Secondary 3 Physics Mechanics Quiz

Secondary 3 Physics Quiz - Mechanics (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. [1 mark] — B

Working:
$a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4 \text{ m/s}^2$

Explanation: Acceleration is the rate of change of velocity. The car starts from rest ($u = 0$) and reaches $20 \text{ m/s}$ in $5 \text{ s}$. Using $a = \frac{\Delta v}{t}$ gives $4 \text{ m/s}^2$.

2. [1 mark] — C

Explanation: At maximum height, the vertical velocity of a projectile is momentarily zero. On a velocity-time graph, this corresponds to where the line crosses the time axis (velocity = 0). Point C is at $(2, 0)$, where velocity is zero.

Common mistake: Choosing point A (initial velocity) or point D (maximum downward velocity).

3. [1 mark] — B

Working:
Resultant force $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 10 - 4 = 6 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{6}{2} = 3 \text{ m/s}^2$

Explanation: Newton's Second Law: $F_{\text{net}} = ma$. The net force is the applied force minus the opposing friction force.

4. [1 mark] — B

Explanation: Newton's Third Law states: "For every action, there is an equal and opposite reaction." The action-reaction pair:

• Are equal in magnitude
• Are opposite in direction
• Act on different objects
• Are of the same type (e.g., both gravitational or both contact)

Why others are wrong:
A: They act on different objects.
C: They must be the same type of force.
D: They occur simultaneously, not sequentially.

5. [1 mark] — C

Working:
Loss in GPE = Gain in KE (conservation of energy)
$mgh = \frac{1}{2}mv^2 = \text{KE}$
$\text{KE} = 0.5 \times 10 \times 2 = 10 \text{ J}$

Explanation: The gravitational potential energy at the start ($mgh$) is converted entirely to kinetic energy just before impact (ignoring air resistance). Mass = $0.5 \text{ kg}$, $g = 10 \text{ N/kg}$, $h = 2 \text{ m}$.

6. [1 mark] — B

Working:
$W = F \cos \theta \times s = 20 \times \cos 30^\circ \times 5 = 20 \times 0.866 \times 5 = 86.6 \text{ J}$

Explanation: Work done = force component in direction of displacement × displacement. Only the horizontal component ($F \cos \theta$) does work since displacement is horizontal.

7. [1 mark] — A

Working:
Take direction of $2 \text{ kg}$ object as positive.
Total momentum before = $(2 \times 4) + (3 \times -2) = 8 - 6 = 2 \text{ kg m/s}$
Combined mass = $5 \text{ kg}$
$v = \frac{2}{5} = 0.4 \text{ m/s}$ (positive, so in direction of $2 \text{ kg}$ object)

Explanation: Conservation of momentum for perfectly inelastic collision. The positive result means the combined mass moves in the original direction of the $2 \text{ kg}$ object.

8. [1 mark] — A

Explanation: For a satellite in orbit, the gravitational force between the Earth and the satellite provides the necessary centripetal force to keep it in circular motion. No string, friction, or normal reaction is involved.

9. [1 mark] — B

Working:
$F_c = \frac{mv^2}{r} = \frac{1000 \times 20^2}{50} = \frac{1000 \times 400}{50} = 8000 \text{ N}$

Explanation: Centripetal force formula $F_c = \frac{mv^2}{r}$. The force is provided by friction between tyres and road.

10. [1 mark] — C

Explanation: For a simple pendulum, $T = 2\pi \sqrt{\frac{l}{g}}$, so $T \propto \sqrt{l}$. Graph C shows a curve that increases with decreasing gradient, characteristic of a square root relationship.

Section B: Structured Questions (30 marks)

11. [3 marks]

(a) [1 mark]
$v = u + at = 0 + 1.5 \times 8 = 12 \text{ m/s}$
Answer: $12 \text{ m/s}$

(b) [2 marks]
Distance during acceleration: $s_1 = \frac{1}{2}at^2 = \frac{1}{2} \times 1.5 \times 8^2 = 48 \text{ m}$
Distance at constant speed: $s_2 = vt = 12 \times 10 = 120 \text{ m}$
Distance during deceleration: $s_3 = \frac{1}{2}vt = \frac{1}{2} \times 12 \times 5 = 30 \text{ m}$
Total distance = $48 + 120 + 30 = 198 \text{ m}$
Answer: $198 \text{ m}$

Mark breakdown: 1 mark for correct method (area under v-t graph or equations of motion), 1 mark for correct total.

12. [4 marks]

(a) [1 mark]
$W = mg = 4 \times 10 = 40 \text{ N}$
Answer: $40 \text{ N}$

(b) [1 mark]
$F_{\text{friction(max)}} = \mu R = \mu mg = 0.3 \times 40 = 12 \text{ N}$
Answer: $12 \text{ N}$

(c) [2 marks]
Applied force ($15 \text{ N}$) > Maximum friction ($12 \text{ N}$), so block will move.
Net force = $15 - 12 = 3 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{3}{4} = 0.75 \text{ m/s}^2$
Answer: Yes, it moves. Acceleration = $0.75 \text{ m/s}^2$

Mark breakdown: 1 mark for correct comparison and conclusion, 1 mark for correct acceleration.

