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Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 3
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Questions
TuitionGoWhere Practice Paper - Physics Secondary 3
School: TuitionGoWhere Secondary School (AI)
Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper – Version 3 of 5
Duration: 60 minutes
Total Marks: 50
Name: ________________________
Class: ________________________
Date: ________________________
Instructions to Candidates
- Write your name, class, and date in the spaces provided above.
- Answer all questions in the spaces provided.
- Show all working clearly — marks are awarded for method as well as final answers.
- The number of marks for each question or part-question is shown in brackets [ ].
- You may use a calculator.
- Take g = 10 m/s² unless otherwise stated.
- This paper consists of 20 questions divided into three sections.
Section A – Multiple Choice (Questions 1–5) [10 marks]
For each question, choose the one most clearly correct answer and write its letter in the space provided.
1. Which of the following is a vector quantity?
A. Speed
B. Distance
C. Mass
D. Displacement
Answer: ________ [1]
2. A car travels 120 km in 2 hours, then stops for 30 minutes, then travels another 60 km in 1 hour. What is the car's average speed for the entire journey?
A. 45 km/h
B. 51.4 km/h
C. 60 km/h
D. 90 km/h
Answer: ________ [1]
3. A ball is thrown vertically upward. At the highest point of its trajectory, which statement is correct?
A. The velocity and acceleration are both zero.
B. The velocity is zero and the acceleration is 10 m/s² downward.
C. The velocity is zero and the acceleration is zero.
D. The velocity is 10 m/s upward and the acceleration is zero.
Answer: ________ [1]
4. A 5 kg box rests on a horizontal table. The normal contact force exerted by the table on the box is:
A. 0 N
B. 5 N
C. 50 N
D. 500 N
Answer: ________ [1]
5. A force of 20 N acts on a 4 kg object initially at rest on a frictionless surface. What is the acceleration of the object?
A. 0.2 m/s²
B. 5 m/s²
C. 80 m/s²
D. 24 m/s²
Answer: ________ [1]
Section B – Short Answer and Structured Questions (Questions 6–15) [20 marks]
Answer all questions. Show your working where appropriate.
6. Define the following terms:
(a) Speed ________________________________________________________________ [1]
(b) Velocity ________________________________________________________________ [1]
7. A cyclist travels 300 m north in 60 s, then 200 m south in 40 s.
(a) Calculate the cyclist's total distance travelled. [1]
(b) Calculate the cyclist's total displacement. [1]
(c) Calculate the cyclist's average speed. [1]
(d) Calculate the cyclist's average velocity. [1]
8. The velocity-time graph below describes the motion of a car over 10 seconds.
v (m/s)
20 | ___________
| / \
10 | / \
| / \
0 |______/ \______
0 2 4 6 8 10 t (s)
(a) Describe the motion of the car during the first 4 seconds. [1]
(b) Calculate the acceleration of the car between t = 0 and t = 4 s. [2]
(c) Calculate the total distance travelled by the car in 10 seconds. [2]
9. A 2 kg object is acted upon by two forces: 12 N to the right and 5 N to the left.
(a) Calculate the resultant force on the object. [1]
(b) Calculate the acceleration of the object. [1]
(c) If the object starts from rest, calculate its velocity after 3 seconds. [1]
10. State Newton's First Law of Motion. [2]
11. A 10 kg box is pushed across a rough horizontal floor with a constant force of 60 N. The frictional force opposing the motion is 20 N.
(a) Draw a clearly labelled free-body diagram showing all forces acting on the box. [2]
(b) Calculate the acceleration of the box. [2]
12. A ball is dropped from a height of 45 m. Take g = 10 m/s² and ignore air resistance.
(a) Calculate the time taken for the ball to reach the ground. [2]
(b) Calculate the velocity of the ball just before it hits the ground. [2]
13. Explain, using Newton's Third Law, why a person can walk forward on the ground. [2]
14. A car of mass 1000 kg accelerates uniformly from rest to 20 m/s in 5 seconds.
(a) Calculate the acceleration of the car. [1]
(b) Calculate the resultant force acting on the car. [1]
(c) Calculate the momentum of the car at the end of 5 seconds. [1]
15. A 0.5 kg ball moving at 6 m/s collides with a stationary 1.5 kg ball. After the collision, the 0.5 kg ball moves at 2 m/s in the same direction.
(a) State the principle of conservation of momentum. [1]
(b) Calculate the velocity of the 1.5 kg ball after the collision. [2]
Section C – Longer Structured and Application Questions (Questions 16–20) [20 marks]
Answer all questions. Show all working clearly.
16. A student investigates the motion of a trolley on a ramp. The trolley is released from rest at the top of a 2 m long ramp inclined at 30° to the horizontal. The mass of the trolley is 1 kg. Assume no friction and take g = 10 m/s².
