Secondary 3 Physics Practice Paper 3

TuitionGoWhere Practice Paper — Answer Key

Subject: Physics | Level: Secondary 3 | Paper: Mechanics Topic Test — Version 3 of 5
Total Marks: 40

Section A: Multiple Choice Questions [10 marks]

1. C — Velocity [1]
Explanation: Velocity has both magnitude and direction, making it a vector. Speed, distance and time are scalars.

2. B — 60 km/h [1]
Working: Average speed = total distance ÷ total time = 150 ÷ 2.5 = 60 km/h

3. B — A straight line sloping upwards [1]
Explanation: Uniform acceleration means velocity increases at a constant rate, which is represented by a straight line with positive gradient on a v–t graph.

**4.**C — 10 m/s² downwards [1]
Explanation: At the highest point the velocity is momentarily zero, but the acceleration due to gravity (10 m/s² downwards) still acts on the ball throughout the motion.

5. C — 50 N [1]
Working: W = mg = 5 × 10 = 50 N. The normal contact force balances the weight, so it is 50 N upwards.

6. B — net external force [1]
Explanation: Newton's First Law (law of inertia) states that an object remains at rest or in uniform motion in a straight line unless acted upon by a resultant (net) external force.

7. B — 5 m/s² [1]
Working: F = ma → a = F/m = 20 ÷ 4 = 5 m/s²

8. B — F = m × a [1]
Explanation: Newton's Second Law states that the net force on an object is equal to the product of its mass and acceleration.

9. B — −5 m/s² [1]
Working: a = (v − u) ÷ t = (0 − 30) ÷ 6 = −5 m/s². The negative sign indicates deceleration.

10. C — distance travelled [1]
Explanation: The area under a velocity-time graph gives the displacement (or distance if no change in direction).

Section B: Structured Questions [20 marks]

11. [4]

(a) Displacement is the shortest straight-line distance from one point to another, measured in a specified direction. [1]

(b) Uniform acceleration is when the velocity of an object changes by equal amounts in equal intervals of time. [1]

(c) Speed is a scalar quantity (magnitude only) whereas velocity is a vector quantity (magnitude and direction). [2]
Marking: [1] for scalar vs vector distinction, [1] for mentioning direction.

12. [6]

(a) Maximum velocity: [2]
Using v = u + at
v = 0 + 0.5 × 20
v = 10 m/s ✓

(b) Total distance = area under v–t graph: [3]
Phase 1 (acceleration): s₁ = ½ × 20 × 10 = 100 m
Phase 2 (constant velocity): s₂ = 10 × 30 = 300 m
Phase 3 (deceleration): s₃ = ½ × 10 × 10 = 50 m
Total distance = 100 + 300 + 50 = 450 m ✓
Marking: [1] each phase, [1] for total (awarded if method shown).

(c) Sketch: A velocity-time graph with three segments — a straight line rising from (0,0) to (20,10), a horizontal line from (20,10) to (50,10), and a straight line falling from (50,10) to (60,0). Axes labelled "Velocity (m/s)" and "Time (s)". [1]
Marking: Award [1] for correct shape with labelled axes and key values.

13. [4]

(a) Using F = ma: [2]
12 = 2 × a
a = 6 m/s² ✓
Marking: [1] for correct substitution, [1] for correct answer with unit.

(b) Using v = u + at: [2]
v = 0 + 6 × 3
v = 18 m/s ✓
Marking: [1] for correct substitution, [1] for correct answer with unit.

14. [3]
When a person walks, their foot pushes backwards on the ground (action force). By Newton's Third Law, the ground exerts an equal and opposite forward reaction force on the person's foot. This forward reaction force propels the person forward. [3]
Marking: [1] for identifying the action (foot pushes ground backwards), [1] for identifying the reaction (ground pushes foot forwards), [1] for stating that the forces are equal and opposite (Newton's Third Law).

15. [3]

(a) Using s = ut + ½gt² (u = 0): [2]
s = 0 + ½ × 10 × 3²
s = 5 × 9
s = 45 m ✓
Marking: [1] for correct substitution, [1] for correct answer with unit.

(b) Assumption: Air resistance is negligible (or the stone falls freely under gravity alone). [1]

Section C: Application Question [10 marks]

16. [5]

(a) Acceleration = gradient of v–t graph: [2]
a = (8 − 0) ÷ (3 − 0)
a = 8/3 ≈ 2.67 m/s² ✓
Marking: [1] for using gradient method, [1] for correct answer.

(b) Total distance = area under graph: [3]
Area 1 (triangle, 0–3 s): ½ × 3 × 8 = 12 m
Area 2 (trapezium, 3–6 s): ½ × (8 + 4) × 3 = 18 m
Area 3 (rectangle, 6–8 s): 4 × 2 = 8 m
Area 4 (trapezium, 8–10 s): ½ × (4 + 8) × 2 = 12 m
Area 5 (triangle, 10–14 s): ½ × 4 × 8 = 16 m
Total distance = 12 + 18 + 8 + 12 + 16 = 66 m ✓
Marking: [1] for correct method (area calculation), [1] for at least 3 areas correct, [1] for correct total.

17. [4]

(a) Scale reading when stationary: [1]
W = mg = 60 × 10 = 600 N ✓

(b) When accelerating upwards: [3]
The net force on the student: R − mg = ma
R = m(g + a) = 60 × (10 + 2) = 60 × 12
R = 720 N ✓
The scale reads 720 N.
Marking: [1] for correct equation (R − mg = ma or equivalent), [1] for correct substitution, [1] for correct answer with unit.
Common mistake: Students may forget that the scale reads the normal reaction force, not the weight directly.

18. [4]

(a) Vertical motion (horizontal projection, u_y = 0): [2]
s = ½gt² → 45 = ½ × 10 × t²
t² = 9
t = 3 s ✓
Marking: [1] for correct equation, [1] for correct answer.

(b) Horizontal distance: [2]
Range = horizontal velocity × time = 15 × 3
Range = 45 m ✓
Marking: [1] for using horizontal velocity × time, [1] for correct answer with unit.
Note: Horizontal velocity remains constant (no horizontal acceleration).

19. [4]

(a) Using v² = u² + 2as: [2]
0 = 15² + 2 × a × 30
0 = 225 + 60a
a = −225 ÷ 60
a = −3.75 m/s² (deceleration = 3.75 m/s²) ✓
Marking: [1] for correct substitution, [1] for correct answer.

(b) Using F = ma: [2]
F = 2000 × 3.75
F = 7500 N ✓
Marking: [1] for using F = ma, [1] for correct answer with unit.

20. [3]

(a) Resultant force (Pythagoras): [2]
R = √(30² + 40²) = √(900 + 1600) = √2500
R = 50 N ✓
Marking: [1] for correct method (Pythagoras or scaled diagram), [1] for correct answer.

(b) Direction: [1]
θ = tan⁻¹(40 ÷ 30) = tan⁻¹(1.333)
θ ≈ 53.1° north of east (or N 36.9° E) ✓
Marking: [1] for correct angle with direction stated.
Accept: "53° north of east" or equivalent bearing.

— End of Answer Key —