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Secondary 2 Science Semestral Assessment 2 (End of Year) Paper 5

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Questions

TuitionGoWhere Practice Paper - Science Secondary 2

TuitionGoWhere Secondary School (AI)

Subject: Science
Level: Secondary 2 (G3)
Paper: SA2 Version 5
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer all questions.
Write your answers in the spaces provided on the question paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
For calculations, show all working clearly.
Use $g = 10 \text{ N/kg}$ or $10 \text{ m/s}^2$ unless otherwise stated.
The total mark for this paper is 60.

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.

1

A ball of mass 0.5 kg is dropped from a height of 20 m. Ignoring air resistance, what is the kinetic energy of the ball just before it hits the ground? [1]

☐ A. 50 J
☐ B. 100 J
☐ C. 150 J
☐ D. 200 J

2

Which of the following energy conversions takes place in a hydroelectric power station? [1]

☐ A. Electrical energy → Kinetic energy → Potential energy
☐ B. Potential energy → Kinetic energy → Electrical energy
☐ C. Chemical energy → Heat energy → Electrical energy
☐ D. Kinetic energy → Potential energy → Electrical energy

3

A force of 25 N is applied to push a box horizontally across a floor for a distance of 4 m. The work done by the force is: [1]

☐ A. 6.25 J
☐ B. 29 J
☐ C. 100 J
☐ D. 400 J

4

A student lifts a 3 kg book from the floor to a shelf 1.5 m high. The gravitational potential energy gained by the book is: [1]

☐ A. 4.5 J
☐ B. 22.5 J
☐ C. 45 J
☐ D. 67.5 J

5

In which of the following situations is NO work done? [1]

☐ A. A girl pushes a trolley that moves 5 m
☐ B. A boy holds a heavy bag stationary above his head
☐ C. A crane lifts a load vertically upwards
☐ D. A car accelerates along a straight road

6

A pendulum bob is released from rest at position X. At which position does the bob have maximum kinetic energy? [1]

Diagram for Q6 (SEC2 Science)

Generated diagram for Q6.

☐ A. Position X only
☐ B. Position Y only
☐ C. Position Z only
☐ D. Positions X and Z

7

An electric kettle rated 2000 W is used for 10 minutes. The electrical energy consumed is: [1]

☐ A. 20 000 J
☐ B. 1 200 000 J
☐ C. 2 000 000 J
☐ D. 12 000 000 J

8

A car of mass 1000 kg accelerates uniformly from rest to 20 m/s in 10 s. The average power developed by the engine is: [1]

☐ A. 2000 W
☐ B. 20 000 W
☐ C. 40 000 W
☐ D. 200 000 W

9

Which statement about the principle of conservation of energy is correct? [1]

☐ A. Energy can be created but not destroyed
☐ B. Energy can be destroyed but not created
☐ C. The total energy of an isolated system remains constant
☐ D. Kinetic energy is always conserved in any collision

10

A roller coaster car starts from rest at the top of a hill 50 m high. Assuming no energy losses, what is its speed at the bottom of the hill? [1]

☐ A. 10 m/s
☐ B. 22 m/s
☐ C. 32 m/s
☐ D. 50 m/s

Section B: Structured Questions [30 marks]

Answer all questions in the spaces provided.

11

A toy car of mass 0.2 kg is released from rest at the top of a frictionless ramp of height 0.5 m. The car then moves along a horizontal track before compressing a spring.

Image pending generation: diagram for Q11.

(a) Calculate the gravitational potential energy of the car at point A. [1]

(b) State the kinetic energy of the car at point B. Explain your answer. [2]

(d) In reality, the ramp has friction. Explain how this affects the maximum compression of the spring. [1]

12

A weightlifter lifts a barbell of mass 120 kg from the floor to a height of 2.0 m in 1.5 s. He holds it stationary for 3.0 s, then lowers it back to the floor in 1.5 s.

(a) Calculate the work done by the weightlifter in lifting the barbell. [2]

(b) Calculate the average power developed during the lift. [1]

(d) When lowering the barbell, the weightlifter exerts an upward force. Explain whether the work done by the weightlifter is positive, negative, or zero. [2]

13

A hydroelectric dam stores water in a reservoir at an average height of 80 m above the turbines. Water flows at a rate of 500 kg/s through the turbines.

(a) Calculate the gravitational potential energy lost by 500 kg of water as it falls 80 m. [1]

(b) If the electrical power output is 300 kW, calculate the efficiency of the energy conversion. [2]

14

A student investigates the relationship between the height of a ramp and the speed of a trolley at the bottom. She releases the trolley from rest at different heights and measures the speed using a light gate.

