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S4 Physics Paper: Forces, Motion and Newton Laws (100 marks)
S4 Physics – forces and motion
S4 Physics paper: six questions on accelerated motion, Newton second law, momentum, work and power, circular motion and incline experiment, with answers and marking notes.
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Paper details S4 Physics practice paper with six questions on uniformly accelerated motion, Newton second law and friction, conservation of momentum and collisions, work and power, circular motion, and an incline experiment checked with energy conservation. Full answers with marking notes.
Section A: Uniformly Accelerated Motion (14 marks)
A car starts from rest and accelerates uniformly along a straight line at 2.5 m s^-2 for 8.0 s, then travels at constant speed for 12 s. Give units in every answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Find the final speed after accelerating for 8.0 s. | Answer: v equals u plus a t, so the speed is 2.5 times 8.0, that is 20 m s^-1. (Marking: 1 mark for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
| 2 | Find the displacement during the acceleration. | Answer: s equals u t plus half a t squared, so the displacement is 0.5 times 2.5 times 64, that is 80 m. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
| 3 | Find the total displacement over the whole 20 s. | Answer: the constant speed part covers 20 times 12, that is 240 m, so the total is 80 plus 240, that is 320 m. (Marking: 2 marks for the constant speed part, 1 mark for adding, 1 mark for the value with unit.) |
| 4 | Find the average velocity for the whole journey. | Answer: average velocity is total displacement divided by total time, that is 320 divided by 20, so 16 m s^-1. (Marking: 1 mark for the definition, 1 mark for substitution, 1 mark for the value with unit.) |
Section B: Newton Second Law and Friction (16 marks)
A block of mass 4.0 kg rests on a horizontal surface and is pulled by a horizontal force of 12 N. Take the acceleration of free fall as 9.81 m s^-2.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | If the surface is smooth, find the acceleration of the block. | Answer: the net force equals mass times acceleration, so a is 12 divided by 4.0, that is 3.0 m s^-2. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
| 6 | Find the speed 5.0 s after starting from rest. | Answer: the speed is a times t, that is 3.0 times 5.0, so 15 m s^-1. (Marking: 1 mark for the equation, 1 mark for substitution, 2 marks for the value with unit.) |
| 7 | If the surface is rough with a friction force of 4.0 N, find the acceleration and the coefficient of kinetic friction. | Answer: the net force is 12 minus 4.0, that is 8.0 N, so the acceleration is 2.0 m s^-2; the coefficient is 4.0 divided by 4.0 times 9.81, about 0.10. (Marking: 1 mark for the net force, 2 marks for the acceleration, 1 mark for the coefficient.) |
| 8 | In this rough case, find the displacement in the first 5.0 s. | Answer: the displacement is half a t squared, that is 0.5 times 2.0 times 25, so 25 m. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
Section C: Conservation of Momentum and Collisions (16 marks)
Ball A of mass 0.50 kg moves at 4.0 m s^-1 along a straight line and strikes a stationary ball B of mass 0.30 kg. After the collision A moves on in the same direction at 1.0 m s^-1. Both balls stay on the same horizontal line.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Find the total momentum of the system before the collision. | Answer: the total momentum is 0.50 times 4.0, that is 2.0 kg m s^-1, pointing along the original direction of A. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the direction.) |
| 10 | Find the velocity of ball B after the collision. | Answer: the momentum equation gives 0.30 times v equal to 1.5, so v is 5.0 m s^-1 forwards. (Marking: 2 marks for the momentum equation, 1 mark for rearranging, 1 mark for the value with direction.) |
| 11 | Use kinetic energy to decide whether the collision is elastic. | Answer: the kinetic energy is 4.0 J before and also 4.0 J after, so the collision is elastic. (Marking: 1 mark for each kinetic energy, 1 mark for the comparison, 1 mark for the conclusion.) |