(a) Calculate the component of the weight of the trolley acting down the ramp. [2]
(b) Calculate the acceleration of the trolley down the ramp. [1]
(c) Calculate the time taken for the trolley to reach the bottom of the ramp. [2]
(d) Calculate the velocity of the trolley at the bottom of the ramp. [1]
17. A 70 kg skydiver jumps from a plane. After falling freely for some time, the parachute opens and the skydiver reaches a constant terminal velocity.
(a) Explain, in terms of forces, why the skydiver initially accelerates downward. [2]
(b) Explain why the skydiver eventually reaches a constant terminal velocity after the parachute opens. [2]
(c) The skydiver's terminal velocity with the parachute open is 5 m/s. If the skydiver falls at this speed for 20 seconds, calculate the distance fallen during this time. [1]
18. Two objects, A and B, have masses of 2 kg and 8 kg respectively. They are placed 3 m apart on a frictionless surface.
(a) State Newton's Law of Gravitation in words. [1]
(b) Explain how the gravitational force on A due to B compares to the gravitational force on B due to A. [1]
(c) Object A is now given an initial velocity of 4 m/s toward object B. Calculate the momentum of object A. [1]
(d) If the two objects stick together after collision, calculate their common velocity after the collision. [2]
19. A car of mass 1200 kg is travelling at 15 m/s when the driver applies the brakes. The car comes to rest in 3 seconds.
(a) Calculate the deceleration of the car. [1]
(b) Calculate the braking force acting on the car. [2]
(c) Calculate the stopping distance of the car. [2]
(d) If the car's speed had been 30 m/s instead, explain qualitatively how the stopping distance would change (assuming the same braking force). [1]
20. A 3 kg block is placed on a smooth horizontal table and connected by a light inextensible string passing over a smooth pulley to a 2 kg block hanging freely.
┌──────────────┐
│ 3 kg block │
└──────┬───────┘
│ string
▼ pulley
│
2 kg block
(hanging)
(a) Draw a free-body diagram for each block, showing all forces acting on it. [2]
(b) Calculate the acceleration of the system. [2]
(c) Calculate the tension in the string. [2]
END OF PAPER
Answers
TuitionGoWhere Practice Paper – Physics Secondary 3
SA2 Practice Paper – Version 3 of 5
Answer Key and Marking Scheme
Section A – Multiple Choice [10 marks]
1. D – Displacement [1]
Reasoning: Displacement has both magnitude and direction, making it a vector. Speed, distance, and mass are scalars.
2. B – 51.4 km/h [1]
Working:
- Total distance = 120 + 60 = 180 km
- Total time = 2 + 0.5 + 1 = 3.5 h
- Average speed = 180 / 3.5 = 51.4 km/h
Common mistake: Forgetting to include the 30-minute stop in total time.
3. B – The velocity is zero and the acceleration is 10 m/s² downward [1]
Reasoning: At the highest point, the ball momentarily stops (v = 0), but gravity still acts downward, so a = 10 m/s² downward throughout.
4. C – 50 N [1]
Working: Normal force = weight = mg = 5 × 10 = 50 N
Common mistake: Confusing mass (5 kg) with weight (50 N).
5. B – 5 m/s² [1]
Working: F = ma → a = F/m = 20/4 = 5 m/s²
Section B – Short Answer and Structured Questions [20 marks]
6.
(a) Speed is the rate of change of distance with time. [1]
(b) Velocity is the rate of change of displacement with time (or speed in a given direction). [1]
7.
(a) Total distance = 300 + 200 = 500 m [1]
(b) Total displacement = 300 m north − 200 m south = 100 m north [1]
(c) Total time = 60 + 40 = 100 s
Average speed = 500 / 100 = 5 m/s [1]
(d) Average velocity = 100 / 100 = 1 m/s north [1]
8.
(a) The car accelerates uniformly from rest to 20 m/s in the first 4 seconds. [1]
(b) a = (v − u)/t = (20 − 0)/4 = 5 m/s² [2]
(c) Distance = area under graph
= area of triangle (0–4 s) + area of trapezium (4–10 s)
= ½ × 4 × 20 + ½ × (20 + 0) × 6
= 40 + 60 = 100 m [2]
Alternative: Area = ½ × 10 × 20 = 100 m (triangle from 0 to 10 s with peak at 4 s)
9.
(a) Resultant force = 12 − 5 = 7 N to the right [1]
(b) a = F/m = 7/2 = 3.5 m/s² [1]
(c) v = u + at = 0 + 3.5 × 3 = 10.5 m/s [1]
10. Newton's First Law: An object at rest stays at rest, and an object in motion continues in uniform motion in a straight line, unless acted upon by a resultant (unbalanced) external force. [2]
Marking: 1 mark for "no change in motion" idea, 1 mark for "unless resultant force acts" condition.
