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Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 1

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TuitionGoWhere Secondary School (AI)

SA2 Practice Paper - Physics Secondary 3

Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper
Duration: 1 hour 15 minutes
Total Marks: 60
Version: 1 of 5

Name: _________________________
Class: _________________________
Date: _________________________

INSTRUCTIONS TO CANDIDATES

Write your name, class, and date in the spaces provided.

This paper consists of THREE sections: A, B, and C.

Answer ALL questions.

Write your answers in the spaces provided on the question paper.

All working must be shown clearly.

You may use a calculator where appropriate.

Take acceleration due to gravity, $g = 10 \text{ m/s}^2$ .

SECTION A: Multiple Choice Questions (10 marks)

Answer ALL questions. For each question, choose the correct answer and write the letter (A, B, C, or D) in the box provided.

[Total: 10 marks]

1. A student measures the thickness of a textbook using a ruler and obtains a value of $2.85 \text{ cm}$ . Which instrument would allow a more precise measurement of the same quantity?


A	A metre rule with mm divisions
B	A vernier caliper
C	A measuring tape
D	A stopwatch

Answer: [ ]

[1 mark]

2. The velocity-time graph for an object moving in a straight line is shown below.

<image_placeholder> id: Q2-fig1 type: graph linked_question: Q2 description: Velocity-time graph showing motion with three distinct phases labels: time (s) on x-axis, velocity (m/s) on y-axis; points at (0,0), (4,8), (8,8), (12,0) values: Phase 1: 0-4s, velocity increases linearly from 0 to 8 m/s; Phase 2: 4-8s, constant velocity 8 m/s; Phase 3: 8-12s, velocity decreases linearly from 8 to 0 m/s must_show: Three straight-line segments forming a trapezium shape; labelled axes with units; key time and velocity values marked </image_placeholder>

During which phase is the object experiencing a constant non-zero acceleration?


A	Phase 1 only
B	Phase 2 only
C	Phase 1 and Phase 3
D	Phase 2 and Phase 3

Answer: [ ]

[1 mark]

3. A book of weight $5 \text{ N}$ rests on a horizontal table. A student pushes vertically downward on the book with a force of $3 \text{ N}$ . What is the normal contact force exerted by the table on the book?


A	$2 \text{ N}$
B	$5 \text{ N}$
C	$8 \text{ N}$
D	$15 \text{ N}$

Answer: [ ]

[1 mark]

4. A parachutist falls from rest through the air. At first, her acceleration increases. Later, she reaches terminal velocity. Which statement correctly explains why terminal velocity is reached?


A	The gravitational force on her becomes zero.
B	The air resistance becomes equal to her weight.
C	Her mass decreases as she falls.
D	The air pressure above her equals the air pressure below her.

Answer: [ ]

[1 mark]

5. Two forces of $6 \text{ N}$ and $8 \text{ N}$ act on a point mass. Which value cannot be the magnitude of the resultant force?


A	$1 \text{ N}$
B	$8 \text{ N}$
C	$10 \text{ N}$
D	$14 \text{ N}$

Answer: [ ]

[1 mark]

6. A car of mass $800 \text{ kg}$ accelerates from rest to a speed of $20 \text{ m/s}$ in $8 \text{ s}$ . What is the average power developed by the engine, ignoring resistive forces?


A	$10 \text{ kW}$
B	$20 \text{ kW}$
C	$40 \text{ kW}$
D	$80 \text{ kW}$

Answer: [ ]

[1 mark]

7. A ball is thrown vertically upward with an initial velocity of $15 \text{ m/s}$ . What is the maximum height reached by the ball? (Assume $g = 10 \text{ m/s}^2$ and air resistance is negligible.)


A	$7.5 \text{ m}$
B	$11.25 \text{ m}$
C	$15 \text{ m}$
D	$22.5 \text{ m}$

Answer: [ ]

[1 mark]

8. The diagram shows a lever in equilibrium.

<image_placeholder> id: Q8-fig1 type: diagram linked_question: Q8 description: A uniform metre rule pivoted at its centre, with loads on either side labels: Pivot at 50 cm mark; Load A = 4 N at 30 cm mark; Load B at 70 cm mark; distance from pivot to Load A = 20 cm; distance from pivot to Load B = 20 cm values: Load A = 4 N, acting downward at 30 cm; rule is horizontal and in equilibrium; Load B is unknown must_show: Uniform rule marked with cm scale; pivot clearly indicated at centre; arrows for forces; distances labelled from pivot; rule horizontal </image_placeholder>

What is the magnitude of Load B for the rule to remain horizontal and in equilibrium?


A	$2 \text{ N}$
B	$4 \text{ N}$
C	$6 \text{ N}$
D	$8 \text{ N}$

Answer: [ ]

[1 mark]

9. A block slides down a smooth inclined plane from rest. Which quantity increases at a constant rate as the block descends?


A	Kinetic energy
B	Speed
C	Acceleration
D	Distance from the top

Answer: [ ]

[1 mark]

10. Newton's Third Law of Motion states that for every action there is an equal and opposite reaction. Which pair of forces represents an action-reaction pair?


A	The weight of a book and the normal force from a table on the book
B	The friction on a car tyre and the forward force on the car
C	The gravitational pull of Earth on the Moon and the gravitational pull of the Moon on Earth
D	The tension in a rope and the weight of a load hanging from it

Answer: [ ]

[1 mark]

SECTION B: Structured Questions (30 marks)

Answer ALL questions. Write your answers in the spaces provided.

[Total: 30 marks]

11. A student walks from her home to school. The journey consists of three parts:

Part 1: Walks $400 \text{ m}$ due East in $5 \text{ minutes}$
Part 2: Stops at a shop for $2 \text{ minutes}$
Part 3: Walks $300 \text{ m}$ due North in $4 \text{ minutes}$

(a) Calculate the total distance travelled by the student.

