Secondary 4 Pure Physics Mechanics Quiz

Secondary 4 Pure Physics Quiz - Mechanics (Answer Key)

Total Marks: 40

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Section A: Multiple Choice Questions (10 marks)

1. Answer: B (1 mark)

Working:

• Acceleration $a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m/s}^2$
• Resultant force $F = ma = 1200 \times 2.5 = 3000 \text{ N}$

Key concept: Newton's second law $F = ma$. Average acceleration used for uniform acceleration.

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2. Answer: C (1 mark)

Reasoning: Since the block does not move, it is in equilibrium. The frictional force exactly balances the applied horizontal force (static friction). Magnitude = 30 N.

Key concept: Static friction matches applied force up to its maximum value. Since no motion, friction = applied force.

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3. Answer: B (1 mark)

Reasoning: At the highest point, the ball momentarily stops (velocity = 0). However, gravity still acts, so acceleration = $g = 10 \text{ m/s}^2$ downwards.

Common mistake: Thinking acceleration is zero when velocity is zero. Acceleration is rate of change of velocity; gravity continues to act.

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4. Answer: A (1 mark)

Working (Principle of Moments about pivot):

• Anticlockwise moment = $4 \text{ N} \times (30 - 10) \text{ cm} = 4 \times 20 = 80 \text{ N cm}$
• Clockwise moment = $W \times (80 - 30) \text{ cm} = W \times 50 \text{ N cm}$
• For equilibrium: $80 = 50W \Rightarrow W = 1.6 \text{ N}$

Key concept: Principle of moments — sum of clockwise moments = sum of anticlockwise moments about any pivot.

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5. Answer: C (1 mark)

Reasoning: At terminal velocity, net force = 0. Air resistance = weight = $mg = 80 \times 10 = 800 \text{ N}$.

Key concept: Terminal velocity occurs when air resistance equals weight, so resultant force is zero and acceleration is zero.

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6. Answer: C (1 mark)

Working:

• Acceleration $a = \frac{F}{m} = \frac{20}{4} = 5 \text{ m/s}^2$
• Final velocity $v = u + at = 0 + 5 \times 3 = 15 \text{ m/s}$

Alternative: Impulse = change in momentum: $Ft = mv \Rightarrow v = \frac{Ft}{m} = \frac{20 \times 3}{4} = 15 \text{ m/s}$

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7. Answer: C (1 mark)

Working:

• Component of weight down plane = $mg \sin 30° = 5 \times 10 \times 0.5 = 25 \text{ N}$
• Friction = 10 N (down plane, opposing motion up)
• For constant velocity, net force = 0: $F_{\text{pull}} = 25 + 10 = 35 \text{ N}$

Key concept: Constant velocity → equilibrium → pulling force balances weight component down plane + friction.

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8. Answer: C (1 mark)

Reasoning: Neglecting air resistance, all objects fall with the same acceleration $g$, regardless of mass. They hit the ground at the same time with the same speed.

Key concept: Galileo's principle — in vacuum, all objects fall with same acceleration.

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9. Answer: C (1 mark)

Working:

• Centripetal acceleration $a_c = \frac{v^2}{r} = \frac{20^2}{50} = \frac{400}{50} = 8 \text{ m/s}^2$

Key concept: Centripetal acceleration formula $a_c = \frac{v^2}{r}$ for uniform circular motion.

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10. Answer: C (1 mark)

Working:

• Original: $F = ma \Rightarrow a = \frac{F}{m}$
• New: $F' = 2F$, $m' = \frac{m}{2}$
• New acceleration $a' = \frac{F'}{m'} = \frac{2F}{m/2} = 4 \frac{F}{m} = 4a$

Key concept: Acceleration is directly proportional to force and inversely proportional to mass.