Experimental Setup for Q14 (SEC2 Science)

Generated experimental_setup for Q14.

The table shows her results:

Height $h$ (m)	Speed $v$ (m/s)	$v^2$ (m $^2$ /s $^2$ )
0.10	1.4	1.96
0.20	2.0	4.00
0.30	2.4	5.76
0.40	2.8	7.84
0.50	3.2	10.24

(a) Complete the table by calculating the missing $v^2$ value for $h = 0.30$ m. [1]

(b) Plot a graph of $v^2$ (y-axis) against $h$ (x-axis) on the grid below. [2]

Graph for Q14 (SEC2 Science)

Generated graph for Q14.

(d) The theoretical relationship is $v^2 = 2gh$ . Use your gradient to calculate a value for $g$ . [1]

(e) Suggest one reason why the experimental value of $g$ may differ from 10 m/s $^2$ . [1]

15

A 60 W filament lamp is connected to a 240 V mains supply and switched on for 5 hours.

(a) Calculate the current drawn by the lamp. [1]

(b) Calculate the electrical energy consumed in kWh. [1]

(d) The lamp converts electrical energy to light and heat. If the lamp is 10% efficient, calculate the useful light energy output in joules over the 5 hours. [2]

Section C: Longer Structured and Data-Based Questions [20 marks]

Answer all questions in the spaces provided.

16

A roller coaster track has three hills of different heights. The car starts from rest at the top of Hill 1 (height 60 m). The track is frictionless.

Diagram for Q16 (SEC2 Science)

Generated diagram for Q16.

(a) Calculate the total mechanical energy of the car at point A. [1]

(b) Calculate the speed of the car at point B. [2]

(d) The car just reaches the top of Hill 3 (point E) with zero speed. Explain whether this is consistent with the principle of conservation of energy. [2]

(e) In reality, the car does not reach the top of Hill 3. Explain why, and state what happens to the "missing" mechanical energy. [2]

17

A spring-loaded toy gun fires a 10 g pellet vertically upwards. The spring has a spring constant of 400 N/m and is compressed by 0.05 m before firing. Assume no air resistance and the spring has negligible mass.

(a) Calculate the elastic potential energy stored in the spring when compressed. [1]

(b) Calculate the maximum height reached by the pellet above the gun. [2]

(c) The pellet falls back and hits the spring, compressing it by 0.04 m. Explain why the compression is less than the original 0.05 m. [2]

18

A pump lifts 200 kg of water per minute from a well 15 m deep and ejects it through a hose at a speed of 8 m/s.

(a) Calculate the gain in gravitational potential energy of 200 kg of water. [1]

(b) Calculate the kinetic energy given to 200 kg of water as it is ejected at 8 m/s. [1]

(d) The actual power input to the pump is 1.2 kW. Calculate the efficiency of the pump. [1]

19

The diagram shows a simple pendulum of length 1.0 m. The bob of mass 0.2 kg is pulled aside until the string makes an angle of 30° with the vertical, then released from rest.

Image pending generation: diagram for Q19.

(a) Calculate the vertical height $h$ through which the bob is raised. [2]

(b) Calculate the maximum speed of the bob as it passes through the rest position. [2]

(c) The bob eventually comes to rest due to air resistance. Describe the energy conversions from the moment of release until the bob stops. [2]

20

A student carries out an experiment to determine the power of a small electric motor. The motor lifts a 0.5 kg load through a height of 1.2 m in 4.0 s. The motor is connected to a 6.0 V supply and draws a current of 0.8 A.

(a) Calculate the work done on the load. [1]

(b) Calculate the output power of the motor. [1]

(d) Calculate the efficiency of the motor. [1]

(e) Suggest two reasons why the efficiency is less than 100%. [2]

End of Paper

Answers

TuitionGoWhere Practice Paper - Science Secondary 2 (SA2 Version 5) - Answer Key

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

1

Answer: B [1]

Working:

Loss in GPE = Gain in KE (conservation of energy)
GPE = $mgh = 0.5 \times 10 \times 20 = 100 \text{ J}$
KE just before hitting ground = 100 J

Key concept: For a falling object with no air resistance, all gravitational potential energy converts to kinetic energy.

2

Answer: B [1]

Explanation: In a hydroelectric power station, water stored at height has gravitational potential energy → flows down gaining kinetic energy → turns turbines connected to generators producing electrical energy.

