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Secondary 5 Physics Practice Paper: Forces and Motion, Work and Energy, Momentum and Collisions, Waves and Sound, Circuits and Electrical Power, Radioactivity
Physics – Secondary 5 – Practice Paper
S5 physics paper on motion, work and power, momentum, waves and echoes, circuits and radioactivity: 6 sections, 24 parts, 100 marks, 60 minutes, with marking scheme.
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Paper details S5 physics practice paper with 24 parts on motion, work, momentum, waves, circuits and radioactivity; full answers and marking points included.
Section A: Forces and Motion (16 marks)
Data 1: a car starts from rest, accelerates uniformly at 2.5 m s-2 for 6 s, then brakes at 5.0 m s-2 to rest.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) Find the speed after 6 s of acceleration and the distance covered in that stage. | Answer: v = u + at = 0 + 2.5 x 6 = 15.0 m s-1; s = 0.5 a t^2 = 0.5 x 2.5 x 36 = 45.0 m. (Marking: formula 1, speed 1, distance 2) |
| 2 | (b) Find the braking time and the braking distance. | Answer: t = v / a = 15 / 5.0 = 3.0 s; s = v^2 / (2a) = 225 / 10 = 22.5 m. (Marking: time 2, distance 2) |
| 3 | (c) The car has mass 60 kg and g = 10 m s-2. Find its weight and the maximum friction when the friction factor is 0.40. | Answer: W = m g = 60 x 10 = 600.0 N; f = 0.40 x 600.0 = 240.0 N. (Marking: weight 2, friction 2) |
| 4 | (d) If a 300 N horizontal push also acts, find the resultant force, the acceleration and its direction. | Answer: resultant = 300 – 240.0 = 60.0 N; a = 60.0 / 60 = 1.0 m s-2, in the direction of the push. (Marking: resultant 1, acceleration 2, direction 1) |
Section B: Work, Energy and Power (16 marks)
Data 2: a worker pushes a box 8 m with a horizontal force of 50 N in 5 s.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (a) Find the work done and the average power. | Answer: W = F s = 50 x 8 = 400.0 J; P = W / t = 400.0 / 5 = 80.0 W. (Marking: work 2, power 2) |
| 6 | (b) A 4 kg object moves at 10 m s-1. Find its kinetic energy. | Answer: KE = 0.5 m v^2 = 0.5 x 4 x 100 = 200.0 J. (Marking: formula 2, value 2) |
| 7 | (c) A machine takes 1 000 J and delivers 800 J of useful energy. Find its efficiency and the energy wasted. | Answer: efficiency = 800 / 1 000 x 100% = 80.0%; wasted = 1 000 – 800 = 200.0 J. (Marking: efficiency 2, waste 2) |
| 8 | (d) A 25 kg load is lifted 6 m in 10 s. Find the gain in potential energy and the average power (g = 10 m s-2). | Answer: PE = m g h = 25 x 10 x 6 = 1,500.0 J; P = 1,500.0 / 10 = 150.0 W. (Marking: PE 2, power 2) |
Section C: Momentum and Collisions (18 marks)
Data 3: a 0.15 kg ball hits a wall at 20 m s-1 and rebounds at 15 m s-1, contact time 0.05 s.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (a) Find the momentum of the ball. | Answer: p = m v = 0.15 x 20 = 3.0 kg m s-1. (Marking: formula 2, value and unit 2) |
| 10 | (b) Find the change in momentum and the average impact force. | Answer: change = 0.15 x (20 + 15) = 5.25 N s; F = 5.25 / 0.05 = 105.0 N. (Marking: sign handling 1, change 1, force 2) |
| 11 | (c) A 2 kg body at 3 m s-1 hits a stationary 1 kg body and they stick together. Find the common velocity. | Answer: momentum conserved: 2 x 3 = (2 + 1) x v, so v = 2.0 m s-1. (Marking: equation 2, answer 2) |