[1 mark]

(b) Calculate the total time taken for the journey in seconds.

[2 marks]

[3 marks]

(d) Explain why the average velocity for the entire journey is not the same as the average speed.

[2 marks]

[Total: 8 marks]

12. A train of mass $2.0 \times 10^5 \text{ kg}$ is travelling at a constant speed of $25 \text{ m/s}$ . The engine produces a driving force of $4.0 \times 10^4 \text{ N}$ .

(a) Explain how the train can be moving at constant speed when there is a forward driving force acting on it.

[2 marks]

(b) The train accelerates uniformly to a speed of $35 \text{ m/s}$ in $20 \text{ s}$ . Calculate:

(i) the acceleration of the train,

[2 marks]

(ii) the resultant force required for this acceleration,

[2 marks]

(iii) the total driving force the engine must now produce.

[2 marks]

[Total: 8 marks]

13. The diagram shows a pinewood derby car on a ramp inclined at $30°$ to the horizontal.

<image_placeholder> id: Q13-fig1 type: diagram linked_question: Q13 description: A toy car on a ramp inclined at 30 degrees, with forces labelled labels: Ramp at 30° to horizontal; car mass = 0.5 kg; weight W shown vertically downward; component of weight parallel to ramp = W sin θ; component of weight perpendicular to ramp = W cos θ; normal contact force R perpendicular to ramp surface values: Mass of car = 0.5 kg; angle θ = 30°; g = 10 m/s² must_show: Ramp with clear 30° angle indication; car as a block on ramp; weight arrow vertically down from centre; components shown as dashed arrows along and perpendicular to ramp; right-angle triangle indication for components; all forces labelled with symbols </image_placeholder>

(a) Calculate the weight of the car.

[1 mark]

(b) Determine the component of the weight acting parallel to the ramp.

[2 marks]

(c) The car is held stationary on the ramp by a string parallel to the ramp surface. Calculate the tension in the string.

[1 mark]

(d) The string is cut and the car slides down the smooth ramp from rest. Calculate the acceleration of the car down the ramp.

[2 marks]

(e) Explain what would happen to the acceleration if the ramp were rough instead of smooth.

[2 marks]

[Total: 8 marks]

14. A stone of mass $0.4 \text{ kg}$ is thrown vertically upward with an initial kinetic energy of $50 \text{ J}$ .

(a) Calculate the initial speed of the stone.

[3 marks]

(b) Determine the maximum height reached by the stone. (Assume $g = 10 \text{ m/s}^2$ and air resistance is negligible.)

[2 marks]

(c) On the axes below, sketch the variation of kinetic energy with height as the stone rises to its maximum height and falls back to the starting point.

<image_placeholder> id: Q14-fig1 type: graph linked_question: Q14(c) description: Blank axes for sketching kinetic energy against height labels: Height (m) on x-axis; Kinetic energy (J) on y-axis; from h = 0 to h_max values: No pre-drawn curve - student to sketch; initial KE = 50 J at h = 0; maximum height h_max to be determined from part (b) must_show: Labelled axes with units; clear indication that student should draw the curve; enough space to show symmetric curve if instructional </image_placeholder>

[2 marks]

(d) State the kinetic energy of the stone when it returns to its starting point, and explain your answer.

[2 marks]

[Total: 9 marks]

SECTION C: Data Analysis and Application (20 marks)

Answer ALL questions. Write your answers in the spaces provided.

[Total: 20 marks]

15. A student investigates the relationship between the stretching force and the extension of a spring. The results are recorded in the table below.

Stretching force $F$ (N)	0	2.0	4.0	6.0	8.0	10.0	12.0
Extension $e$ (cm)	0	1.6	3.2	4.8	6.4	8.0	10.4

(a) Plot a graph of stretching force against extension on the grid below.

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15(a) description: Grid for plotting force-extension graph labels: Extension e (cm) on x-axis from 0 to 12; Force F (N) on y-axis from 0 to 14; grid lines every 1 cm and 2 N values: Data points to be plotted: (0,0), (1.6, 2.0), (3.2, 4.0), (4.8, 6.0), (6.4, 8.0), (8.0, 10.0), (10.4, 12.0); scale: 1 cm represents 1 cm extension horizontally, 1 cm represents 2 N vertically must_show: Clearly labelled axes with units and scales; grid lines; title "Graph of stretching force against extension"; space for best-fit line </image_placeholder>

[3 marks]

(b) Determine the spring constant $k$ of the spring in $\text{N/cm}$ . Show your working clearly.

[3 marks]

(c) The student continues to extend the spring beyond $10.0 \text{ cm}$ . Describe and explain what would happen to the shape of the graph.

[3 marks]

[Total: 9 marks]

16. The diagram shows an apparatus used to determine the acceleration due to gravity using a simple pendulum.

<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Simple pendulum setup for measuring g labels: Retort stand with clamp; string of length L; bob at bottom; protractor for measuring angle of swing; stopwatch; ruler measuring L from pivot to centre of bob; displacement angle θ shown as small angle (< 10°) values: Typical setup with adjustable string length L; small angle displacement indicated must_show: Clear pendulum bob on string; pivot point at top; measurement of L from pivot to centre of bob with ruler; protractor indicating small angle; retort stand support; stopwatch nearby </image_placeholder>

(a) State two precautions the student should take to ensure accurate results.

[2 marks]

(b) The period $T$ of a simple pendulum is given by the formula:

$T = 2\pi\sqrt{\frac{L}{g}}$

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Rearrange this formula to make $g$ the subject.

[2 marks]

(c) For a pendulum of length $0.81 \text{ m}$ , the student measures the time for 20 complete oscillations to be $36.0 \text{ s}$ . Calculate:

(i) the period $T$ of the pendulum,

[1 mark]

(ii) the experimental value of $g$ .