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Section B: Structured Questions (30 marks)

11. Toy Car Motion Analysis

(a) Graph plotting (2 marks)

• 1 mark: Axes labelled with units, appropriate scales (Time 0–4 s, Distance 0–9 m), points plotted correctly
• 1 mark: Smooth curve through all points (parabolic shape for uniform acceleration)

Expected graph: Curve passing through (0,0), (1,0.5), (2,2.0), (3,4.5), (4,8.0) — shape of $s = \frac{1}{2}at^2$

(b) Speed at $t = 2.5 \text{ s}$ (2 marks)

• 1 mark: Tangent drawn at $t = 2.5 \text{ s}$ on graph
• 1 mark: Correct gradient calculation from tangent
• Expected gradient ≈ 2.5 m/s (since $v = at = 2.5 \times 2.5 = 6.25 \text{ m/s}$ — wait, let's recalculate)

Recalculation from data:

• $s = \frac{1}{2}at^2$ → at $t=2$, $s=2.0$ → $2.0 = \frac{1}{2}a(4) \Rightarrow a = 1.0 \text{ m/s}^2$
• At $t=2.5 \text{ s}$, $v = at = 1.0 \times 2.5 = 2.5 \text{ m/s}$
• Answer: 2.5 m/s (accept 2.4–2.6 m/s from graph)

(c) Acceleration (2 marks)

• Method 1: From $s = \frac{1}{2}at^2$, using $t=2 \text{ s}, s=2.0 \text{ m}$: $2.0 = \frac{1}{2}a(4) \Rightarrow a = 1.0 \text{ m/s}^2$
• Method 2: Gradient of velocity-time graph (if constructed)
• Answer: 1.0 m/s² (2 marks for correct value with working; 1 mark for correct method with arithmetic error)

(d) Resultant force (1 mark)

• $F = ma = 0.2 \times 1.0 = 0.2 \text{ N}$
• Answer: 0.2 N (ecf from (c))

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12. Beam and Cable Equilibrium

(a) Principle of moments (1 mark)

For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.

(b) Tension in cable (3 marks) Taking moments about A (hinge):

• Clockwise moments:
  • Beam weight: $200 \text{ N} \times 2.0 \text{ m} = 400 \text{ N m}$
  • Load: $300 \text{ N} \times 1.5 \text{ m} = 450 \text{ N m}$
  • Total clockwise = $850 \text{ N m}$
• Anticlockwise moment:
  • Vertical component of tension: $T \sin 30° \times 4.0 \text{ m} = T \times 0.5 \times 4.0 = 2.0T \text{ N m}$
• Equilibrium: $2.0T = 850 \Rightarrow T = 425 \text{ N}$

Mark breakdown:

• 1 mark: Correct identification of perpendicular distances
• 1 mark: Correct moment equation setup
• 1 mark: Correct final answer with unit

Answer: 425 N

(c) Vertical component of hinge force (2 marks)

• Vertical forces equilibrium: Upward forces = Downward forces
• $R_{Ay} + T \sin 30° = 200 + 300$
• $R_{Ay} + 425 \times 0.5 = 500$
• $R_{Ay} + 212.5 = 500$
• $R_{Ay} = 287.5 \text{ N}$

Answer: 287.5 N (or 288 N) — 1 mark for equation, 1 mark for answer

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13. Skydiver Velocity-Time Graph

(a) Motion between 0–10 s (2 marks)

• The skydiver accelerates from rest.
• Velocity increases non-linearly (curved graph), indicating decreasing acceleration.
• Speed increases from 0 to 40 m/s.

Key points: Accelerating, but acceleration decreases (curve flattens), due to increasing air resistance.

(b) Why acceleration decreases (0–30 s) (2 marks)

• As speed increases, air resistance increases.
• Net force = Weight – Air resistance decreases.
• Since $a = F_{\text{net}}/m$, acceleration decreases.
• At terminal velocity (≈50 m/s at 30 s), air resistance = weight, net force = 0, acceleration = 0.

(c) Acceleration at $t = 5 \text{ s}$ (2 marks)

• 1 mark: Tangent drawn at $t = 5 \text{ s}$ on graph
• 1 mark: Gradient calculated correctly
• Expected gradient ≈ 7–8 m/s² (steep initial slope)
• Accept range: 6.5–8.5 m/s² with correct tangent method

(d) Why velocity decreases rapidly after 35 s (2 marks)

• Parachute opens → large increase in surface area → large increase in air resistance.
• Air resistance becomes much greater than weight.
• Large net upward force → large deceleration (negative acceleration).
• Velocity decreases until new, lower terminal velocity reached.

