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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
TuitionGoWhere Secondary School (AI)
SA2 Examination – Version 3
| Subject: | Physics (Pure) |
| Level: | Secondary 3 |
| Paper: | SA2 Practice Paper |
| Duration: | 1 hour 15 minutes |
| Total Marks: | 60 |
Name: _________________________
Class: _________________________
Date: _________________________
Instructions to Candidates
- This paper consists of three sections: Section A, Section B, and Section C.
- Answer all questions.
- Write your answers in the spaces provided.
- Show all working clearly for calculation questions. Marks are awarded for correct method, even if the final answer is wrong.
- Take g = 10 m/s² unless otherwise stated.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- You are advised to spend about 45 minutes on Section A, 20 minutes on Section B, and 10 minutes on Section C.
Section A: Structured Questions (35 marks)
Answer all questions in this section.
1. A student investigates the motion of a toy car on a straight track. The velocity-time graph below shows the car's motion over 12 seconds.
Velocity / m s⁻¹
8 | _______
| / \
6 | / \
| / \
4 | / \
| / \
2 | / \
| / \
0 |_____________________________ Time / s
0 2 4 6 8 10 12
(a) Describe the motion of the car between t = 0 s and t = 4 s. [1]
(b) Calculate the acceleration of the car during the first 4 seconds. [2]
(c) Determine the total distance travelled by the car in the 12 seconds. [2]
(d) State the time interval(s) during which the resultant force on the car is zero. Explain your answer. [2]
2. A wooden crate of mass 25 kg is pulled across a rough horizontal floor by a horizontal force of 120 N. The crate moves at a constant velocity.
(a) Explain what is meant by "constant velocity" in terms of the forces acting on the crate. [2]
(b) Calculate the weight of the crate. [1]
(c) Determine the magnitude of the frictional force acting on the crate. [1]
(d) The pulling force is increased to 160 N. Calculate the acceleration of the crate, assuming the frictional force remains unchanged. [3]
3. A uniform metre rule is pivoted at its 50 cm mark. A 200 g mass is hung at the 20 cm mark, and a 150 g mass is hung at the 80 cm mark.
(a) Draw a diagram showing all the forces acting on the metre rule. Label the pivot, the two masses, and the weight of the rule. [2]
(b) State the principle of moments. [1]
(c) Calculate the moment due to the 200 g mass about the pivot. State whether it is clockwise or anticlockwise. [2]
(d) Determine whether the metre rule is in equilibrium. Show your working. [3]
4. A hydraulic lift is used to raise a car of mass 1200 kg. The lift has a small piston of area 0.02 m² and a large piston of area 0.80 m².
(a) State Pascal's principle. [1]
(b) Calculate the weight of the car. [1]
(c) Calculate the minimum force that must be applied to the small piston to lift the car. [3]
(d) Explain one advantage of using a hydraulic system rather than a single piston to lift heavy loads. [1]
5. A ball of mass 0.50 kg is thrown vertically upwards with an initial speed of 15 m/s. Air resistance is negligible.
(a) Calculate the initial kinetic energy of the ball. [2]
(b) State the principle of conservation of energy. [1]
(c) Determine the maximum height reached by the ball. [2]
(d) Explain what happens to the kinetic energy and gravitational potential energy of the ball as it rises. [2]
Section B: Diagram and Data Interpretation (15 marks)
Answer all questions in this section.
6. The diagram below shows a block of mass 8.0 kg being pulled up a rough inclined plane at constant speed by a force of 60 N parallel to the plane. The block moves 5.0 m along the plane and rises through a vertical height of 2.0 m.
/|
/ |
/ |
/ | 2.0 m
/ |
/_____|
5.0 m
(a) Calculate the work done by the 60 N force. [1]
(b) Calculate the gain in gravitational potential energy of the block. [2]
(c) Determine the amount of energy dissipated as heat due to friction. [2]
(d) Calculate the frictional force acting on the block. [2]
7. A student investigates the relationship between force and extension for a spring. The results are shown in the table below.
| Force / N | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 | |-----------|---|---|-----|-----|-----|-----|-----|-----| | Extension / cm | 0 | 2.5 | 5.0 | 7.5 | 10.0 | 13.0 | 16.5 |
(a) Plot a graph of force (y-axis) against extension (x-axis) on the grid below. [3]
Force / N
6 |
|
5 |
|
4 |
|
3 |
|
2 |
|
1 |
|
0 |________________________________ Extension / cm
0 2 4 6 8 10 12 14 16 18
(b) Using your graph, determine the spring constant for the linear region of the graph. [2]
(c) State the extension at which the spring exceeds its limit of proportionality. Explain how you identified this from the graph. [2]
(d) Calculate the elastic potential energy stored in the spring when the extension is 7.5 cm. [1]
Section C: Free-Response Questions (10 marks)
Answer all questions in this section.