[3 marks]

(d) Suggest one reason why the student's value of $g$ might differ from the accepted value of $10 \text{ m/s}^2$ , and state whether this error would cause the calculated value to be too high or too low.

[2 marks]

[Total: 10 marks]

17. A $60 \text{ kg}$ gymnast is practising on a trampoline.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: Trampoline with gymnast at two positions labels: Trampoline surface at equilibrium position; Position A: gymnast at maximum height above trampoline (4.0 m above surface); Position B: gymnast at lowest point, trampoline depressed by 0.5 m below equilibrium; height reference shown values: Mass of gymnast = 60 kg; maximum height above trampoline = 4.0 m; maximum depression of trampoline = 0.5 m below equilibrium; g = 10 m/s² must_show: Two distinct positions of gymnast; height measurement from trampoline surface at Position A; depression measurement at Position B; clear labels A and B; trampoline shape deformed at Position B </image_placeholder>

(a) Calculate the gravitational potential energy gained by the gymnast relative to the trampoline surface when she is at Position A ( $4.0 \text{ m}$ above the surface).

[2 marks]

(b) The gymnast falls from rest at Position A. Calculate her speed just as she reaches the trampoline surface. (Assume air resistance is negligible.)

[3 marks]

(c) At Position B, the gymnast is momentarily at rest with the trampoline depressed by $0.5 \text{ m}$ below its equilibrium position. Calculate the spring constant of the trampoline, assuming it behaves like a spring and energy losses are negligible.

[4 marks]

(d) Explain why, in practice, the gymnast would not reach the same maximum height of $4.0 \text{ m}$ on her rebound.

[2 marks]

[Total: 11 marks]

END OF PAPER

Total Marks: 60

Answers

TuitionGoWhere Secondary School (AI)

SA2 Practice Paper - Physics Secondary 3: Answer Key

Version: 1 of 5
Total Marks: 60
Duration: 1 hour 15 minutes

SECTION A: Multiple Choice Questions (10 marks)

1. B — A vernier caliper

Explanation: A vernier caliper can measure to a precision of $0.01 \text{ cm}$ ( $0.1 \text{ mm}$ ), compared to a ruler's typical precision of $0.1 \text{ cm}$ ( $1 \text{ mm}$ ). The question shows $2.85 \text{ cm}$ , suggesting the measurement was estimated to the nearest $0.01 \text{ cm}$ . A vernier caliper is the only instrument listed with this precision. A metre rule with mm divisions still requires estimation between marks. A measuring tape is less precise, and a stopwatch measures time, not length.

Common mistake: Choosing A because "mm divisions" sounds precise, but reading between mm marks still only gives about $0.5 \text{ mm}$ precision with estimation.

[1 mark]

2. A — Phase 1 only

Explanation: Acceleration is the gradient of a velocity-time graph. In Phase 1 (0–4 s), the graph is a straight line with constant positive gradient ( $\frac{8-0}{4-0} = 2 \text{ m/s}^2$ ), so acceleration is constant and non-zero. In Phase 2 (4–8 s), the gradient is zero (horizontal line, constant velocity), so acceleration is zero. In Phase 3 (8–12 s), the gradient is constant but negative ( $\frac{0-8}{12-8} = -2 \text{ m/s}^2$ ), so this is also constant non-zero acceleration — deceleration.

Re-evaluation: The question asks for "constant non-zero acceleration." Phase 1 has $a = +2 \text{ m/s}^2$ and Phase 3 has $a = -2 \text{ m/s}^2$ . Both are constant and non-zero. However, the answer should be C — Phase 1 and Phase 3 based on this analysis.

Correction to answer key: The correct answer is C — Phase 1 and Phase 3.

Explanation: Both Phase 1 and Phase 3 show straight-line segments on the v-t graph, indicating constant acceleration (positive in Phase 1, negative/deceleration in Phase 3). Phase 2 shows zero acceleration. The original answer key contained an error.

[1 mark]

3. C — $8 \text{ N}$

Explanation: The normal contact force balances the total downward force on the book. Downward forces: weight of book $= 5 \text{ N}$ (gravitational pull of Earth), plus pushing force $= 3 \text{ N}$ . Total downward force $= 5 + 3 = 8 \text{ N}$ . By Newton's Third Law / equilibrium, the table pushes upward with equal force: $R = 8 \text{ N}$ .

Common mistake: Choosing B by forgetting the additional $3 \text{ N}$ push, or A by incorrectly subtracting.

[1 mark]

4. B — The air resistance becomes equal to her weight

Explanation: Terminal velocity occurs when forces are balanced. Initially, with low speed, air resistance $<$ weight, so net force is downward and she accelerates. As speed increases, air resistance increases (often proportional to $v$ or $v^2$ ). When air resistance exactly equals her weight, the net force becomes zero, so by Newton's First Law, she continues at constant velocity — terminal velocity. Her mass does not change significantly (eliminating C), and gravity doesn't become zero (eliminating A).

[1 mark]

5. A — $1 \text{ N}$

Explanation: For two forces $F_1$ and $F_2$ , the resultant $F_R$ satisfies: $|F_1 - F_2| \leq F_R \leq F_1 + F_2$ . Here: $|6-8| = 2 \text{ N} \leq F_R \leq 14 \text{ N}$ . So the resultant can be any value from 2 N to 14 N. The value $1 \text{ N}$ is outside this range and therefore impossible.

Verification: $14 \text{ N}$ occurs when forces act in same direction (6+8). $2 \text{ N}$ occurs when opposite (8-6). $8 \text{ N}$ occurs at some intermediate angle. $10 \text{ N}$ would occur if perpendicular ( $\sqrt{6^2+8^2} = 10$ ).