Energy conversion chain: Potential energy → Kinetic energy → Electrical energy

3

Answer: C [1]

Working:

Work done = Force × Distance moved in direction of force
$W = 25 \times 4 = 100 \text{ J}$

4

Answer: C [1]

Working:

GPE gained = $mgh = 3 \times 10 \times 1.5 = 45 \text{ J}$

5

Answer: B [1]

Explanation: Work done = Force × Distance moved in direction of force. When holding a bag stationary, there is no displacement in the direction of the upward force, so no work is done (even though the person gets tired - this is due to internal muscular work, not mechanical work on the bag).

Common mistake: Confusing "effort" with "work done" in the physics sense.

6

Answer: B [1]

Explanation: At position Y (lowest point), all gravitational potential energy has been converted to kinetic energy (maximum speed). At positions X and Z, the bob is momentarily at rest (zero kinetic energy, maximum potential energy).

7

Answer: B [1]

Working:

Power = 2000 W = 2000 J/s
Time = 10 minutes = 600 s
Energy = Power × Time = 2000 × 600 = 1,200,000 J

8

Answer: B [1]

Working:

Final KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200,000 \text{ J}$
Average power = Work done / Time = 200,000 / 10 = 20,000 W

9

Answer: C [1]

Explanation: The principle of conservation of energy states that energy cannot be created or destroyed; the total energy of an isolated system remains constant. Energy can only be converted from one form to another.

10

Answer: C [1]

Working:

$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 50} = \sqrt{1000} \approx 31.6 \text{ m/s} \approx 32 \text{ m/s}$

Section B: Structured Questions [30 marks]

11

(a) [1] Answer: 1.0 J

Working:

GPE = $mgh = 0.2 \times 10 \times 0.5 = 1.0 \text{ J}$

(b) [2] Answer: 1.0 J

Explanation: By conservation of energy on a frictionless ramp, all GPE at A converts to KE at B. Since GPE at A = 1.0 J, KE at B = 1.0 J.

Mark breakdown:

1 mark for correct value (1.0 J)
1 mark for correct explanation (conservation of energy / no energy loss on frictionless ramp)

(c) [2] Answer: 0.1 m (or 10 cm)

Working:

At maximum compression, all KE converts to elastic potential energy in spring
$\frac{1}{2}kx^2 = 1.0$
$\frac{1}{2} \times 200 \times x^2 = 1.0$
$100x^2 = 1.0$
$x^2 = 0.01$
$x = 0.1 \text{ m}$

Mark breakdown:

1 mark for correct equation $\frac{1}{2}kx^2 = \text{KE}$
1 mark for correct calculation and answer with unit

(d) [1] Answer: Friction does negative work on the car, converting some mechanical energy to heat/sound. Less energy is available to compress the spring, so maximum compression decreases.

Key concept: Non-conservative forces (friction) dissipate mechanical energy.

12

(a) [2] Answer: 2400 J

Working:

Work done = Force × Distance = Weight × Height
$W = mg \times h = 120 \times 10 \times 2.0 = 2400 \text{ J}$

Mark breakdown:

1 mark for correct force (weight = 1200 N)
1 mark for correct work calculation with unit

(b) [1] Answer: 1600 W

Working:

Power = Work / Time = 2400 / 1.5 = 1600 W

(c) [1] Answer: 0 J. No displacement occurs while holding the barbell stationary, so no work is done on the barbell.

Key concept: Work requires displacement in the direction of the force.

(d) [2] Answer: Negative work.

Explanation: The weightlifter exerts an upward force while the barbell moves downward. The force and displacement are in opposite directions, so work done by the weightlifter is negative. The weightlifter is controlling the descent, absorbing energy.

Mark breakdown:

1 mark for "negative"
1 mark for correct explanation (force opposite to displacement)

13

(a) [1] Answer: 400,000 J (or $4.0 \times 10^5 \text{ J}$ )

Working:

GPE lost = $mgh = 500 \times 10 \times 80 = 400,000 \text{ J}$

(b) [2] Answer: 75%

Working:

Input power (rate of GPE loss) = 400,000 J/s = 400 kW
Output power = 300 kW
Efficiency = $\frac{\text{Output}}{\text{Input}} \times 100\% = \frac{300}{400} \times 100\% = 75\%$

Mark breakdown:

1 mark for calculating input power (400 kW)
1 mark for correct efficiency calculation with %

(c) [1] Answer: Heat and sound (or kinetic energy of water turbulence, heat in bearings/generator)

Acceptable answers: Heat, sound, kinetic energy of splashing water, electrical resistance losses in generator.