8. A student claims that "if an object is moving, there must be a net force acting on it."
Discuss whether this statement is correct, using Newton's laws of motion to support your answer. Include examples in your discussion. [4]
9. A spacecraft is travelling from Earth to the Moon. At a certain point between the Earth and the Moon, the gravitational forces from the Earth and the Moon on the spacecraft are equal in magnitude but opposite in direction.
(a) Explain why the gravitational force from the Earth on the spacecraft decreases as the spacecraft moves away from Earth. [2]
(b) The mass of the Earth is about 81 times the mass of the Moon. Explain why the point where the gravitational forces are equal is closer to the Moon than to the Earth. [2]
(c) At this point, the spacecraft's engines are turned off. Describe and explain the subsequent motion of the spacecraft. [2]
END OF PAPER
Check your work carefully. Ensure all questions are attempted.
Answers
TuitionGoWhere Practice Paper – Physics Secondary 3
SA2 Examination – Version 3 – ANSWER KEY AND MARKING SCHEME
Total Marks: 60
Section A: Structured Questions (35 marks)
Question 1 (7 marks)
(a) Describe the motion of the car between t = 0 s and t = 4 s. [1]
Answer: The car accelerates uniformly from rest to a velocity of 8 m/s. [1]
Accept: constant acceleration / velocity increases at a constant rate.
(b) Calculate the acceleration of the car during the first 4 seconds. [2]
Answer: a = (v - u) / t = (8 - 0) / 4 = 2 m/s² [2]
Award [1] for correct formula/substitution, [1] for correct answer with unit.
(c) Determine the total distance travelled by the car in the 12 seconds. [2]
Answer: Distance = area under velocity-time graph = area of trapezium = ½ × (12 + 4) × 8 [1] = ½ × 16 × 8 = 64 m [1]
Accept: splitting into triangle + rectangle + triangle: (½ × 4 × 8) + (4 × 8) + (½ × 4 × 8) = 16 + 32 + 16 = 64 m.
(d) State the time interval(s) during which the resultant force on the car is zero. Explain your answer. [2]
Answer: Time intervals: t = 4 s to t = 8 s [1]
Explanation: During this interval, the velocity is constant (8 m/s), so acceleration is zero. By Newton's Second Law (F = ma), if a = 0, then the resultant force F = 0. [1]
Accept: "resultant force is zero when velocity is constant" or equivalent reasoning.
Question 2 (7 marks)
(a) Explain what is meant by "constant velocity" in terms of the forces acting on the crate. [2]
Answer: Constant velocity means the crate is moving in a straight line at a steady speed. [1] This implies the resultant force acting on the crate is zero; the forward pulling force is exactly balanced by the opposing frictional force. [1]
Accept: "forces are balanced" / "net force = 0" / "no acceleration".
(b) Calculate the weight of the crate. [1]
Answer: Weight = mg = 25 × 10 = 250 N [1]
(c) Determine the magnitude of the frictional force acting on the crate. [1]
Answer: Since the crate moves at constant velocity, resultant force = 0. Frictional force = pulling force = 120 N [1]
(d) The pulling force is increased to 160 N. Calculate the acceleration of the crate, assuming the frictional force remains unchanged. [3]
Answer: Resultant force = 160 - 120 = 40 N [1] F = ma → a = F / m = 40 / 25 [1] a = 1.6 m/s² [1]
Award [1] for correct resultant force, [1] for correct substitution, [1] for correct answer with unit.
Question 3 (8 marks)
(a) Draw a diagram showing all the forces acting on the metre rule. Label the pivot, the two masses, and the weight of the rule. [2]
Answer: Diagram should show:
- Pivot at 50 cm mark (upward reaction force) [0.5]
- 200 g mass at 20 cm mark (downward force = 2.0 N) [0.5]
- 150 g mass at 80 cm mark (downward force = 1.5 N) [0.5]
- Weight of rule at 50 cm mark (downward, but acts through pivot so produces no moment) [0.5]
(b) State the principle of moments. [1]
Answer: For an object in equilibrium, the sum of clockwise moments about any pivot equals the sum of anticlockwise moments about the same pivot. [1]
Accept: "total clockwise moment = total anticlockwise moment".