(e) Terminal velocity after parachute opens (1 mark)

• From graph: horizontal section from 35–60 s at 10 m/s
• Answer: 10 m/s

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14. Block on Inclined Plane (Work-Energy)

(a) Loss in GPE (2 marks)

• Vertical height $h = 5.0 \times \sin 30° = 5.0 \times 0.5 = 2.5 \text{ m}$
• Loss in GPE = $mgh = 2.0 \times 10 \times 2.5 = 50 \text{ J}$

Answer: 50 J

(b) Kinetic energy at bottom (1 mark)

• $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \times 4.0^2 = 16 \text{ J}$

Answer: 16 J

(c) Work done against friction (2 marks)

• Work-energy principle: Loss in GPE = Gain in KE + Work against friction
• $50 = 16 + W_{\text{friction}}$
• $W_{\text{friction}} = 34 \text{ J}$

Answer: 34 J (1 mark for principle, 1 mark for answer)

(d) Average frictional force (2 marks)

• Work = Force × distance (along plane)
• $34 = F \times 5.0$
• $F = 6.8 \text{ N}$

Answer: 6.8 N (ecf from (c))

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15. Banked Track

(a) Free-body diagram (2 marks)

• 1 mark: Weight $mg$ vertically down from centre of car
• 1 mark: Normal reaction $N$ perpendicular to track surface (at 15° to vertical)
• Optional but good: Components shown — $N \cos 15°$ vertical, $N \sin 15°$ horizontal toward centre

(b) Derivation of $v = \sqrt{rg \tan \theta}$ (3 marks) Vertical equilibrium: $N \cos \theta = mg$ (1) Horizontal (centripetal) force: $N \sin \theta = \frac{mv^2}{r}$ (2) Divide (2) by (1): $\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg}$ $\tan \theta = \frac{v^2}{rg}$ $v^2 = rg \tan \theta$ $v = \sqrt{rg \tan \theta}$

Mark breakdown:

• 1 mark: Vertical resolution correct
• 1 mark: Horizontal resolution correct
• 1 mark: Correct algebraic manipulation to final formula

(c) Speed calculation (2 marks)

• $v = \sqrt{80 \times 10 \times \tan 15°}$
• $\tan 15° \approx 0.268$
• $v = \sqrt{800 \times 0.268} = \sqrt{214.4} \approx 14.6 \text{ m/s}$

Answer: 14.6 m/s (accept 14.5–14.7 m/s)

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16. Rocket Launch

(a) Initial thrust (2 marks)

• Thrust = rate of mass ejection × exhaust speed relative to rocket
• $F_{\text{thrust}} = \frac{dm}{dt} \times v_{\text{exhaust}} = 2.0 \times 800 = 1600 \text{ N}$

Answer: 1600 N

(b) Initial acceleration (2 marks)

• Net force = Thrust – Weight = $1600 - (500 \times 10) = 1600 - 5000 = -3400 \text{ N}$
• Wait — this gives negative acceleration! The rocket wouldn't lift off.
• Correction: For a realistic rocket, thrust must exceed weight. Let me adjust the numbers in the answer key to match a realistic scenario, or note the issue.

Actually, let's recalculate with the given numbers:

• Thrust = 1600 N
• Weight = 5000 N
• Net force = -3400 N → acceleration = -6.8 m/s² (downwards)
• This means the rocket doesn't lift off with these parameters.

For the answer key, I'll note this and provide the calculation:

• Acceleration $a = \frac{F_{\text{net}}}{m} = \frac{1600 - 5000}{500} = -6.8 \text{ m/s}^2$
• Note: With given values, thrust < weight, so rocket cannot lift off. In reality, thrust must exceed weight.

(c) Why acceleration increases as fuel mass decreases → acceleration increases (2 marks)

• Thrust remains constant (constant mass flow rate and exhaust speed).
• Mass of rocket decreases as fuel is burnt.
• From $a = \frac{F_{\text{net}}}{m} = \frac{\text{Thrust} - mg}{m}$, as $m$ decreases:
  • Denominator decreases
  • Weight $mg$ decreases
  • Both effects increase acceleration

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17. Centre of Gravity Experiment

(a) Method using plumb line (3 marks)

1. Make three small holes near the edge of the lamina (at A, B, C).
2. Suspend the lamina from a pin through hole A, allowing it to swing freely.
3. Hang a plumb line from the same pin. When stationary, draw a vertical line on the lamina along the plumb line.
4. Repeat steps 2–3 for holes B and C.
5. The intersection point of the three lines is the centre of gravity.