13. [3 marks]

(a) [1 mark]
$v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 1.8 = 36$
$v = 6 \text{ m/s}$ (downwards)
Answer: $6 \text{ m/s}$

(b) [1 mark]
$v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 1.2 = 24$
$v = \sqrt{24} = 4.9 \text{ m/s}$ (upwards)
Answer: $4.9 \text{ m/s}$ (or $\sqrt{24} \text{ m/s}$)

(c) [1 mark]
Take upward as positive.
Change in momentum = $m(v - u) = 0.2 \times (4.9 - (-6)) = 0.2 \times 10.9 = 2.18 \text{ kg m/s}$
Average force = $\frac{\Delta p}{\Delta t} = \frac{2.18}{0.05} = 43.6 \text{ N}$ (upwards)
Answer: $43.6 \text{ N}$ upwards

Alternative using impulse: $F_{\text{avg}} \Delta t = m(v - u)$

14. [4 marks]

(a) [1 mark]
$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = 375\,000 \text{ J}$
Answer: $375\,000 \text{ J}$ (or $3.75 \times 10^5 \text{ J}$)

(b) [2 marks]
Work done by braking force = Loss in KE
$F \times d = 375\,000$
$F \times 50 = 375\,000$
$F = 7500 \text{ N}$
Answer: $7500 \text{ N}$

Mark breakdown: 1 mark for work-energy principle, 1 mark for correct calculation.

(c) [1 mark]
The kinetic energy is converted into heat energy (and some sound energy) due to friction between the brake pads and discs/drums, and between tyres and road.

15. [3 marks]

(a) [1 mark]
Weight = $mg = 500 \times 10 = 5000 \text{ N}$ (downwards)
Resultant force = Thrust - Weight = $8000 - 5000 = 3000 \text{ N}$ (upwards)
Answer: $3000 \text{ N}$ upwards

(b) [1 mark]
$a = \frac{F_{\text{net}}}{m} = \frac{3000}{500} = 6 \text{ m/s}^2$ (upwards)
Answer: $6 \text{ m/s}^2$ upwards

(c) [1 mark]
$v = u + at = 0 + 6 \times 10 = 60 \text{ m/s}$
Answer: $60 \text{ m/s}$ upwards

16. [4 marks]

(a) [1 mark]
$h = l(1 - \cos \theta) = 1.0 \times (1 - \cos 30^\circ) = 1.0 \times (1 - 0.866) = 0.134 \text{ m}$
Answer: $0.134 \text{ m}$ (or $13.4 \text{ cm}$)

(b) [1 mark]
Max KE = Loss in GPE = $mgh = 0.5 \times 10 \times 0.134 = 0.67 \text{ J}$
Answer: $0.67 \text{ J}$

(c) [1 mark]
$\frac{1}{2}mv^2 = 0.67$
$v^2 = \frac{2 \times 0.67}{0.5} = 2.68$
$v = \sqrt{2.68} = 1.64 \text{ m/s}$
Answer: $1.64 \text{ m/s}$

(d) [1 mark]
Gravitational potential energy → Kinetic energy

17. [3 marks]

(a) [1 mark]
The total momentum of a closed system remains constant if no external resultant force acts on the system.

(b) [2 marks]
Initial momentum = 0 (both at rest)
Final momentum = $m_A v_A + m_B v_B = 0$
$60 v_A + 40 \times 3 = 0$
$60 v_A = -120$
$v_A = -2 \text{ m/s}$
Answer: $2 \text{ m/s}$ in the opposite direction to skater B (i.e., away from skater B)

Mark breakdown: 1 mark for conservation of momentum equation, 1 mark for correct magnitude and direction.

18. [3 marks]

(a) [1 mark]
Clockwise moment = $3 \text{ N} \times 0.30 \text{ m} = 0.9 \text{ N m}$
Answer: $0.9 \text{ N m}$

(b) [1 mark]
Anticlockwise moment = $2 \text{ N} \times 0.30 \text{ m} = 0.6 \text{ N m}$
Answer: $0.6 \text{ N m}$

(c) [1 mark]
Not in equilibrium. Clockwise moment ($0.9 \text{ N m}$) > Anticlockwise moment ($0.6 \text{ N m}$).
The metre rule will rotate clockwise.

19. [3 marks]

(a) [1 mark]
$F = mg = 200 \times 8.7 = 1740 \text{ N}$
Answer: $1740 \text{ N}$ (towards Earth's centre)

(b) [1 mark]
Centripetal acceleration = gravitational field strength = $8.7 \text{ m/s}^2$
Answer: $8.7 \text{ m/s}^2$ (towards Earth's centre)

Explanation: For a satellite, the gravitational force provides the centripetal force: $F = ma_c$, so $mg = ma_c$, giving $a_c = g$.

(c) [1 mark]
Orbital radius $r = 6400 + 400 = 6800 \text{ km} = 6.8 \times 10^6 \text{ m}$
$a_c = \frac{v^2}{r}$
$v^2 = a_c \times r = 8.7 \times 6.8 \times 10^6 = 5.916 \times 10^7$
$v = \sqrt{5.916 \times 10^7} = 7690 \text{ m/s}$
Answer: $7690 \text{ m/s}$ (or $7.69 \text{ km/s}$)

20. [3 marks]

(a) [1 mark]
Component down plane = $mg \sin \theta = 5 \times 10 \times \sin 30^\circ = 50 \times 0.5 = 25 \text{ N}$
Answer: $25 \text{ N}$

(b) [1 mark]
Normal reaction $R = mg \cos \theta = 5 \times 10 \times \cos 30^\circ = 50 \times 0.866 = 43.3 \text{ N}$
Friction $f = \mu R = 0.2 \times 43.3 = 8.66 \text{ N}$
Answer: $8.66 \text{ N}$ (acting down the plane)

(c) [1 mark]
Constant speed → resultant force = 0
$F = \text{component down plane} + \text{friction} = 25 + 8.66 = 33.66 \text{ N}$
Answer: $33.7 \text{ N}$ (or $33.66 \text{ N}$) up the plane

Mark breakdown: 1 mark for resolving forces correctly and equating to zero net force.

End of Answer Key