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Secondary 4 Pure Physics Mechanics Quiz
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Questions
Secondary 4 Pure Physics Quiz - Mechanics
Name: ________________________
Class: ________________________
Date: ________________________
Score: _____ / 40
Duration: 50 minutes
Total Marks: 40
Instructions:
- Answer all questions in the spaces provided.
- Show all working clearly for calculation questions. Marks are awarded for correct method even if the final answer is incorrect.
- Include units in your final answers where appropriate.
- The number of marks for each question or part-question is shown in brackets [ ].
- You may use a calculator.
Section A: Multiple Choice (Questions 1–5)
Each question carries 2 marks. Choose the most accurate answer.
1. A car accelerates uniformly from rest to 24 m/s in 8.0 s. What is the acceleration of the car?
A. 2.0 m/s²
B. 3.0 m/s²
C. 4.0 m/s²
D. 6.0 m/s²
Answer: _______________
2. Which of the following is a vector quantity?
A. Speed
B. Distance
C. Velocity
D. Time
Answer: _______________
3. A ball is thrown vertically upwards. At the highest point of its trajectory, what is the acceleration of the ball? (Ignore air resistance.)
A. 0 m/s²
B. 9.8 m/s² upwards
C. 9.8 m/s² downwards
D. 19.6 m/s² downwards
Answer: _______________
4. A 5.0 kg object is acted upon by a net force of 20 N. What is the acceleration of the object?
A. 0.25 m/s²
B. 4.0 m/s²
C. 15 m/s²
D. 100 m/s²
Answer: _______________
5. A 60 kg student stands on a scale inside a lift. The scale reads 660 N. What is the acceleration of the lift? (Take g = 10 m/s².)
A. 1.0 m/s² upwards
B. 1.0 m/s² downwards
C. 2.0 m/s² upwards
D. 2.0 m/s² downwards
Answer: _______________
Section B: Short Answer and Structured Questions (Questions 6–15)
6. Define the following terms:
(a) Displacement. [1]
(b) Acceleration. [1]
7. State Newton's First Law of Motion. [2]
8. A car travelling at 15 m/s brakes uniformly and comes to rest in 3.0 s.
(a) Calculate the deceleration of the car. [2]
(b) Calculate the distance travelled by the car during braking. [2]
9. A 0.50 kg ball is dropped from a height of 20 m. (Take g = 10 m/s². Ignore air resistance.)
(a) Calculate the speed of the ball just before it hits the ground. [3]
(b) State the principle used in your calculation. [1]
10. A 4.0 kg block is pushed along a horizontal frictionless surface by a constant force of 12 N.
(a) Calculate the acceleration of the block. [2]
(b) If the block starts from rest, calculate its velocity after 5.0 s. [2]
11. A 70 kg skydiver falls vertically downwards. The air resistance acting on the skydiver is 500 N. (Take g = 10 m/s².)
(a) Calculate the net force acting on the skydiver. [2]
(b) Calculate the acceleration of the skydiver. [2]
(c) State what happens to the acceleration of the skydiver as their speed increases. Explain your answer. [2]
12. A 2.0 kg object is placed on a frictionless inclined plane at an angle of 30° to the horizontal. (Take g = 10 m/s².)
(a) Calculate the component of the object's weight acting parallel to the slope. [2]
(b) Calculate the acceleration of the object down the slope. [2]
13. A trolley of mass 3.0 kg moving at 4.0 m/s collides and sticks to a stationary trolley of mass 2.0 kg.
(a) Calculate the total momentum before the collision. [2]
(b) Calculate the velocity of the combined trolleys after the collision. [2]
(c) State the principle used in your calculation. [1]
14. A 0.15 kg tennis ball is struck by a racket. The ball approaches the racket at 10 m/s and leaves at 15 m/s in the opposite direction. The contact time between the ball and racket is 0.020 s.
(a) Calculate the change in momentum of the ball. [3]
(b) Calculate the average force exerted by the racket on the ball. [2]
15. A 60 kg student stands on a scale in a lift. The lift accelerates upwards at 2.0 m/s². (Take g = 10 m/s².)
(a) Draw a free-body diagram showing the weight of the student and the normal force (reaction force) from the scale. Label the forces clearly. [2]
[Diagram space]
(b) Calculate the reading on the scale. [3]
Section C: Application and Extended Response (Questions 16–20)
16. A ball is projected horizontally at 8.0 m/s from the top of a cliff 45 m high. (Take g = 10 m/s². Ignore air resistance.)
(a) Calculate the time taken for the ball to reach the ground. [3]
(b) Calculate the horizontal distance from the base of the cliff where the ball lands. [2]
(c) Calculate the speed of the ball just before it hits the ground. [3]
17. A 1200 kg car travelling at 25 m/s collides head-on with a stationary 800 kg car. After the collision, the two cars move together.