[1 mark]

6. B — $20 \text{ kW}$

Explanation:

Method 1 (using work and time): Final $KE = \frac{1}{2}mv^2 = \frac{1}{2}(800)(20)^2 = 160000 \text{ J}$ . Average power $= \frac{\text{work done}}{\text{time}} = \frac{160000}{8} = 20000 \text{ W} = 20 \text{ kW}$ .
Method 2 (using $P = Fv$ average): Average velocity $= \frac{0+20}{2} = 10 \text{ m/s}$ . Acceleration $= \frac{20}{8} = 2.5 \text{ m/s}^2$ . Force $ma = 800 \times 2.5 = 2000 \text{ N}$ . Average power $= 2000 \times 10 = 20000 \text{ W} = 20 \text{ kW}$ .

Common mistake: Using final velocity instead of average gives $2000 \times 20 = 40 \text{ kW}$ (answer C), which is incorrect for average power. However, since the question asks for "average power," Method 1 is most reliable.

[1 mark]

7. B — $11.25 \text{ m}$

Explanation: Using $v^2 = u^2 + 2as$ with $v = 0$ (at top), $u = 15 \text{ m/s}$ , $a = -g = -10 \text{ m/s}^2$ :

$0 = 15^2 + 2(-10)s$ $0 = 225 - 20s$ $s = \frac{225}{20} = 11.25 \text{ m}$

Alternative: Using conservation of energy: $\frac{1}{2}mu^2 = mgh$ , so $h = \frac{u^2}{2g} = \frac{225}{20} = 11.25 \text{ m}$ .

Common mistake: Using $s = ut$ with $t = 1.5 \text{ s}$ (time to top) gives $s = 15 \times 1.5 = 22.5$ (answer D), but this is wrong because velocity is not constant.

[1 mark]

8. B — $4 \text{ N}$

Explanation: For rotational equilibrium, sum of clockwise moments = sum of anticlockwise moments about the pivot.

Anticlockwise moment from Load A: $4 \text{ N} \times 20 \text{ cm} = 80 \text{ N cm}$
Clockwise moment from Load B: $F_B \times 20 \text{ cm}$

Setting equal: $F_B \times 20 = 80$ , so $F_B = 4 \text{ N}$ .

The rule is uniform and pivoted at centre, so its weight creates no moment. The situation is symmetric.

[1 mark]

9. C — Acceleration

Explanation: On a smooth inclined plane, the only force along the ramp is the component of weight $mg\sin\theta$ , which is constant. So $a = g\sin\theta = \text{constant}$ .

A (kinetic energy): Increases, but rate increases since $KE = \frac{1}{2}mv^2$ and $v$ increases non-linearly with time ( $v = at$ )
B (speed): Increases linearly with time ( $v = u + at = 0 + at$ ), so rate of increase of speed is constant, not speed itself
C (acceleration): Constant — correct
D (distance from top): Increases as $s = \frac{1}{2}at^2$ , so rate of increase of distance itself increases with time

[1 mark]

10. C — The gravitational pull of Earth on the Moon and the gravitational pull of the Moon on Earth

Explanation: Newton's Third Law requires: (1) same type of force, (2) equal magnitude, (3) opposite directions, (4) acting on different bodies.

A: Weight and normal force are different types (gravitational vs electromagnetic), and both act on the same body (the book). These balance but are not Third Law pair.
B: Friction and forward force same type but complicated system — not simple action-reaction.
C: Both gravitational, equal magnitude, opposite, Earth ↔ Moon. Correct.
D: Tension and weight are different types, both act on same body (the load).

[1 mark]

SECTION B: Structured Questions (30 marks)

11(a) $700 \text{ m}$

Working: Total distance = sum of all path lengths = $400 + 0 + 300 = 700 \text{ m}$

Note: Distance is scalar; the stop doesn't contribute to distance but does to time. Don't forget the $300 \text{ m}$ North leg.

[1 mark]

11(b) $660 \text{ s}$ (or $11 \text{ minutes}$ )

Working:

Time walking: $5 + 4 = 9$ minutes $= 9 \times 60 = 540 \text{ s}$
Time stopped: $2$ minutes $= 120 \text{ s}$
Total time: $540 + 120 = 660 \text{ s}$

Or: $5 + 2 + 4 = 11$ minutes $= 11 \times 60 = 660 \text{ s}$

[2 marks]

11(c) $500 \text{ m}$ (or $5.0 \times 10^2 \text{ m}$ )

Working:

Method — Pythagoras theorem: The displacement forms the hypotenuse of a right triangle with sides $400 \text{ m}$ (East) and $300 \text{ m}$ (North).

$|\vec{s}| = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \text{ m}$

Scale diagram method: Choose scale (e.g., $1 \text{ cm}: 100 \text{ m}$ ), draw $4 \text{ cm}$ East then $3 \text{ cm}$ North, measure resultant as $5 \text{ cm}$ → $500 \text{ m}$ .

Direction: The displacement is at angle $\theta = \tan^{-1}(\frac{300}{400}) = 36.9°$ North of East (or bearing $053°$ ), but the question asks for magnitude only.

Marking breakdown:

[1] Correct method (Pythagoras or scale diagram with stated scale)
[1] Correct substitution and arithmetic
[1] Final answer with unit

[3 marks]

11(d)

Key points:

Average speed = $\frac{\text{total distance}}{\text{total time}} = \frac{700}{660} = 1.06 \text{ m/s}$ (scalar)
Average velocity = $\frac{\text{displacement}}{\text{total time}} = \frac{500 \text{ m at } 36.9° \text{ N of E}}{660 \text{ s}} = 0.758 \text{ m/s}$ in that direction (vector)

Explanation: Average velocity depends on displacement (the straight-line change in position from start to finish), while average speed depends on total path length. Since the student does not travel in a straight line ( $700 \text{ m} \neq 500 \text{ m}$ ), and velocity includes direction while speed does not, they are different. Even their magnitudes differ: $1.06 \text{ m/s} \neq 0.758 \text{ m/s}$ .