| 12 | (d) From (c), find the total kinetic energy before and after, the energy lost and the percentage lost. Also find the recoil speed when a 3 kg gun fires a 0.02 kg bullet at 400 m s-1. | Answer: KE before = 0.5 x 2 x 9 = 9.0 J; after = 0.5 x 3 x 4 = 6.0 J; loss = 3.0 J, i.e. 33.3%, so the collision is inelastic. Recoil: 0.02 x 400 = 3 x v, v = 2.67 m s-1 opposite to the bullet. (Marking: two KE 1 each, loss 1, percentage 1, recoil 2) |
Section D: Waves and Sound (16 marks)
Data 4: a sound wave has frequency 250 Hz and wavelength 1.36 m. Take the speed of sound in air as 340 m s-1.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | (a) Find the speed of the sound wave and its period. | Answer: v = f L = 250 x 1.36 = 340.0 m s-1; T = 1 / f = 1 / 250 = 0.004 s. (Marking: speed 2, period 2) |
| 14 | (b) An echo returns 0.6 s after the sound is made. Find the distance to the obstacle. | Answer: 2 d = v t = 340 x 0.6 = 204 m, so d = 102.0 m. (Marking: two way path 2, distance 2) |
| 15 | (c) Find the distance travelled by sound in 2.5 s. | Answer: s = v t = 340 x 2.5 = 850.0 m. (Marking: formula 2, value 2) |
| 16 | (d) A 512 Hz tuning fork sounds in air. Find the wavelength and state how it changes if the frequency doubles. | Answer: L = v / f = 340 / 512 = 0.664 m; at constant speed, doubling the frequency halves the wavelength to 0.332 m. (Marking: wavelength 2, relation 2) |
Section E: Circuits and Electrical Power (18 marks)
Data 5: resistors of 6 ohm and 4 ohm are connected in series to a 12 V supply.
| # | Question | Answer and marking notes |
|---|---|---|
| 17 | (a) Find the total resistance, the current and the total power. | Answer: R = 6 + 4 = 10.0 ohm; I = 12 / 10.0 = 1.2 A; P = V I = 12 x 1.2 = 14.4 W. (Marking: R 1, I 2, P 1) |
| 18 | (b) If 6 ohm and 3 ohm are connected in parallel to 12 V, find the equivalent resistance and the total current. | Answer: R = (6 x 3) / (6 + 3) = 2.0 ohm; I = 12 / 2.0 = 6.0 A. (Marking: parallel formula 2, current 2) |
| 19 | (c) A 2 000 W kettle runs for 0.25 h. Find the energy used in kWh and the cost at 1.20 per kWh. | Answer: energy = 2 kW x 0.25 h = 0.5 kWh; cost = 0.5 x 1.20 = 0.60. (Marking: kWh 2, cost 2) |
| 20 | (d) An appliance is rated 2 200 W at 220 V. Find the current and explain why a 10 A fuse is chosen rather than 5 A. | Answer: I = P / V = 2 200 / 220 = 10.0 A; the current reaches 10 A, so a 5 A fuse would blow in normal use, while 10 A carries it without nuisance blowing. (Marking: current 3, fuse reason 3) |
Section F: Radioactive Decay (16 marks)
Data 6: a sample of 80 g has a half life of 5 days. A count rate falls from 640 Bq to 40 Bq in 12 hours.
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | (a) Find the mass remaining after 15 days. | Answer: 15 / 5 = 3 half lives; remaining = 80 x (0.5)^3 = 10.0 g. (Marking: number of half lives 2, mass 2) |
| 22 | (b) From the count rate data, find the half life of the nuclide. | Answer: 640 to 320 to 160 to 80 to 40 is 4 half lives; half life = 12 / 4 = 3.0 hours. (Marking: count 2, half life 2) |
| 23 | (c) Write the alpha decay equation of radium 226 (proton number 88) and state the mass and proton numbers of the daughter. | Answer: 226/88 Ra decays to 222/86 Rn + 4/2 He; daughter mass number 222, proton number 86. (Marking: equation 3, two numbers 1) |
| 24 | (d) What fraction remains after two half lives, and which of alpha, beta and gamma radiation is most penetrating? | Answer: fraction remaining = (0.5)^2 = 25.0%; gamma is uncharged and most penetrating, needing several centimetres of lead, while alpha is stopped by paper. (Marking: fraction 2, comparison 2) |
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