| 12 | If the two balls stick together after the collision, find the common velocity and the loss of kinetic energy. | Answer: the common velocity is 2.0 divided by 0.80, that is 2.5 m s^-1, leaving 2.5 J of kinetic energy, so 1.5 J is lost. (Marking: 2 marks for the common velocity, 1 mark for the kinetic energy, 1 mark for the loss.) |
Section D: Work, Energy and Power (18 marks)
A student of mass 60 kg runs up a staircase of vertical height 12 m in 20 s. Take g as 9.81 m s^-2 and ignore the work done in horizontal movement.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | Find the gain in gravitational potential energy of the student. | Answer: the gain is m g h, that is 60 times 9.81 times 12, so about 7.06 times 10^3 J. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
| 14 | Find the average output power of the student. | Answer: power is energy divided by time, that is 7063 divided by 20, about 353 W. (Marking: 2 marks for the equation, 2 marks for substitution, 1 mark for the value with unit.) |
| 15 | If the efficiency of the muscles is 25 per cent, find the input power needed. | Answer: the input power is 353 divided by 0.25, about 1.41 times 10^3 W, that is 1.4 kW. (Marking: 2 marks for the efficiency relation, 1 mark for substitution, 1 mark for the value with unit.) |
| 16 | A person of mass 75 kg climbs the same staircase at the same average output power. Find the time needed. | Answer: the energy needed is 75 times 9.81 times 12, that is 8829 J, so the time is 8829 divided by 353, about 25 s. (Marking: 2 marks for the energy, 2 marks for the power relation, 1 mark for the value with unit.) |
Section E: Circular Motion (18 marks)
A ball of mass 0.20 kg moves at a steady speed of 3.0 m s^-1 in a horizontal circle of radius 0.50 m.
| # | Question | Answer and marking notes |
|---|---|---|
| 17 | Find the centripetal acceleration of the ball. | Answer: the acceleration is v squared divided by r, that is 9.0 divided by 0.50, so 18 m s^-2 towards the centre. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the direction.) |
| 18 | Find the centripetal force acting on the ball. | Answer: the force is mass times acceleration, that is 0.20 times 18, so 3.6 N. (Marking: 2 marks for the equation, 1 mark for substitution, 1 mark for the value with unit.) |
| 19 | Find the time for one complete revolution. | Answer: the period is two pi r divided by v, that is 3.142 divided by 3.0, so about 1.05 s. (Marking: 2 marks for the equation, 2 marks for substitution, 1 mark for the value with unit.) |
| 20 | If the speed rises to 6.0 m s^-1 with the same radius, find the new centripetal force and its ratio to the original value. | Answer: the new force is 0.20 times 36 divided by 0.50, that is 14.4 N, and the ratio is 4.0, so doubling the speed makes the force four times larger. (Marking: 2 marks for the new force, 2 marks for the ratio, 1 mark for the conclusion.) |
Section F: Incline Experiment and Energy Check (18 marks)
A student lets a ball of mass 0.50 kg slide from rest down a smooth incline of length 2.0 m and height 0.80 m. Take g as 9.81 m s^-2.
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | Find the sine of the angle of the incline and the acceleration of the ball along it. | Answer: the sine is 0.80 divided by 2.0, that is 0.40, so the acceleration is 9.81 times 0.40, about 3.92 m s^-2. (Marking: 1 mark for the sine, 2 marks for the component equation, 1 mark for the value.) |
| 22 | Use a kinematics equation to find the speed of the ball at the bottom. | Answer: v squared is two a s, that is 2 times 3.92 times 2.0, so the speed is about 3.96 m s^-1. (Marking: 2 marks for the equation, 2 marks for substitution, 1 mark for the value with unit.) |
| 23 | Check the same speed independently with conservation of energy. | Answer: setting m g h equal to half m v squared gives v as the square root of 2 times 9.81 times 0.80, that is 3.96 m s^-1, the same as before. (Marking: 2 marks for the energy equation, 1 mark for cancelling the mass, 1 mark for the value and comparison.) |
| 24 | If the measured speed at the bottom is only 3.6 m s^-1, find the percentage loss of kinetic energy. | Answer: the expected kinetic energy is about 3.92 J and the measured value is 3.24 J, so about 0.68 J is lost, that is 17 per cent. (Marking: 1 mark for each kinetic energy, 1 mark for the loss, 2 marks for the percentage.) |
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