks), speed with direction (4 marks). |
| 3 | A block of mass 5 kg rests on a horizontal table. A horizontal force of 15 N is applied. Take g = 10 m/s2. (a) Find the acceleration if the table is smooth. (b) Find the acceleration if the coefficient of kinetic friction between the block and the table is 0.20. |
(a) a = F / m = 15 / 5 = 3.0 m/s2. (b) friction f = umg = 0.20 x 5 x 10 = 10.0 N; net force = 15 – 10.0 = 5.0 N, so a = 5.0 / 5 = 1.0 m/s2. Marking notes: Newton second law (2 marks), smooth case (2 marks), friction (2 marks), net force and acceleration (2 marks). |
| 4 | A car of mass 1200 kg accelerates from 10 m/s to 20 m/s on a straight level road. (a) Find the increase in kinetic energy, in kJ. (b) If the acceleration takes 5 seconds, find the average power delivered by the net force. |
(a) dKE = 0.5m(v^2 – u^2) = 0.5 x 1200 x (20^2 – 10^2) = 180000 J = 180.0 kJ. (b) Work done by the net force equals the change in kinetic energy, 180000 J; P = W / t = 180000 / 5 = 36000 W. Marking notes: kinetic energy formula (2 marks), energy change with unit conversion (3 marks), work and power relation (3 marks). |
Section B: Work, Energy and Waves (Q6-Q8)
Take g = 10 m/s2 and neglect air resistance unless stated otherwise.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A ball of mass 0 kg hits a wall perpendicularly at 20 m/s and rebounds along the same line at 15 m/s. (a) Find the magnitude of the change in momentum of the ball. (b) State the direction of the impulse exerted by the wall on the ball. |
(a) Taking the direction away from the wall as positive: dp = m(v + u) = 0 x (15 + 20) = 5.25 kg m/s. (b) The change in momentum points away from the wall, so the impulse from the wall also points perpendicularly away from the wall. Marking notes: momentum definition (2 marks), sign convention and value (4 marks), impulse direction (3 marks). |
| 6 | A crane lifts a 60 kg load at constant speed through a height of 12 m in 15 s. Take g = 10 m/s2. (a) Find the work done by the crane on the load. (b) Find the average power of the crane. |
(a) At constant speed the tension equals the weight: F = mg = 60 x 10 = 600 N; W = Fs = 600 x 12 = 7200 J. (b) P = W / t = 7200 / 15 = 480 W. Marking notes: force balance (2 marks), work done (3 marks), power with unit (3 marks). |
| 7 | A sound source of frequency 250 Hz produces a wave of wavelength 1.36 m in air. (a) Find the wave speed. (b) Find the period of the wave. (c) If the wave enters water where the speed is 1500 m/s, does the frequency change? Find the wavelength in water. |
(a) v = f x wavelength = 250 x 1.36 = 340.0 m/s. (b) T = 1 / f = 1 / 250 = 0.004 s. (c) Frequency is set by the source and does not change, so it stays 250 Hz; in water the wavelength is 1500 / 250 = 6.00 m. Marking notes: wave equation (2 marks), period (2 marks), frequency unchanged explained (3 marks), wavelength in water (2 marks). |
| 8 | A beam of light passes from air into a transparent medium with an angle of incidence of 45 degrees and an angle of refraction of 30 degrees. (a) Find the refractive index of the medium. (b) Find the critical angle for light travelling from the medium to air. (c) What happens if the angle of incidence exceeds the critical angle? |
(a) n = sin i / sin r = sin 45 / sin 30 = 1.41. (b) sin c = 1 / n = 1 / 1.41, so c = 45.0 degrees. (c) The light is no longer refracted into the air but is totally reflected back into the medium; this is total internal reflection. Marking notes: refractive index formula (2 marks), value (2 marks), critical angle (3 marks), total internal reflection (2 marks). |
Section C: Electricity and Heat (Q9-Q12)
Keep significant figures consistent with the data; state the formula used in circuit questions.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Two resistors of 6 ohm and 3 ohm are connected to a battery of e.m.f. 12 V of negligible internal resistance. (a) If they are in series, find the total resistance and the current. (b) If they are in parallel, find the total resistance and the total current from the battery. (c) Which connection dissipates more total power? |
(a) Series: R = 6 + 3 = 9.0 ohm; I = V / R = 12 / 9 = 1.33 A. (b) Parallel: R = (6 x 3) / (6 + 3) = 2.0 ohm; I = 12 / 2 = 6.0 A. (c) The parallel circuit draws a larger total current, so its total power P = VI is larger. Marking notes: series resistance and current (3 marks), parallel resistance and current (4 marks), power comparison (2 marks). |
| 10 | An electric kettle rated at 2000 W is switched on for 5 minutes. (a) Find the electrical energy consumed, in kJ. (b) If electricity costs 1.2 dollars per kWh, find the cost of this heating. (c) Why is the energy actually transferred to the water less than the answer in (a)? |
(a) E = Pt = 2000 x 300 = 600000 J = 600.0 kJ. (b) Energy = 600000 / 3600000 = 0.167 kWh; cost = 0.167 x 1.2 = 0.20 dollars. (c) Some electrical energy is lost as heat to the surroundings and the body of the kettle, so not all of it reaches the water. Marking notes: E = Pt with unit conversion (3 marks), kWh and cost (4 marks), energy loss (2 marks). |
| 11 | Water of mass 0 kg is heated from 20 degrees C to 80 degrees C. The specific heat capacity of water is 4200 J kg-1 K-1. (a) Find the energy absorbed by the water, in kJ. (b) If a heater of power 700 W supplies all this energy, find the time needed. |
(a) E = mc(dT) = 0 x 4200 x 60 = 126000 J = 126.0 kJ. (b) t = E / P = 126000 / 700 = 180 s, that is 3 minutes. Marking notes: specific heat capacity equation (2 marks), energy with unit (3 marks), time (3 marks). |
| 12 | A spring of stiffness 200 N/m is stretched by 0.15 m within its elastic limit. (a) Find the tension in the spring. (b) Find the elastic potential energy stored. (c) If the extension is doubled, by what factor does the stored energy change? |
(a) F = kx = 200 x 0.15 = 30 N. (b) E = 0.5kx^2 = 0.5 x 200 x 0.15^2 = 2.25 J. (c) The energy is proportional to the square of the extension, so doubling the extension gives 4 times the energy. Marking notes: Hooke law (2 marks), elastic potential energy (3 marks), square relation (3 marks). |
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