(b) Why swing freely (1 mark)

• So that the lamina can rotate until its centre of gravity is directly below the pivot point, ensuring the plumb line passes through the centre of gravity.

(c) Reason for different intersection points (1 mark)

• Parallax error when drawing lines / marking plumb line position.
• Pin not perfectly horizontal / friction at pivot preventing free rotation.
• Lamina not uniform thickness / density variations.
• Any one valid reason.

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18. Ball Bouncing (Impulse)

(a) Speed before impact (2 marks)

• $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 20 = 400$
• $v = 20 \text{ m/s}$ (downwards)

Answer: 20 m/s

(b) Speed after rebound (2 marks)

• $v^2 = u^2 + 2gh$ (upwards, $v=0$ at max height)
• $0 = u^2 - 2 \times 10 \times 12$
• $u^2 = 240$
• $u = \sqrt{240} \approx 15.5 \text{ m/s}$ (upwards)

Answer: 15.5 m/s (accept 15.5–15.6 m/s)

(c) Average force during contact (3 marks)

• Take upward as positive.
• Change in momentum = $m(v - u) = 0.5 \times [15.5 - (-20)] = 0.5 \times 35.5 = 17.75 \text{ kg m/s}$
• Impulse = Force × time = Change in momentum
• $F \times 0.05 = 17.75$
• $F = \frac{17.75}{0.05} = 355 \text{ N}$

Answer: 355 N (accept 350–360 N)

• 1 mark: Correct momentum change (signs handled correctly)
• 1 mark: Impulse equation used correctly
• 1 mark: Correct final answer with unit

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19. Ladder Against Wall

(a) Why wall reaction is horizontal (1 mark)

• The wall is smooth (frictionless), so it can only exert a force perpendicular to its surface (normal reaction). No

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Secondary 4 Pure Physics Quiz - Mechanics (Answer Key)

Total Marks: 40

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Section A: Multiple Choice Questions (10 marks)

Question	Answer	Explanation
1	C	$a = \frac{v-u}{t} = \frac{25-0}{10} = 2.5 \text{ m/s}^2$; $F = ma = 1200 \times 2.5 = 3000 \text{ N}$
2	C	Block doesn't move → static friction equals applied force = 30 N
3	B	At highest point: velocity = 0, acceleration = $g = 10 \text{ m/s}^2$ downwards
4	A	Clockwise moment = $4 \times (30-10) = 80 \text{ Ncm}$; Anticlockwise moment = $W \times (80-30) = 50W$; $50W = 80 \Rightarrow W = 1.6 \text{ N}$
5	C	At terminal velocity: air resistance = weight = $mg = 80 \times 10 = 800 \text{ N}$
6	C	$a = \frac{F}{m} = \frac{20}{4} = 5 \text{ m/s}^2$; $v = u + at = 0 + 5 \times 3 = 15 \text{ m/s}$
7	C	Component of weight down plane = $mg \sin 30° = 5 \times 10 \times 0.5 = 25 \text{ N}$; Constant velocity → pulling force = friction + weight component = $10 + 25 = 35 \text{ N}$
8	C	Neglecting air resistance, all objects fall with same acceleration $g$
9	C	$a_c = \frac{v^2}{r} = \frac{20^2}{50} = \frac{400}{50} = 8 \text{ m/s}^2$
10	D	$F = ma$; New force = $2F$, new mass = $\frac{m}{2}$; New acceleration = $\frac{2F}{m/2} = 4 \frac{F}{m} = 4a$

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Section B: Structured Questions (30 marks)

11. Toy Car Motion Analysis

(a) Graph plotting: [2 marks]

• Axes labelled with units (Time/s, Distance/m) [1]
• Points plotted correctly and smooth curve drawn through points [1]

(b) Speed at $t = 2.5 \text{ s}$: [2 marks]

• Draw tangent at $t = 2.5 \text{ s}$ on graph
• Gradient of tangent = speed
• Expected value: 5.0 m/s (accept 4.8–5.2 m/s)

(c) Acceleration calculation: [2 marks]

• Using $s = ut + \frac{1}{2}at^2$ with $u=0$: $8.0 = \frac{1}{2}a(4)^2 \Rightarrow a = 1.0 \text{ m/s}^2$
• Or from graph: gradient of velocity-time graph derived from distance-time graph
• Answer: 1.0 m/s²

(d) Resultant force: [1 mark]

• $F = ma = 0.2 \times 1.0 = \textbf{0.2 N}$

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12. Beam and Cable

(a) Principle of moments: [1 mark]

• For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.

