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S4 Physics Paper: Uniform Acceleration, Newton Laws, Momentum, Work and Power, Projectile and Circular Motion (100 marks)
S4 Physics – mechanics
S4 Physics paper: 6 questions, 24 parts, 100 marks, 60 minutes. Covers kinematics, Newton laws with friction, momentum, work and power, projectiles and circular motion.
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Paper details S4 Physics practice paper with 6 questions on kinematics, forces, momentum, work and power, projectiles and circular motion, with full answers.
Section A: Uniform Acceleration (18 marks)
A car starts from rest and accelerates uniformly at 2.50 m/s2 for 8.0 s. It then travels at the speed reached for 12 s, and finally decelerates uniformly to rest in 5.0 s.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Find the speed at the end of 8.0 s and the displacement during acceleration. | Answer: speed = 0 + 2.50 x 8.0 = 20.0 m/s; displacement = 0.5 x 2.50 x 8.0^2 = 80.0 m. (Marking: speed 2, displacement 2) |
| 2 | Find the displacement during the 12 s of constant speed. | Answer: 20.0 x 12 = 240.0 m. (Marking: formula 1, substitution 2, answer 1) |
| 3 | Find the magnitude of the deceleration and the displacement in that stage. | Answer: deceleration = 20.0 / 5.0 = 4.0 m/s2; displacement = 0.5 x (20.0 + 0) x 5.0 = 50.0 m. (Marking: deceleration 2, displacement 3) |
| 4 | Find the average speed for the whole journey. | Answer: total displacement = 80.0 + 240.0 + 50.0 = 370.0 m; total time = 25.0 s; average speed = 370.0 / 25.0 = 14.8 m/s. (Marking: total 3, average 2) |
Section B: Newton Second Law and Friction (18 marks)
A block of mass 5.0 kg rests on a rough horizontal table and is pulled by a horizontal force of 21 N. The friction between the block and the table is 6.0 N. Take g = 9.8 m/s2.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Find the weight of the block and the normal reaction from the table. | Answer: weight = 5.0 x 9.8 = 49.0 N; normal reaction = 49.0 N. (Marking: weight 2, reaction 2) |
| 6 | Find the acceleration of the block. | Answer: net force = 21 – 6.0 = 15.0 N; acceleration = 15.0 / 5.0 = 3.0 m/s2. (Marking: net force 2, acceleration 3) |
| 7 | Starting from rest, find its speed and displacement after 4.0 s. | Answer: speed = 3.0 x 4.0 = 12.0 m/s; displacement = 0.5 x 3.0 x 4.0^2 = 24.0 m. (Marking: speed 3, displacement 2) |
| 8 | The pulling force then disappears. Find how far the block slides before stopping. | Answer: deceleration = 6.0 / 5.0 = 1.2 m/s2; distance = 12.0^2 / (2 x 1.2) = 60.0 m. (Marking: deceleration 2, distance 2) |
Section C: Momentum and Collision (16 marks)
Car A of mass 1200 kg travels at 15 m/s and hits car B of mass 800 kg which is at rest. After the collision the two cars lock together and move in the original direction.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Find the momentum of car A before the collision. | Answer: momentum = 1200 x 15 = 18000.0 kg m/s. (Marking: formula 2, answer 2) |
| 10 | Find the common velocity of the two cars after the collision. | Answer: by conservation of momentum, 18000.0 = (1200 + 800) x v, so v = 9.0 m/s. (Marking: equation 2, answer 2) |
| 11 | Find the total kinetic energy before and after the collision and state where the energy goes. | Answer: before = 0.5 x 1200 x 15^2 = 135000 J; after = 0.5 x 2000 x 9.0^2 = 81000 J; 54000 J is lost to sound, internal energy and deformation. (Marking: energies 2, explanation 2) |
| 12 | Find the magnitude of the impulse acting on car B during the collision. | Answer: impulse = change in momentum = 800 x 9.0 = 7200 N s. (Marking: formula 2, answer 2) |
Section D: Work, Energy and Power (18 marks)
A person of mass 60 kg climbs a staircase to a floor 12 m higher and takes 20 s. Take g = 9.8 m/s2.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | Find the weight of the person. | Answer: weight = 60 x 9.8 = 588 N. (Marking: formula 2, answer 2) |
| 14 | Find the work done against gravity. | Answer: work = 588 x 12 = 7056 J (about 7.1 x 10^3 J). (Marking: formula 2, substitution 2, answer 1) |
| 15 | Find his average power output in climbing. | Answer: power = 7056 / 20 = 352.8 W (about 353 W). (Marking: formula 2, answer 3) |
| 16 | If the same stairs are climbed in 15 s, what is the power and the percentage rise in power? | Answer: power = 7056 / 15 = 470.4 W; rise = (470.4 – 352.8) / 352.8 x 100 = 33.3 per cent. (Marking: power 2, percentage 2) |
Section E: Horizontal Projectile (16 marks)
A ball is projected horizontally at 12 m/s from the edge of a platform 44.1 m above the ground. Take g = 9.8 m/s2 and ignore air resistance.
| # | Question | Answer and marking notes |
|---|---|---|
| 17 | Find the time the ball takes to reach the ground. | Answer: vertical motion is free fall, 44.1 = 0.5 x 9.8 x t^2, so t = 3.0 s. (Marking: equation 2, answer 2) |
| 18 | Find the horizontal range. | Answer: range = 12 x 3.0 = 36 m. (Marking: formula 2, answer 2) |
| 19 | Find the vertical velocity component and the speed of the ball on landing. | Answer: vertical component = 9.8 x 3.0 = 29.4 m/s; speed = square root of (12^2 + 29.4^2) = 31.75 m/s (about 31.8 m/s). (Marking: component 2, speed 2) |
| 20 | Find the angle between the landing velocity and the horizontal. | Answer: tan(theta) = 29.4 / 12 = 2.45, theta = 67.8 degrees, about 68 degrees below the horizontal. (Marking: ratio 2, angle 2) |
Section F: Circular Motion (14 marks)
A ball of mass 0.50 kg moves at 2.0 m/s in a horizontal circle of radius 0.80 m. A string provides the centripetal force.
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | Find the centripetal acceleration of the ball. | Answer: a = 2.0^2 / 0.80 = 5.0 m/s2, directed towards the centre. (Marking: formula 2, value and direction 2) |
| 22 | Find the tension in the string. | Answer: tension = 0.50 x 5.0 = 2.5 N. (Marking: formula 2, answer 2) |
| 23 | Find the period of the motion. | Answer: T = 2pi x 0.80 / 2.0 = 2.51 s. (Marking: formula 2, answer 1) |
| 24 | If the speed doubles and the radius is unchanged, by what factor does the tension change? Find the new tension. | Answer: tension varies as the square of the speed, so the factor is 4; new tension = 0.50 x 4.0^2 / 0.80 = 10.0 N. (Marking: factor 2, new tension 1) |
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