O Level Physics Mechanics Quiz

O-Level Physics Quiz - Mechanics: Answer Key

Total Marks: 40
Level: O-Level
Topic: Mechanics

Section A Answers (Q1–10, 2 marks each)

Q1. Average speed = distance / time = $240 / 20 = 12 \text{ m/s}$.
Teaching note: Speed is a scalar; use total distance over total time. [2]

Q2. A scalar has magnitude only (e.g., mass, speed); a vector has magnitude and direction (e.g., force, velocity).
Teaching note: Direction is the key difference. [2]

Q3. $a = (v - u)/t = (15 - 5)/5 = 2 \text{ m/s}^2$.
Working: substitution shown. [2]

Q4. Weight is the gravitational force acting on an object. $W = mg$ where $m$ = mass, $g$ = gravitational field strength.
Teaching note: Weight is a vector measured in N. [2]

Q5. Moment = $F \times d = 12 \times 0.5 = 6 \text{ N m}$.
Teaching note: Perpendicular distance must be used. [2]

Q6. Normal force = weight = $mg = 2 \times 10 = 20 \text{ N}$.
Teaching note: On horizontal table, normal equals weight if no vertical acceleration. [2]

Q7. $P = F/A = 20 / 0.004 = 5000 \text{ Pa}$.
Teaching note: 1 Pa = 1 N/m². [2]

Q8. For equilibrium, sum of clockwise moments = sum of anticlockwise moments about any pivot.
Teaching note: Principle of moments. [2]

Q9. $E_k = \frac{1}{2}mv^2 = 0.5 \times 0.5 \times 4^2 = 4 \text{ J}$.
Working: $0.5 \times 0.5 = 0.25$; $0.25 \times 16 = 4$. [2]

Q10. $P = h \rho g$; $h$ = height of column, $\rho$ = density of liquid, $g$ = gravitational field strength.
Teaching note: Pressure increases linearly with depth. [2]

Section B Answers (Q11–16, 3 marks each)

Q11.
(a) Input pressure = $F/A = 40 / (4.0 \times 10^{-4}) = 1.0 \times 10^5 \text{ Pa}$. [1]
(b) Output force = $P \times A_{out} = 1.0 \times 10^5 \times (160 \times 10^{-4}) = 1600 \text{ N}$. [2]
Note: Convert cm² to m² ($1 \text{ cm}^2 = 10^{-4} \text{ m}^2$).

Q12.
(a) Resultant = $30 - 10 = 20 \text{ N}$ forward. [1]
(b) $a = F/m = 20 / 10 = 2.0 \text{ m/s}^2$. [2]

Q13. Area under graph = distance.
Stage1 triangle: $\frac{1}{2} \times 5 \times 10 = 25 \text{ m}$. [1]
Stage2 rectangle: $10 \times 10 = 100 \text{ m}$. [1]
Stage3 triangle: $\frac{1}{2} \times 5 \times 10 = 25 \text{ m}$. [1]
Total = 150 m.

Q14.
(a) Moment = $2 \times (0.50 - 0.20) = 2 \times 0.30 = 0.6 \text{ N m}$. [1]
(b) 4 N at distance $d$: $4d = 0.6 \Rightarrow d = 0.15 \text{ m}$ from pivot on opposite side → at 65 cm mark. [2]

Q15. $E_p = mgh = 0.2 \times 10 \times 5 = 10 \text{ J}$. [1.5]
Just before ground, $E_k = E_p = 10 \text{ J}$ (energy conservation). [1.5]

Q16. Power = $W/t = 600 / 3 = 200 \text{ W}$. [1.5]
Height: $W = Fd \Rightarrow d = 600 / 150 = 4 \text{ m}$. [1.5]

Section C Answers (Q17–20, 5 marks each)

Q17.
(a) $v = 0 + 2.0 \times 5.0 = 10.0 \text{ m/s}$. [1]
(b) Stage1: $s_1 = \frac{1}{2} \times 2.0 \times 5.0^2 = 25.0 \text{ m}$. [1]
Stage2: $s_2 = 10.0 \times 10.0 = 100.0 \text{ m}$. [1]
Stage3: $s_3 = \frac{1}{2} \times 10.0 \times 4.0 = 20.0 \text{ m}$. [1]
(c) Total = $25 + 100 + 20 = 145.0 \text{ m}$. [1]

Q18.
(a) Weight = $mg = 4 \times 10 = 40 \text{ N}$. [1]
(b) Max pressure = smallest area = $0.1 \times 0.05 = 0.005 \text{ m}^2$: $40/0.005 = 8000 \text{ Pa}$. [1]
Min pressure = largest area = $0.2 \times 0.1 = 0.02 \text{ m}^2$: $40/0.02 = 2000 \text{ Pa}$. [1]
(c) Pressure = F/A, so smaller area gives higher pressure for same force. [2]

Q19.
(a) Perp distance = $(4/2) \times \cos\theta$, where $\cos\theta = 1/4 = 0.25$; $2 \times 0.25 = 0.5 \text{ m}$. [2]
(b) Moment = $80 \times 0.5 = 40 \text{ N m}$. [1]
(c) Horizontal force at top, away from wall, equal moment (friction/normal) to balance. [2]

Q20.
(a) Work = $mgh = 500 \times 10 \times 20 = 100\,000 \text{ J}$. [1.5]
(b) Power = $100\,000 / 10 = 10\,000 \text{ W}$. [1.5]
(c) Input = useful / efficiency = $100\,000 / 0.5 = 200\,000 \text{ J}$. [2]