(b) Tension $T$ (moments about A): [3 marks]

• Clockwise moments: $(200 \times 2.0) + (300 \times 1.5) = 400 + 450 = 850 \text{ Nm}$
• Anticlockwise moment: $T \sin 30° \times 4.0 = T \times 0.5 \times 4.0 = 2.0T$
• $2.0T = 850 \Rightarrow T = \textbf{425 N}$

(c) Vertical component at hinge: [2 marks]

• Vertical forces: $R_{Ay} + T \sin 30° = 200 + 300$
• $R_{Ay} + 425 \times 0.5 = 500$
• $R_{Ay} + 212.5 = 500 \Rightarrow R_{Ay} = \textbf{287.5 N}$ (upwards)

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13. Skydiver Velocity-Time Graph

(a) Motion between $t = 0$ and $t = 10 \text{ s}$: [2 marks]

• Skydiver accelerates from rest
• Velocity increases non-uniformly (curved graph)
• Acceleration decreases with time

(b) Why acceleration decreases: [2 marks]

• As velocity increases, air resistance increases
• Net force ($mg - \text{air resistance}$) decreases
• Since $a = F_{net}/m$, acceleration decreases

(c) Acceleration at $t = 5 \text{ s}$: [2 marks]

• Draw tangent at $t = 5 \text{ s}$ on graph
• Gradient ≈ $\frac{30-10}{6-2} = 5 \text{ m/s}^2$ (accept 4.5–5.5 m/s²)
• Answer: ~5.0 m/s²

(d) Velocity decrease after parachute opens: [2 marks]

• Parachute greatly increases air resistance
• Air resistance > weight → net force upwards
• Upward acceleration (deceleration) causes rapid velocity decrease

(e) Terminal velocity after parachute opens: [1 mark]

• From graph: horizontal section after 35 s at 10 m/s

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14. Block on Inclined Plane

(a) Loss in GPE: [2 marks]

• Vertical height $h = 5.0 \times \sin 30° = 2.5 \text{ m}$
• $\Delta GPE = mgh = 2.0 \times 10 \times 2.5 = \textbf{50 J}$

(b) Kinetic energy at bottom: [1 mark]

• $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \times 4.0^2 = \textbf{16 J}$

(c) Work done against friction: [2 marks]

• Work-energy: Loss in GPE = Gain in KE + Work against friction
• $50 = 16 + W_f \Rightarrow W_f = \textbf{34 J}$

(d) Average frictional force: [2 marks]

• $W_f = f \times d \Rightarrow 34 = f \times 5.0 \Rightarrow f = \textbf{6.8 N}$

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15. Banked Track

(a) Free-body diagram: [2 marks]

• Weight $mg$ vertically downwards
• Normal reaction $N$ perpendicular to track surface
• No friction force shown (smooth/no sideways friction)

(b) Derivation of $v = \sqrt{rg \tan \theta}$: [3 marks]

• Vertical equilibrium: $N \cos \theta = mg$ (1)
• Horizontal (centripetal): $N \sin \theta = \frac{mv^2}{r}$ (2)
• Divide (2) by (1): $\tan \theta = \frac{v^2}{rg}$
• $v^2 = rg \tan \theta \Rightarrow v = \sqrt{rg \tan \theta}$

(c) Speed calculation: [2 marks]

• $v = \sqrt{80 \times 10 \times \tan 15°} = \sqrt{800 \times 0.2679} = \sqrt{214.3} = \textbf{14.6 m/s}$

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16. Rocket Launch

(a) Initial thrust: [2 marks]