(c) Calculate the moment due to the 200 g mass about the pivot. State whether it is clockwise or anticlockwise. [2]
Answer: Weight of 200 g mass = 0.200 × 10 = 2.0 N [0.5] Distance from pivot = 50 - 20 = 30 cm = 0.30 m [0.5] Moment = F × d = 2.0 × 0.30 = 0.60 N m [0.5] Direction: Anticlockwise [0.5]
Accept: moment in N cm (60 N cm).
(d) Determine whether the metre rule is in equilibrium. Show your working. [3]
Answer: Moment due to 150 g mass: Weight = 0.150 × 10 = 1.5 N [0.5] Distance from pivot = 80 - 50 = 30 cm = 0.30 m [0.5] Moment = 1.5 × 0.30 = 0.60 N m (clockwise) [0.5]
Anticlockwise moment (0.60 N m) = Clockwise moment (0.60 N m) [0.5] Weight of rule acts through pivot (zero moment). [0.5] Therefore, the metre rule is in equilibrium. [0.5]
Question 4 (6 marks)
(a) State Pascal's principle. [1]
Answer: Pressure applied to an enclosed fluid is transmitted equally and undiminished to all parts of the fluid and to the walls of the container. [1]
(b) Calculate the weight of the car. [1]
Answer: Weight = mg = 1200 × 10 = 12 000 N [1]
(c) Calculate the minimum force that must be applied to the small piston to lift the car. [3]
Answer: Pressure on large piston = Force / Area = 12 000 / 0.80 = 15 000 Pa [1] By Pascal's principle, pressure on small piston = 15 000 Pa [1] Force on small piston = Pressure × Area = 15 000 × 0.02 = 300 N [1]
Alternative method: F₁/A₁ = F₂/A₂ → F₁ = F₂ × (A₁/A₂) = 12 000 × (0.02/0.80) = 300 N.
(d) Explain one advantage of using a hydraulic system rather than a single piston to lift heavy loads. [1]
Answer: A hydraulic system allows a small force applied to the small piston to lift a much heavier load on the large piston (force multiplication). [1]
Accept: "the force is multiplied" / "a smaller effort can lift a larger load".
Question 5 (7 marks)
(a) Calculate the initial kinetic energy of the ball. [2]
Answer: KE = ½mv² = ½ × 0.50 × (15)² [1] = 0.25 × 225 = 56.25 J [1]
Accept: 56.3 J.
(b) State the principle of conservation of energy. [1]
Answer: Energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy of an isolated system remains constant. [1]
(c) Determine the maximum height reached by the ball. [2]
Answer: At maximum height, all KE is converted to GPE. GPE = mgh = 56.25 J [1] h = 56.25 / (0.50 × 10) = 56.25 / 5 = 11.25 m [1]
Accept: 11.3 m.
(d) Explain what happens to the kinetic energy and gravitational potential energy of the ball as it rises. [2]
Answer: As the ball rises, its speed decreases, so kinetic energy decreases. [1] The height increases, so gravitational potential energy increases. Kinetic energy is converted into gravitational potential energy. The total mechanical energy remains constant (assuming no air resistance). [1]
Section B: Diagram and Data Interpretation (15 marks)
Question 6 (7 marks)
(a) Calculate the work done by the 60 N force. [1]
Answer: Work = Force × distance (along direction of force) = 60 × 5.0 = 300 J [1]
(b) Calculate the gain in gravitational potential energy of the block. [2]
Answer: GPE = mgh = 8.0 × 10 × 2.0 [1] = 160 J [1]
(c) Determine the amount of energy dissipated as heat due to friction. [2]
Answer: Energy dissipated = Work done by applied force - Gain in GPE [1] = 300 - 160 = 140 J [1]
(d) Calculate the frictional force acting on the block. [2]
Answer: Work done against friction = Frictional force × distance along plane [1] 140 = f × 5.0 → f = 140 / 5.0 = 28 N [1]
Question 7 (8 marks)
(a) Plot a graph of force (y-axis) against extension (x-axis) on the grid below. [3]
Answer:
- Correct axes labelled with units [0.5]
- Appropriate scales chosen [0.5]
- All 7 points plotted correctly [1.5]
- Best-fit straight line through first 5 points (0 to 4.0 N), curve beyond [0.5]
Points: (0,0), (2.5, 1.0), (5.0, 2.0), (7.5, 3.0), (10.0, 4.0), (13.0, 5.0), (16.5, 6.0)
(b) Using your graph, determine the spring constant for the linear region of the graph. [2]
Answer: Spring constant k = F / x (gradient of linear region) [1] Using points from linear region: k = 4.0 / 0.100 = 40 N/m [1]
Accept: values in range 38–42 N/m depending on graph reading. Accept N/cm (0.40 N/cm).
