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S6 Physics: Circular Motion and Gravitation – Practice Paper 3 (Challenge, 10 Questions)
S6 Physics: Circular Motion and Gravitation – Practice Paper 3 (Challenge, 10 Questions)
Paper three combines both topics: period ratio from a radius ratio, gravitational force at a given orbital radius, the angle between acceleration and velocity, the large acceleration inside a centrifuge, the speed limit set by friction on a bend, string tension, surface gravity on Mars, the period change from r to 2r, and finally the Moon orbit data used for both speed and acceleration, closing with the error that the centripetal force points along the velocity.
Quick answer A radius ratio of 1 to 4 gives a period ratio of 8 to 1; a 500 kg satellite feels about 4.1 times 10 cubed N; the acceleration points to the centre at 90 degrees to the velocity; 3000 rpm at 0.2 m gives 1.97 times 10 to the 4 m/s squared; friction 0.6 on a 40 m bend allows 15.3 m/s; the string tension is 10 N; Mars surface gravity is about 3.69 m/s squared; r to 2r changes the period by 2.83 times; the Moon moves at 1.02 km/s with acceleration 2.7 times 10 to the minus 3 m/s squared; the centripetal force is perpendicular to the velocity.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Circular Motion and Gravitation: Centripetal Force, Angular Speed and Orbital Period |
| Level | S6 |
02 Questions (10 items)
- FoundationTwo satellites have orbital radii in the ratio 1 to 4. Find the ratio of their periods.
- FoundationA 500 kg satellite orbits at radius 7.0 times 10 to the 6 m. Find the gravitational force on it.
- FoundationIn uniform circular motion, which way does the acceleration point, and what angle does it make with the velocity?
- FoundationA centrifuge of radius 0.2 m spins at 3000 rpm. Find its centripetal acceleration.
- IntermediateA car takes a bend of radius 40 m with a friction coefficient of 0.6. Find the maximum speed without skidding.
- IntermediateA 0.5 kg ball moves at 4 m/s on a horizontal circle of radius 0.8 m. Find the tension in the string.
- IntermediateMars has mass 6.4 times 10 to the 23 kg and radius 3.4 times 10 to the 6 m. Find the surface gravity.
- ChallengeA satellite orbital radius increases from r to 2r. By what factor does its period change?
- ChallengeThe Moon orbits at about 3.84 times 10 to the 8 m with a period of about 2.36 times 10 to the 6 s. Find its average speed and centripetal acceleration.
- Error redoA student says the centripetal force points along the velocity, so the body speeds up. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The period ratio squared is 4 cubed, which is 64, so the periods are in the ratio 8 to 1 | The period ratio is the radius ratio raised to the power 1.5. |
| 2 | F = GMm over r squared = 4.0 times 10 to the 14 times 500 divided by 4.9 times 10 to the 13, about 4.1 times 10 cubed N | This gravitational force is exactly the centripetal force the satellite needs. |
| 3 | Towards the centre, at 90 degrees to the velocity, so it changes the direction of motion without changing the speed | Acceleration perpendicular to velocity turns the motion without speeding it up. |
| 4 | f = 50 Hz; a = 4 pi squared f squared r, about 1.97 times 10 to the 4 m/s squared | Convert the frequency first, then apply a = 4 pi squared f squared r, which is roughly two thousand g. |
| 5 | Setting friction equal to the required centripetal force gives v equal to the square root of 0.6 times 9.8 times 40, about 15.3 m/s | Friction supplies the centripetal force, and the car skids once that limit is exceeded. |
| 6 | Tension = 0.5 times 16 divided by 0.8, which is 10 N | For horizontal circular motion the tension is the source of the centripetal force. |
| 7 | g = GM over R squared, about 3.69 m/s squared | A smaller radius helps but a much smaller mass wins, so surface gravity is about a third of the Earth value. |
| 8 | The period goes as r to the power 1.5, so 2 to the power 1.5, about 2.83 times the original | Doubling the radius does not double the period; use the power 1.5. |
| 9 | v = 2 pi r over T, about 1.02 times 10 cubed m/s; a = v squared over r, about 2.7 times 10 to the minus 3 m/s squared | The Moon centripetal acceleration is far below surface gravity, exactly as the inverse square law predicts. |
| 10 | The centripetal force points along the radius towards the centre, perpendicular to the velocity; a perpendicular force only turns the motion and does no work, so the speed stays the same | Only a force along the velocity changes the speed; a perpendicular one only changes direction. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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