• Thrust = rate of mass ejection × exhaust speed = $2.0 \times 800 = \textbf{1600 N}$

(b) Initial acceleration: [2 marks]

• Weight = $500 \times 10 = 5000 \text{ N}$
• Net force = Thrust - Weight = $1600 - 5000 = -3400 \text{ N}$
• Acceleration = $\frac{-3400}{500} = \textbf{-6.8 m/s}^2$ (rocket doesn't lift off initially!)
• Note: Thrust < Weight, so rocket remains on launch pad initially

(c) Why acceleration increases: [2 marks]

• Mass decreases as fuel is burnt ($m$ decreases)
• Thrust remains constant
• $a = \frac{F_{net}}{m} = \frac{\text{Thrust} - mg}{m}$; as $m$ decreases, $a$ increases
• Also weight $mg$ decreases, increasing net force

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17. Centre of Gravity Experiment

(a) Plumb line method: [3 marks]

1. Make 3 small holes near edge of lamina
2. Suspend lamina from hole A using pin; hang plumb line from same pin
3. When lamina swings freely and comes to rest, draw vertical line along plumb line
4. Repeat for holes B and C
5. Intersection of three lines = centre of gravity

(b) Why lamina must swing freely: [1 mark]

• So that centre of gravity is vertically below the suspension point (equilibrium position)

(c) Reason for different intersection points: [1 mark]

• Parallax error when drawing lines / pin not horizontal / lamina not uniform thickness / holes too large

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18. Ball Bouncing

(a) Speed before hitting floor: [2 marks]

• $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 20 = 400$
• $v = \textbf{20 m/s}$ (downwards)

(b) Speed after leaving floor: [2 marks]

• $v^2 = u^2 + 2gh$; at max height $v=0$: $0 = u^2 - 2 \times 10 \times 12$
• $u^2 = 240 \Rightarrow u = \textbf{15.5 m/s}$ (upwards)

(c) Average force during contact: [3 marks]

• Change in momentum = $m(v - u) = 0.5 \times [15.5 - (-20)] = 0.5 \times 35.5 = 17.75 \text{ kg m/s}$
• $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{17.75}{0.05} = \textbf{355 N}$ (upwards)

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19. Ladder Problem

(a) Wall reaction is horizontal: [1 mark]

• Wall is smooth → no friction → reaction force perpendicular to wall (horizontal)

(b) Wall reaction (moments about foot): [3 marks]

• Clockwise moments: $(200 \times 3.0 \cos 60°) + (600 \times 4.0 \cos 60°) = (200 \times 1.5) + (600 \times 2.0) = 300 + 1200 = 1500 \text{ Nm}$
• Anticlockwise moment: $R_w \times 6.0 \sin 60° = R_w \times 6.0 \times 0.866 = 5.196 R_w$
• $5.196 R_w = 1500 \Rightarrow R_w = \textbf{289 N}$

(c) Frictional force at foot: [1 mark]

• Horizontal equilibrium: $F = R_w = \textbf{289 N}$

(d) Will ladder slip? [2 marks]

• Vertical equilibrium: $R_f = 200 + 600 = 800 \text{ N}$
• Max friction = $\mu R_f = 0.4 \times 800 = 320 \text{ N}$
• Required friction = 289 N < 320 N → Ladder will not slip

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20. Particle on Horizontal Circle

(a) Centripetal force provider: [1 mark]

• Tension in the string (horizontal component)

(b) Tension in string: [1 mark]

• $T = mg = 0.5 \times 10 = \textbf{5.0 N}$

(c) Speed of particle: [2 marks]

• $T = \frac{mv^2}{r}$; but mass of particle not given!
• Assuming particle mass = 0.5 kg (same as hanging mass):
• $5.0 = \frac{0.5 \times v^2}{0.8} \Rightarrow v^2 = 8.0 \Rightarrow v = \textbf{2.83 m/s}$
• If particle mass not given, answer in terms of m: $v = \sqrt{\frac{Tr}{m}}$

(d) Effect of increasing hanging mass: [1 mark]

• Tension increases → centripetal force increases → speed increases (for same radius)
• Or: radius decreases if speed constant

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Marking Summary

Section	Questions	Marks
A	1–10	10
B	11–20	30
Total		40

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End of Answer Key