(c) State the extension at which the spring exceeds its limit of proportionality. Explain how you identified this from the graph. [2]
Answer: Extension: 10.0 cm (or between 10.0 cm and 13.0 cm) [1] Explanation: Beyond this point, the graph is no longer a straight line / the force is no longer directly proportional to extension. [1]
(d) Calculate the elastic potential energy stored in the spring when the extension is 7.5 cm. [1]
Answer: EPE = ½Fx = ½ × 3.0 × 0.075 = 0.1125 J [1]
Accept: 0.11 J or 0.113 J. Accept EPE = ½kx² = ½ × 40 × (0.075)² = 0.1125 J.
Section C: Free-Response Questions (10 marks)
Question 8 (4 marks)
A student claims that "if an object is moving, there must be a net force acting on it." Discuss whether this statement is correct, using Newton's laws of motion to support your answer. Include examples in your discussion.
Marking Scheme:
| Mark | Descriptor |
|---|---|
| 4 | Clear, well-structured discussion that correctly identifies the statement as incorrect. Uses Newton's First Law to explain that an object can move at constant velocity with no net force. Uses Newton's Second Law to explain when net force is required. Provides at least one clear, relevant example for each case. |
| 3 | Correctly identifies the statement as incorrect. Uses Newton's First and/or Second Law with reasonable explanation. Provides at least one relevant example. Minor omissions or lack of clarity. |
| 2 | Identifies the statement as incorrect but explanation is partial or contains minor errors. May only reference one law or provide weak examples. |
| 1 | Makes a relevant point about forces and motion but does not clearly address the statement or contains significant errors. |
| 0 | No attempt or completely incorrect physics. |
Model Answer:
The student's statement is incorrect. According to Newton's First Law of Motion, an object will remain at rest or continue moving at constant velocity in a straight line unless acted upon by a resultant (net) external force. This means that if an object is already moving and no net force acts on it, it will continue moving at constant velocity indefinitely.
For example, a spacecraft moving through deep space far from any gravitational fields will continue moving at constant velocity without any net force acting on it.
A net force is only required to change the velocity of an object (i.e., to accelerate it, decelerate it, or change its direction), as stated by Newton's Second Law (F = ma). For example, a car accelerating from rest requires a net forward force from the engine to overcome friction and air resistance.
Therefore, motion does not require a net force; only a change in motion does.
Question 9 (6 marks)
(a) Explain why the gravitational force from the Earth on the spacecraft decreases as the spacecraft moves away from Earth. [2]
Answer: According to Newton's Law of Gravitation, F = Gm₁m₂/r². [1] As the spacecraft moves away from Earth, the distance r between the centres of mass of the Earth and the spacecraft increases. Since F is inversely proportional to r², the gravitational force decreases. [1]
(b) The mass of the Earth is about 81 times the mass of the Moon. Explain why the point where the gravitational forces are equal is closer to the Moon than to the Earth. [2]
Answer: The gravitational force depends on both the mass of the attracting body and the distance (F ∝ M/r²). [1] Since the Earth's mass is 81 times the Moon's mass, for the forces to be equal, the spacecraft must be much closer to the Moon (so the smaller mass is compensated by a much smaller distance, since force increases as distance decreases). [1]
Accept: "the Earth's greater mass means its gravitational influence extends further, so the balance point is nearer the Moon" or equivalent reasoning.
(c) At this point, the spacecraft's engines are turned off. Describe and explain the subsequent motion of the spacecraft. [2]
Answer: At the point where the gravitational forces are equal, the net gravitational force on the spacecraft is zero. [1] However, the spacecraft still has velocity. According to Newton's First Law, it will continue moving in a straight line at constant velocity (assuming no other forces act). In reality, as it moves slightly away from this point, one gravitational force will become slightly larger, and the spacecraft will begin to accelerate toward the body with the stronger net attraction. [1]
Accept: discussion of how the equilibrium is unstable and the spacecraft will eventually be pulled toward one body.
END OF ANSWER KEY
Total: 60 marks