(a) Calculate the total momentum before the collision. [2]
(b) Calculate the common velocity of the two cars after the collision. [2]
(c) Calculate the kinetic energy lost during the collision. [3]
(d) Explain what happens to the kinetic energy that is lost. [2]
18. A 50 kg boy sits on a 10 kg trolley initially at rest on a smooth horizontal surface. The boy jumps off the trolley horizontally at 2.0 m/s relative to the ground.
(a) State the principle of conservation of momentum and explain why it applies in this situation. [2]
(b) Calculate the velocity of the trolley immediately after the boy jumps off. [3]
(c) If the boy was in contact with the trolley for 0.50 s during the jump, calculate the average force the boy exerted on the trolley. [2]
19. A 0.20 kg stone is tied to a string and whirled in a vertical circle of radius 0.80 m. The stone moves at a constant speed of 4.0 m/s. (Take g = 10 m/s².)
(a) Calculate the centripetal acceleration of the stone. [2]
(b) Calculate the tension in the string when the stone is at the lowest point of the circle. [3]
(c) Explain why the tension in the string is greatest at the lowest point of the circle. [2]
20. A 1500 kg car starts from rest and accelerates uniformly along a straight road. After travelling 100 m, the car reaches a speed of 20 m/s.
(a) Calculate the acceleration of the car. [3]
(b) Calculate the net force acting on the car. [2]
(c) The engine of the car provides a driving force of 4000 N. Calculate the average frictional force acting on the car during this motion. [3]
(d) On the axes below, sketch a velocity-time graph for the motion of the car. Label the axes with appropriate values. [2]
[Graph space]
End of Quiz
Answers
Secondary 4 Pure Physics Quiz - Mechanics
Answer Key
Section A: Multiple Choice
1. B [2]
Using v = u + at:
24 = 0 + a(8.0)
a = 24 / 8.0 = 3.0 m/s²
2. C [2]
Velocity is a vector quantity because it has both magnitude and speed. Speed, distance, and time are scalar quantities.
3. C [2]
At the highest point, the ball's velocity is momentarily zero, but the acceleration due to gravity is still acting downwards at 9.8 m/s² (or 10 m/s² depending on convention used). Acceleration is constant throughout the motion (ignoring air resistance).
4. B [2]
Using F = ma:
20 = 5.0 × a
a = 20 / 5.0 = 4.0 m/s²
5. A [2]
Weight of student = mg = 60 × 10 = 600 N
Scale reading (normal force) N = 660 N
Since N > mg, the net force is upwards.
Net force = 660 − 600 = 60 N
Using F = ma: 60 = 60 × a → a = 1.0 m/s² upwards
Section B: Short Answer and Structured Questions
6. (a) Displacement is the shortest distance from one point to another in a specified direction. [1]
(b) Acceleration is the rate of change of velocity. [1]
7. Newton's First Law of Motion states that an object will remain at rest or continue to move at a constant velocity unless acted upon by a resultant (net) external force. [2]
8. (a) Using v = u + at:
0 = 15 + a(3.0)
a = −15 / 3.0 = −5.0 m/s² (deceleration = 5.0 m/s²) [2]
(b) Using s = (u + v)t / 2:
s = (15 + 0) × 3.0 / 2 = 22.5 m (or 23 m to 2 s.f.) [2]
9. (a) Using conservation of energy (or v² = u² + 2as):
v² = 0 + 2(10)(20) = 400
v = 20 m/s [3]
Award marks for: correct equation [1], correct substitution [1], correct answer with unit [1].
(b) The principle of conservation of energy (or conservation of mechanical energy). [1]
10. (a) Using F = ma:
12 = 4.0 × a
a = 3.0 m/s² [2]
(b) Using v = u + at:
v = 0 + 3.0 × 5.0 = 15 m/s [2]
11. (a) Weight = mg = 70 × 10 = 700 N (downwards)
Net force = Weight − Air resistance = 700 − 500 = 200 N (downwards) [2]
(b) Using F = ma:
200 = 70 × a
a = 200 / 70 = 2.86 m/s² (or 2.9 m/s² to 2 s.f.) [2]
(c) As the skydiver's speed increases, air resistance increases. [1] This reduces the net force acting on the skydiver, so the acceleration decreases. [1] (Eventually, when air resistance equals weight, acceleration becomes zero and terminal velocity is reached.)
