Secondary 3 Physics Mechanics Quiz

Secondary 3 Physics Quiz - Mechanics (Answer Key)

Section A: Multiple Choice [10 marks]

1. C - The distance between the centers of the two masses Explanation: In Newton's Law of Gravitation, r represents the center-to-center distance between the two masses, not surface-to-surface distance.

2. B - Velocity is zero but acceleration is 10 m/s² downward Explanation: At the highest point, the ball momentarily stops (v = 0) but gravity still acts downward (a = g = 10 m/s²).

3. B - The friction force equals the component of weight parallel to the plane Explanation: At constant velocity, net force = 0, so friction force balances the component of weight down the plane.

4. B - Distance traveled Explanation: The area under a velocity-time curve gives the displacement/distance traveled.

5. B - Ball B reaches the ground first Explanation: Ball A experiences air resistance which slows it down, while Ball B falls freely in vacuum.

Section B: Structured Questions [25 marks]

6. Car acceleration problem [4 marks]

(a) Calculate acceleration [2 marks] Using v = u + at 20 = 0 + a(8) a = 20/8 = 2.5 m/s²

Marking: 1 mark for correct formula, 1 mark for correct answer with unit

(b) Calculate distance [2 marks] Using s = ut + ½at² s = 0(8) + ½(2.5)(8)² s = ½(2.5)(64) = 80 m

Marking: 1 mark for correct formula, 1 mark for correct answer with unit

7. Velocity-time graph [5 marks]

(a) Acceleration in first 2 seconds [2 marks] Acceleration = gradient = (8-0)/(2-0) = 4 m/s²

Marking: 1 mark for identifying gradient method, 1 mark for correct calculation

(b) Distance between t = 4s and t = 8s [2 marks] Distance = area under curve = velocity × time = 8 × 4 = 32 m

Marking: 1 mark for identifying area method, 1 mark for correct calculation

(c) Motion during final 2 seconds [1 mark] The cyclist decelerates uniformly from 8 m/s to 0 m/s

Marking: 1 mark for correct description of uniform deceleration

8. Child sliding down rope [3 marks]

Apply Newton's second law: Net force = ma = 40 × 8 = 320 N (downward) Weight = mg = 40 × 10 = 400 N (downward) Friction force = Weight - Net force = 400 - 320 = 80 N (upward)

Marking: 1 mark for identifying forces, 1 mark for correct application of F = ma, 1 mark for correct answer

9. Ball thrown upward [8 marks]

(a) Initial kinetic energy [2 marks] KE = ½mv² = ½ × 0.5 × 15² = 56.25 J

Marking: 1 mark for correct formula, 1 mark for correct calculation

(b) Maximum height [3 marks] At maximum height, all KE converts to additional PE: mgh = 56.25 J h = 56.25/(0.5 × 10) = 11.25 m above throwing point Maximum height above ground = 2 + 11.25 = 13.25 m

Marking: 1 mark for energy conservation principle, 1 mark for calculation of additional height, 1 mark for total height above ground

(c) Speed when hitting ground [3 marks] Using energy conservation: Total energy at ground = mgh = 0.5 × 10 × 13.25 = 66.25 J ½mv² = 66.25 v² = 2 × 66.25/0.5 = 265 v = 16.3 m/s

Marking: 1 mark for energy conservation approach, 1 mark for correct energy calculation, 1 mark for final speed

10. Block on inclined plane [5 marks]

(a) Component parallel to plane [2 marks] F∥ = mg sin 30° = 5 × 10 × 0.5 = 25 N

Marking: 1 mark for correct formula, 1 mark for correct calculation

(b) Maximum friction force [2 marks] Normal force N = mg cos 30° = 5 × 10 × 0.866 = 43.3 N Maximum friction = μN = 0.3 × 43.3 = 13.0 N

Marking: 1 mark for calculating normal force, 1 mark for friction calculation

(c) Will block slide? [1 mark] Since 25 N > 13.0 N, the driving force exceeds maximum friction, so the block will slide down.

Marking: 1 mark for correct comparison and conclusion

Total: 35 marks