Marking breakdown:

[1] Identifies that displacement $\neq$ distance (or velocity is vector, speed is scalar)
[1] Explains consequence: different numerical values result

[2 marks]

12(a)

Explanation: For constant velocity, net force must be zero (Newton's First Law). The forward driving force of $4.0 \times 10^4 \text{ N}$ is exactly balanced by resistive forces (air resistance, rolling friction, etc.) totalling $4.0 \times 10^4 \text{ N}$ in the opposite direction. These cancel out, giving zero resultant force and thus zero acceleration — constant speed.

[2 marks]

12(b)(i) $0.50 \text{ m/s}^2$ (or $0.5 \text{ m/s}^2$ )

Working: $a = \frac{v-u}{t} = \frac{35-25}{20} = \frac{10}{20} = 0.50 \text{ m/s}^2$

[2 marks]

12(b)(ii) $1.0 \times 10^5 \text{ N}$

Working: Using Newton's Second Law: $F = ma$

$F = (2.0 \times 10^5)(0.50) = 1.0 \times 10^5 \text{ N}$

[2 marks]

12(b)(iii) $1.4 \times 10^5 \text{ N}$ (or $1.40 \times 10^5 \text{ N}$ )

Working:

Resultant force needed: $1.0 \times 10^5 \text{ N}$ (forward)
Resistive force (still present, assumed constant): $4.0 \times 10^4 \text{ N}$ (backward)
Total driving force required: $F_{\text{driving}} = F_{\text{resultant}} + F_{\text{resistive}} = 1.0 \times 10^5 + 4.0 \times 10^4 = 1.4 \times 10^5 \text{ N}$

Explanation: The engine must overcome both the resistive forces (which haven't changed) AND provide the extra force for acceleration. So it must produce more than just the resultant force.

Common mistake: Forgetting to add the resistive force, giving $1.0 \times 10^5 \text{ N}$ .

[2 marks]

13(a) $5 \text{ N}$

Working: $W = mg = 0.5 \times 10 = 5 \text{ N}$

[1 mark]

13(b) $2.5 \text{ N}$ (or accept $2.50 \text{ N}$ )

Working: Component parallel to ramp:

$W_{\parallel} = W\sin\theta = mg\sin\theta = 5 \times \sin(30°) = 5 \times 0.5 = 2.5 \text{ N}$

Or from first principles: $W_{\parallel} = 0.5 \times 10 \times \sin(30°) = 2.5 \text{ N}$

[2 marks]

13(c) $2.5 \text{ N}$

Working: For equilibrium, tension balances the component of weight down the ramp:

$T = W_{\parallel} = 2.5 \text{ N}$

Since the car is stationary, by Newton's First Law, forces up the ramp = forces down the ramp.

[1 mark]

13(d) $5.0 \text{ m/s}^2$ (or $5 \text{ m/s}^2$ )

Working:

Smooth ramp: no friction
Resultant force down ramp $= W_{\parallel} = 2.5 \text{ N}$ (only force along ramp)
Using $F = ma$ :

$a = \frac{F}{m} = \frac{2.5}{0.5} = 5.0 \text{ m/s}^2$

Or: $a = g\sin\theta = 10 \times \sin(30°) = 5.0 \text{ m/s}^2$

[2 marks]

13(e)

Explanation: If the ramp were rough, friction would oppose the motion of the car sliding down. This friction force would act up the ramp, reducing the resultant force down the ramp.

From $F_{\text{net}} = W_{\parallel} - f$ (where $f$ is friction), the net force is less than $2.5 \text{ N}$ .

Since $a = \frac{F_{\text{net}}}{m}$ , a smaller net force means smaller acceleration.

The car would still accelerate down (if $W_{\parallel} > f$ ), but more slowly than on the smooth ramp. If friction were large enough ( $f \geq W_{\parallel}$ ), the car might not move at all or would move at constant velocity.

Marking breakdown:

[1] Identifies that friction opposes motion / acts up the ramp
[1] Explains consequence: reduced net force → reduced acceleration

[2 marks]

14(a) $15.8 \text{ m/s}$ (or $15.81 \text{ m/s}$ , or $\sqrt{250} \text{ m/s} \approx 15.8 \text{ m/s}$ )

Working: Using $KE = \frac{1}{2}mv^2$ :

$50 = \frac{1}{2}(0.4)v^2$ $50 = 0.2v^2$ $v^2 = \frac{50}{0.2} = 250$ $v = \sqrt{250} = 15.811... \approx 15.8 \text{ m/s}$

Marking breakdown:

[1] Correct formula stated
[1] Correct substitution with mass $0.4 \text{ kg}$
[1] Final answer with unit (accept $5\sqrt{10}$ or equivalent)

[3 marks]

14(b) $12.5 \text{ m}$

Working: Using conservation of energy: initial $KE \rightarrow$ final $GPE$ at maximum height.

$mgh_{\max} = \frac{1}{2}mu^2 = 50 \text{ J}$

Or using $v^2 = u^2 + 2as$ with $v = 0$ at top:

$0 = 250 + 2(-10)h$ $h = \frac{250}{20} = 12.5 \text{ m}$

Alternative: $h = \frac{u^2}{2g} = \frac{250}{20} = 12.5 \text{ m}$

Or from $GPE = mgh$ : $50 = 0.4 \times 10 \times h$ , so $h = \frac{50}{4} = 12.5 \text{ m}$ .

[2 marks]

14(c)

Expected sketch description:

Feature	Description
Shape	Straight line with negative gradient from $(0, 50)$ to $(12.5, 0)$ rising phase; then straight line with positive gradient from $(12.5, 0)$ back to $(25, 50)$ if height measured from starting point, OR the curve should show KE decreasing as height increases, reaching zero at $h = 12.5$ m, then increasing again as the stone falls.