12. (a) Component of weight parallel to slope = mg sin θ
= 2.0 × 10 × sin 30°
= 2.0 × 10 × 0.50
= 10 N [2]
(b) Using F = ma along the slope:
10 = 2.0 × a
a = 5.0 m/s² [2]
13. (a) Total momentum before = m₁u₁ + m₂u₂
= 3.0 × 4.0 + 2.0 × 0
= 12 kg·m/s [2]
(b) By conservation of momentum:
12 = (3.0 + 2.0) × v
v = 12 / 5.0 = 2.4 m/s [2]
(c) The principle of conservation of momentum. [1]
14. (a) Taking the initial direction as positive:
Initial momentum = 0.15 × 10 = 1.5 kg·m/s
Final momentum = 0.15 × (−15) = −2.25 kg·m/s
Change in momentum = final − initial = −2.25 − 1.5 = −3.75 kg·m/s
Magnitude = 3.75 kg·m/s (or 3.8 kg·m/s to 2 s.f.) [3]
Award marks for: correct direction convention [1], correct substitution [1], correct answer with unit [1].
(b) Using F = Δp / Δt:
F = 3.75 / 0.020 = 187.5 N (or 190 N to 2 s.f.) [2]
15. (a) Free-body diagram should show:
- Weight (W = mg = 600 N) acting downwards from the centre of mass [1]
- Normal force / reaction force (N) acting upwards from the scale [1]
Both forces should be clearly labelled with magnitudes or labels.
(b) Using Newton's Second Law (upwards positive):
N − mg = ma
N = m(g + a) = 60 × (10 + 2.0) = 60 × 12 = 720 N [3]
Award marks for: correct equation [1], correct substitution [1], correct answer with unit [1].
Section C: Application and Extended Response
16. (a) Vertical motion: s = ½gt²
45 = ½ × 10 × t²
t² = 9.0
t = 3.0 s [3]
(b) Horizontal distance = horizontal velocity × time
= 8.0 × 3.0 = 24 m [2]
(c) Vertical component of velocity: v_y = gt = 10 × 3.0 = 30 m/s
Horizontal component: v_x = 8.0 m/s
Resultant speed = √(v_x² + v_y²) = √(64 + 900) = √964 = 31.0 m/s (or 31 m/s to 2 s.f.) [3]
17. (a) Total momentum before = 1200 × 25 + 800 × 0 = 30 000 kg·m/s [2]
(b) By conservation of momentum:
30 000 = (1200 + 800) × v
v = 30 000 / 2000 = 15 m/s [2]
(c) KE before = ½ × 1200 × 25² = ½ × 1200 × 625 = 375 000 J
KE after = ½ × 2000 × 15² = ½ × 2000 × 225 = 225 000 J
KE lost = 375 000 − 225 000 = 150 000 J (or 1.5 × 10⁵ J) [3]
(d) The lost kinetic energy is converted into other forms of energy, such as thermal energy (heat) and sound energy, due to deformation of the cars during the collision. [2]
18. (a) The principle of conservation of momentum states that the total momentum of a system remains constant if no external resultant force acts on the system. [1] In this situation, the system (boy + trolley) is on a smooth horizontal surface, so there is no external horizontal force acting on the system. [1]
(b) Total momentum before = 0 (system at rest)
Total momentum after = 0
0 = 50 × 2.0 + 10 × v_trolley
10 × v_trolley = −100
v_trolley = −10 m/s
The trolley moves at 10 m/s in the opposite direction to the boy. [3]
(c) Change in momentum of trolley = 10 × 10 = 100 kg·m/s
Average force = Δp / Δt = 100 / 0.50 = 200 N [2]
19. (a) Centripetal acceleration a = v² / r = 4.0² / 0.80 = 16 / 0.80 = 20 m/s² [2]
(b) At the lowest point, the tension acts upwards and weight acts downwards. The net force towards the centre (upwards) provides the centripetal force:
T − mg = mv² / r
T = mg + mv² / r = 0.20 × 10 + 0.20 × 20 = 2.0 + 4.0 = 6.0 N [3]
Award marks for: correct equation [1], correct substitution [1], correct answer with unit [1].
(c) At the lowest point, the tension must support the weight of the stone AND provide the centripetal force. [1] At other points in the circle, only a component of the weight acts along the string, so the tension required is less. [1]
20. (a) Using v² = u² + 2as:
20² = 0 + 2 × a × 100
400 = 200a
a = 2.0 m/s² [3]
(b) Using F = ma:
F = 1500 × 2.0 = 3000 N [2]
(c) Net force = Driving force − Frictional force
3000 = 4000 − f
f = 4000 − 3000 = 1000 N [3]
(d) Velocity-time graph:
- Straight line starting from (0, 0) with constant positive gradient [1]
- Ending at approximately v = 20 m/s
- Time axis: t = s/v_avg = 100/10 = 10 s, so line ends at (10 s, 20 m/s) [1]
- Axes labelled: "Velocity (m/s)" and "Time (s)" with appropriate values
End of Answer Key