11.
(a) Free-body diagram should show:
- Weight (W = mg = 100 N) acting downward [½]
- Normal force (N = 100 N) acting upward [½]
- Applied force (60 N) acting horizontally in direction of motion [½]
- Frictional force (20 N) acting horizontally opposite to motion [½]
[2]
(b) Resultant force = 60 − 20 = 40 N
a = F/m = 40/10 = 4 m/s² [2]
12.
(a) s = ½gt² → 45 = ½ × 10 × t² → t² = 9 → t = 3 s [2]
(b) v = gt = 10 × 3 = 30 m/s (or v² = u² + 2gs = 0 + 2×10×45 = 900 → v = 30 m/s) [2]
13. When a person walks, their foot pushes backward on the ground (action). By Newton's Third Law, the ground pushes forward on the person's foot (reaction) with an equal and opposite force. This forward reaction force propels the person forward. [2]
Marking: 1 mark for identifying action-reaction pair, 1 mark for explaining how this causes forward motion.
14.
(a) a = (v − u)/t = (20 − 0)/5 = 4 m/s² [1]
(b) F = ma = 1000 × 4 = 4000 N [1]
(c) p = mv = 1000 × 20 = 20,000 kg·m/s [1]
15.
(a) Principle of conservation of momentum: In a closed system with no external forces, the total momentum before a collision equals the total momentum after the collision. [1]
(b) Total momentum before = 0.5 × 6 + 1.5 × 0 = 3 kg·m/s
Total momentum after = 0.5 × 2 + 1.5 × v = 1 + 1.5v
By conservation: 3 = 1 + 1.5v → 1.5v = 2 → v = 1.33 m/s (or 4/3 m/s) [2]
Section C – Longer Structured and Application Questions [20 marks]
16.
(a) Component of weight down ramp = mg sin θ = 1 × 10 × sin 30° = 10 × 0.5 = 5 N [2]
(b) a = F/m = 5/1 = 5 m/s² [1]
(c) s = ½at² → 2 = ½ × 5 × t² → t² = 0.8 → t = 0.894 s (or √0.8 ≈ 0.89 s) [2]
(d) v² = u² + 2as = 0 + 2 × 5 × 2 = 20 → v = 4.47 m/s (or √20 ≈ 4.5 m/s) [1]
17.
(a) Initially, the only significant force acting on the skydiver is weight (gravity) acting downward. Since air resistance is small at low speeds, the resultant force is downward, causing downward acceleration. [2]
(b) As speed increases, air resistance (drag) increases. Eventually, the upward air resistance equals the downward weight. The resultant force becomes zero, so acceleration becomes zero and the skydiver falls at constant terminal velocity. [2]
(c) Distance = speed × time = 5 × 20 = 100 m [1]
18.
(a) Newton's Law of Gravitation: Every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. [1]
(b) The gravitational force on A due to B is equal in magnitude to the gravitational force on B due to A (Newton's Third Law — action and reaction are equal and opposite). [1]
(c) Momentum of A = mv = 2 × 4 = 8 kg·m/s [1]
(d) By conservation of momentum:
Total momentum before = 8 + 0 = 8 kg·m/s
Total mass after = 2 + 8 = 10 kg
Common velocity = 8/10 = 0.8 m/s [2]
19.
(a) a = (v − u)/t = (0 − 15)/3 = −5 m/s² (deceleration = 5 m/s²) [1]
(b) F = ma = 1200 × (−5) = −6000 N (braking force = 6000 N opposite to motion) [2]
(c) s = ut + ½at² = 15 × 3 + ½ × (−5) × 9 = 45 − 22.5 = 22.5 m
Alternative: s = (u + v)/2 × t = (15 + 0)/2 × 3 = 22.5 m [2]
(d) If speed doubles to 30 m/s, the stopping distance would be four times greater (since v² = 2as, distance is proportional to the square of speed for constant deceleration). [1]
20.
(a) Free-body diagrams:
- 3 kg block (on table): Weight (30 N) downward, Normal force (30 N) upward, Tension (T) horizontally toward pulley [1]
- 2 kg block (hanging): Weight (20 N) downward, Tension (T) upward [1]
[2]
(b) For the 2 kg block: 20 − T = 2a (weight − tension = ma, downward positive)
For the 3 kg block: T = 3a (tension = ma, horizontal)
Adding: 20 = 5a → a = 4 m/s² [2]
(c) T = 3a = 3 × 4 = 12 N [2]
Check: For 2 kg block: 20 − 12 = 8 N = 2 × 4 ✓
END OF ANSWER KEY
Mark Summary
| Section | Marks |
|---|---|
| A (Q1–5) | 10 |
| B (Q6–15) | 20 |
| C (Q16–20) | 20 |
| Total | 50 |