Correct interpretation for "rising to max height and falling back to starting point":

The graph shows KE versus height:

At $h = 0$ : $KE = 50 \text{ J}$ (initial)
As $h$ increases: $KE$ decreases linearly (since $KE + GPE = \text{constant}$ , so $KE = 50 - mgh = 50 - 4h$ )
At $h = 12.5 \text{ m}$ : $KE = 0$ (at maximum height, momentarily at rest)
As $h$ decreases back to 0: $KE$ increases linearly back to $50 \text{ J}$

So the graph is a V-shape or two straight lines forming an inverted V (Λ shape), with maximum at $h=0$ and $h=25$ ...

Wait — re-reading: The x-axis is "height" from 0 to $h_{\max} = 12.5$ m. As the stone rises then falls, it passes through each height twice (except maximum). So for a given height $h$ , the KE is the same going up and coming down (energy conservation, no air resistance).

So the graph of $KE$ vs $h$ is: a single straight line from $(0, 50)$ to $(12.5, 0)$ with negative gradient. Each height (except $h_{\max}$ ) corresponds to one KE value. The stone passes through that KE at two different times, but the graph $KE(h)$ is single-valued — a decreasing straight line.

Acceptable answer: Straight line from $(0, 50)$ to $(12.5, 0)$ , with $KE$ in $\text{J}$ and $h$ in $\text{m}$ .

Marking breakdown:

[1] Correct shape: straight line with negative gradient
[1] Correct intercepts: $(0, 50)$ and approximately $(12.5, 0)$ — accept 12–13 m range

[2 marks]

14(d) $50 \text{ J}$

Explanation: When the stone returns to its starting point, it is at the same height as its initial position. Since air resistance is negligible, mechanical energy is conserved. The total energy (KE + GPE) remains constant at $50 \text{ J}$ throughout. At the starting height, $GPE = 0$ (by reference), so $KE = 50 \text{ J}$ — the same as initially.

Alternative sound explanation: By symmetry of motion under constant acceleration (time up = time down, speed at any height is same going up and down), the speed at return is equal to initial speed ( $15.8 \text{ m/s}$ ), so $KE = \frac{1}{2}m(15.8)^2 = 50 \text{ J}$ .

Marking breakdown:

[1] $50 \text{ J}$
[1] Explanation referencing conservation of energy or symmetry of motion

[2 marks]

SECTION C: Data Analysis and Application (20 marks)

15(a)

Graph marking criteria:

Criterion	Requirement
Axes	Correctly labelled: x-axis "Extension e (cm)", y-axis "Stretching force F (N)"
Scales	Uniform, sensible, covering all data; at least half the grid used
Points	All 7 points plotted correctly (within half a small square)
Line	Best-fit straight line through origin for first 6 points; last point identified as anomaly or line continued correctly

Expected line: Straight line through origin with positive gradient, passing through or very near: $(0,0), (1.6, 2), (3.2, 4), (4.8, 6), (6.4, 8), (8.0, 10)$ . The point $(10.4, 12)$ lies above the line (Hooke's Law limit exceeded).

Marking breakdown:

[1] Correct axes with labels and units
[1] Correct scales and all points plotted accurately
[1] Appropriate best-fit line (straight for Hooke's law region, noting deviation)

[3 marks]

15(b) $1.25 \text{ N/cm}$ (accept range $1.20–1.30 \text{ N/cm}$ or $125 \text{ N/m}$ )

Working: Using Hooke's Law: $F = ke$ , so $k = \frac{F}{e} = \text{gradient}$

From first 6 points (linear region): $k = \frac{\Delta F}{\Delta e} = \frac{10.0 - 0}{8.0 - 0} = \frac{10.0}{8.0} = 1.25 \text{ N/cm}$

Or using any point in linear region: $k = \frac{2.0}{1.6} = 1.25 \text{ N/cm}$

Conversion to SI: $1.25 \text{ N/cm} = 125 \text{ N/m}$

Note on anomaly: The last point $(10.4, 12)$ gives $\frac{12}{10.4} = 1.15 \text{ N/cm}$ , which is lower — this is beyond the elastic limit.

Marking breakdown:

[1] Correct method (gradient calculation or using $F = ke$ )
[1] Correct data selection (using linear region, not including anomalous point)
[1] Correct arithmetic and unit

[3 marks]

15(c)

Description: The graph would cease to be a straight line and would curve (bend toward the extension axis, i.e., force increases more slowly than before, or the line becomes less steep).

Explanation: Hooke's Law ( $F \propto e$ ) is only valid within the elastic limit. Beyond this limit, the spring undergoes plastic deformation — the material's structure begins to change permanently. The spring becomes permanently stretched and does not return to its original length. The force required to produce additional extension becomes proportionally less than predicted by Hooke's Law, so the $F$ - $e$ relationship becomes non-linear. If stretched too far, the spring may break.

Key terms: elastic limit, plastic deformation, permanent deformation, non-linear, no longer proportional.

Marking breakdown:

[1] Identifies shape change: curve / no longer straight line / bends
[1] Identifies cause: exceeds elastic limit / enters plastic region
[1] Explains consequence: permanent extension / spring does not return to original length / proportionality lost

[3 marks]

16(a) Any two valid precautions:

Precaution	Explanation
Use a small angle of swing (less than $10°$ )	Ensures motion approximates simple harmonic motion; period formula $T = 2\pi\sqrt{L/g}$ is derived for small angles
Time multiple oscillations (e.g., 20 or 50) and divide	Reduces random error in period measurement; reaction time error spread over many swings
Measure time from when the bob passes through the equilibrium position (centre)	Easier to judge than extreme positions; moving fastest so less uncertainty
Ensure pendulum swings in one plane only (no conical motion / no wobbling)	Ensures true simple pendulum motion
Measure length $L$ from pivot to centre of mass of bob	Correct definition of $L$ in formula; measuring to top or bottom of bob introduces systematic error
Count oscillations carefully (complete to-and-fro swings)	Avoids error of counting half-swings or miscounting

Marking breakdown:

[1] Any valid precaution with brief explanation
[1] Second valid precaution with brief explanation

[2 marks]

16(b) $g = \frac{4\pi^2 L}{T^2}$

Working: $T = 2\pi\sqrt{\frac{L}{g}}$

Square both sides: $T^2 = 4\pi^2 \times \frac{L}{g}$

Multiply by $g$ : $gT^2 = 4\pi^2 L$

Divide by $T^2$ : $g = \frac{4\pi^2 L}{T^2}$

Marking breakdown:

[1] Correct squaring step
[1] Correct isolation of $g$

[2 marks]

16(c)(i) $1.80 \text{ s}$ (or $1.8 \text{ s}$ )

Working: $T = \frac{\text{time for } n \text{ oscillations}}{n} = \frac{36.0}{20} = 1.80 \text{ s}$

[1 mark]

16(c)(ii) $9.87 \text{ m/s}^2$ (or approximately $9.9 \text{ m/s}^2$ ; accept $9.60–10.2 \text{ m/s}^2$ )

Working: $g = \frac{4\pi^2 L}{T^2} = \frac{4 \times \pi^2 \times 0.81}{(1.80)^2}$

$g = \frac{4 \times 9.87 \times 0.81}{3.24} = \frac{31.98}{3.24} = 9.87 \text{ m/s}^2$

Or with $\pi^2 \approx 9.87$ : $g = \frac{4 \times 9.87 \times 0.81}{3.24} = \frac{31.97}{3.24} = 9.87 \text{ m/s}^2$

Marking breakdown:

[1] Correct substitution into formula with $L = 0.81$ and $T = 1.80$
[1] Correct arithmetic (accept slight rounding differences)
[1] Unit $\text{m/s}^2$ (or $\text{ms}^{-2}$ )

[3 marks]

16(d)

Example answer: The student's measurement of length $L$ might be incorrect — for example, measuring to the bottom of the bob instead of its centre of mass would give $L$ too large, leading to $g$ calculated as too high. Or measuring to the top of the bob would give $L$ too small, giving $g$ too low.

Other valid answers:

Source of error	Effect on $g$
Counted fewer oscillations than actual (e.g., counted 19 as 20)	Time per oscillation appears smaller, so $T$ smaller, $g$ too high
String stretched during experiment, true $L$ longer than measured	$L$ underestimated, $g$ too low
Large angle of swing (greater than $10°$ )	Actual period longer than small-angle formula predicts, $T$ measured is larger than "true" simple pendulum value, $g$ too low
Reaction time — started/stopped timer early or late	Random error; effect depends on direction; on average may balance out
Air resistance / swinging in an arc with conical motion	Effective period altered; typically $g$ too low

Marking breakdown:

[1] Valid source of error identified
[1] Correct direction of effect on $g$ (too high / too low) with explanation

[2 marks]

17(a) $2400 \text{ J}$ (or $2.4 \times 10^3 \text{ J}$ )

Working: $GPE = mgh = 60 \times 10 \times 4.0 = 2400 \text{ J}$

[2 marks]

17(b) $8.94 \text{ m/s}$ (or $8.9 \text{ m/s}$ , or $\sqrt{80} \approx 8.94 \text{ m/s}$ )

Working: Using conservation of energy or kinematics:

Method 1 — Energy: $GPE_{\text{lost}} = KE_{\text{gained}}$ $mgh = \frac{1}{2}mv^2$ $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 4.0} = \sqrt{80} = 8.944... \approx 8.94 \text{ m/s}$

Method 2 — Kinematics: $v^2 = u^2 + 2as = 0 + 2(10)(4.0) = 80$ $v = \sqrt{80} = 8.94 \text{ m/s}$

Marking breakdown:

[1] Correct formula ( $v = \sqrt{2gh}$ or $v^2 = u^2 + 2as$ )
[1] Correct substitution
[1] Final answer with unit

[3 marks]

17(c) $19200 \text{ N/m}$ (or $1.92 \times 10^4 \text{ N/m}$ , or $1920 \text{ N/cm}$ if using cm; but standard unit is $\text{N/m}$ )

Working: Using conservation of energy: total $GPE$ lost from maximum height to lowest point equals elastic potential energy stored in trampoline at maximum depression.

Total height fallen from Position A to Position B: $h_{\text{total}} = 4.0 + 0.5 = 4.5 \text{ m}$

(She falls $4.0 \text{ m}$ to the surface, then another $0.5 \text{ m}$ as the trampoline depresses.)

$GPE_{\text{lost}} = mg \times h_{\text{total}} = 60 \times 10 \times 4.5 = 2700 \text{ J}$

At Position B, momentarily at rest, so $KE = 0$ . This $GPE$ lost becomes elastic potential energy:

$EPE = \frac{1}{2}kx^2$

where $x = 0.5 \text{ m}$ (depression of trampoline).

Setting equal: $\frac{1}{2}k(0.5)^2 = 2700$ $\frac{1}{2}k \times 0.25 = 2700$ $0.125k = 2700$ $k = \frac{2700}{0.125} = \frac{2700 \times 8}{1} = 21600 \text{ N/m}$

Wait — let me recheck: The gymnast at Position A has $GPE = 2400 \text{ J}$ relative to trampoline surface. At Position B, she is $0.5 \text{ m}$ BELOW the surface. The reference point for zero GPE needs care.

Better approach — using displacement from equilibrium:

Actually, using the equilibrium position of the unstretched trampoline as reference:

At Position A: height = $4.0 + 0.5 = 4.5 \text{ m}$ above lowest point, or $4.0 \text{ m}$ above surface. If we take the equilibrium position as $h=0$ , then at Position A, $h = 4.0$ m (but this is above equilibrium), and at Position B, $h = -0.5 \text{ m}$ (below equilibrium).

The gravitational potential energy change from A to B: $mg \times \Delta h = 60 \times 10 \times (4.0 - (-0.5)) = 60 \times 10 \times 4.5$ — no wait, she's falling, so potential energy decreases.

Actually, simplest: take lowest point (Position B) as the reference where we measure energies.

At Position A (relative to Position B): total height $= 4.0 + 0.5 = 4.5 \text{ m}$

$GPE_{\text{at A (rel B)}} = 60 \times 10 \times 4.5 = 2700 \text{ J}$

At Position B: $GPE = 0$ (by our choice), $KE = 0$ (momentarily at rest), $EPE = \frac{1}{2}k(0.5)^2$

By conservation: $\frac{1}{2}k(0.5)^2 = 2700$ $k = \frac{5400}{0.25} = 21600 \text{ N/m}$

But wait — this assumes the trampoline has zero spring energy at its equilibrium position. When the gymnast first contacts the trampoline at the surface level, it starts to stretch. At the surface level (undeformed), $EPE = 0$ . So the compression from surface to $0.5 \text{ m}$ below is $x = 0.5 \text{ m}$ .

The energy conservation from surface contact point to lowest point:

At contact: $KE = \frac{1}{2}mv^2 = 60 \times 10 \times 4.0 = 2400 \text{ J}$ (from part b), and this is all from GPE lost from height 4.0 m.
From contact to lowest: loses another $GPE = 60 \times 10 \times 0.5 = 300 \text{ J}$
Total energy stored as EPE at lowest point: $2400 + 300 = 2700 \text{ J} = \frac{1}{2}k(0.5)^2$

So $k = 21600 \text{ N/m}$ appears correct.

But let me re-read the question — it says "energy losses are negligible," implying we use conservation throughout.

However, looking at this again, the answer seems high. Let me check if the spring constant should be calculated differently.

Actually, I realize I should verify: is $x = 0.5 \text{ m}$ the maximum depression? The problem states "trampoline depressed by 0.5 m below equilibrium" at Position B.

Yes, so $x = 0.5 \text{ m}$ is the spring compression from natural length.

The calculation stands: $k = 21600 \text{ N/m}$ or $2.16 \times 10^4 \text{ N/m}$ .

But wait — I want to double-check this is reasonable. Trampolines do have high spring constants. Actually, using $F = kx$ at maximum depression: the force at lowest point would be $kx = 21600 \times 0.5 = 10800 \text{ N}$ , and the acceleration would be $\frac{10800 - 600}{60} = 170 \text{ m/s}^2$ upward, which is $17g$ . This seems high but briefly sustainable for a trampoline's purpose.

However, I want to re-verify my energy approach. When she falls from A to B:

Initial energy at A (taking B as reference): $GPE = mg(4.5) = 2700 \text{ J}$ , $KE = 0$ , $EPE = 0$ (trampoline not engaged)
Final energy at B: $GPE = 0$ , $KE = 0$ , $EPE = \frac{1}{2}k(0.5)^2$

This is correct. So $k = 21600 \text{ N/m}$ .

But actually, I realize there may be an error: at Position A, the trampoline is not stretched, so its spring potential is zero. But the gymnast is not in contact with the trampoline. The spring only starts to compress when she hits the trampoline surface (at $0.5$ m above Position B).

Actually, re-reading: the trampoline surface is "at equilibrium position" normally, and at Position B it is "depressed by 0.5 m below equilibrium". So the equilibrium of the empty trampoline is at some level, and when the gymnast stands or falls on it, it depresses.

For energy conservation from A to B, we need to account for the fact that the spring only starts to compress during the last 0.5 m of fall. The energy before contacting the trampoline (at the surface, 4.0 m below A) is all kinetic: $KE = 2400 \text{ J}$ (from part b, $v = \sqrt{80}$ ).

Then from surface to B (0.5 m further down):

She loses more $GPE = mg \times 0.5 = 300 \text{ J}$
This 300 J + the 2400 J KE gets converted to EPE at B

So total EPE = 2700 J, and $k = 21600 \text{ N/m}$ .

I confirm this answer.

Alternative acceptable answer: Some might calculate using $F = kx$ where $F = mg$ at equilibrium, but that's not the maximum force. At maximum depression, the acceleration is upward so net force is upward, meaning spring force exceeds weight. So $kx > mg$ at B, and using $kx = mg$ would be incorrect for this dynamic situation.

Final answer: $k = 21600 \text{ N/m}$ or $2.16 \times 10^4 \text{ N/m}$

Marking breakdown:

[1] Correct total energy at Position A or correct approach using conservation of energy throughout
[1] Correct total height or correct energy at contact point with trampoline
[1] Correct equation $\frac{1}{2}kx^2 =$ total energy available
[1] Correct final answer with unit

Note: If a student uses $mg = kx$ (static equilibrium approach), this is incorrect for this dynamic problem and should receive [0] for method, though for a 60 kg person stationary on a trampoline at equilibrium, it would give $k = \frac{600}{0.5} = 1200 \text{ N/m}$ , which is wrong for this context.

[4 marks]

17(d)

Explanation: In practice, some energy is lost to non-conservative forces:

Air resistance during the fall and rebound dissipates mechanical energy as thermal energy in the air
Internal friction in the trampoline material (stretching fibers, air resistance in the canvas, heating of springs) converts kinetic/spring energy to thermal energy in the trampoline
Sound energy is produced on impact

Therefore, on rebound, the mechanical energy available to convert back to kinetic and then gravitational potential energy is less than the initial 2400 J. She cannot regain the original height of 4.0 m. After several bounces, she would need to add energy by pushing with her legs to continue.

Key terms: air resistance, friction, thermal energy, sound energy, dissipation, non-conservative forces.

Marking breakdown:

[1] Identifies energy loss mechanism(s)
[1] Explains consequence: less mechanical energy available for rebound → lower maximum height

[2 marks]

END OF ANSWER KEY

Total Marks: 60