14

(a) [1] Answer: 5.76 (already filled in table)

Working: $v^2 = 2.4^2 = 5.76 \text{ m}^2/\text{s}^2$

(b) [2] Answer: Correctly plotted points with best-fit straight line through origin.

Mark breakdown:

1 mark for all 5 points plotted correctly (± half a small square)
1 mark for best-fit straight line passing through origin (0,0)

(c) [1] Answer: 20 m/s² (or 20.0 m/s²)

Working:

Gradient = $\frac{\Delta v^2}{\Delta h} = \frac{10.24 - 0}{0.5 - 0} = 20.48 \approx 20 \text{ m/s}^2$
Accept range 20–21 m/s² depending on plotted points

(d) [1] Answer: 10 m/s²

Working:

Theory: $v^2 = 2gh$ , so gradient = $2g$
$g = \frac{\text{gradient}}{2} = \frac{20}{2} = 10 \text{ m/s}^2$

(e) [1] Answer: Friction between trolley and ramp / air resistance / rotational kinetic energy of wheels not accounted for / measurement errors in height or speed.

Any one valid reason accepted.

15

(a) [1] Answer: 0.25 A

Working:

$P = VI \Rightarrow I = \frac{P}{V} = \frac{60}{240} = 0.25 \text{ A}$

(b) [1] Answer: 0.3 kWh

Working:

Energy = Power × Time = 60 W × 5 h = 300 Wh = 0.3 kWh

(c) [1] Answer: $0.084 (or 8.4 cents)

Working:

Cost = 0.3 kWh × $0.28/kWh =$ 0.084

(d) [2] Answer: 648,000 J (or $6.48 \times 10^5 \text{ J}$ )

Working:

Total electrical energy = Power × Time = 60 W × (5 × 3600 s) = 60 × 18,000 = 1,080,000 J
Useful light energy = 10% × 1,080,000 = 108,000 J

Wait, let me recalculate:

5 hours = 5 × 3600 = 18,000 s
Total energy = 60 × 18,000 = 1,080,000 J
10% efficient → Light energy = 0.10 × 1,080,000 = 108,000 J

Mark breakdown:

1 mark for total electrical energy in joules (1,080,000 J)
1 mark for correct useful energy (108,000 J)

Section C: Longer Structured and Data-Based Questions [20 marks]

16

(a) [1] Answer: 300,000 J (or $3.0 \times 10^5 \text{ J}$ )

Working:

Total mechanical energy at A = GPE (since KE = 0 at rest)
$E = mgh = 500 \times 10 \times 60 = 300,000 \text{ J}$

(b) [2] Answer: 34.6 m/s (or $\sqrt{1200} \approx 34.6 \text{ m/s}$ )

Working:

At B (height 0), all energy is KE
$\frac{1}{2}mv^2 = 300,000$
$\frac{1}{2} \times 500 \times v^2 = 300,000$
$250v^2 = 300,000$
$v^2 = 1200$
$v = \sqrt{1200} \approx 34.6 \text{ m/s}$

Mark breakdown:

1 mark for correct energy equation
1 mark for correct calculation and unit

(c) [2] Answer: 100,000 J (or $1.0 \times 10^5 \text{ J}$ )

Working:

At C (height 40 m): GPE = $mgh = 500 \times 10 \times 40 = 200,000 \text{ J}$
Total energy = 300,000 J (conserved)
KE at C = Total - GPE = 300,000 - 200,000 = 100,000 J

Mark breakdown:

1 mark for GPE at C (200,000 J)
1 mark for KE = 100,000 J

(d) [2] Answer: Yes, it is consistent. At point E (height 30 m), GPE = $500 \times 10 \times 30 = 150,000 \text{ J}$ . Since speed is zero, KE = 0. Total mechanical energy = 150,000 J, which is less than the initial 300,000 J. This means the car cannot reach point E with zero speed if energy is conserved — it would need 300,000 J total energy to reach 60 m height.

Wait, let me re-read the question: "The car just reaches the top of Hill 3 (point E) with zero speed. Explain whether this is consistent with the principle of conservation of energy."

If the car reaches height 30 m with zero speed, its total energy is 150,000 J. But initial energy was 300,000 J. This violates conservation of energy unless energy was lost. So the statement "car just reaches top of Hill 3 with zero speed" is NOT consistent with conservation of energy on a frictionless track.

Corrected Answer: No, it is not consistent. On a frictionless track, total mechanical energy is conserved (300,000 J). At point E (30 m), GPE would be 150,000 J. With zero speed (KE = 0), total energy would be 150,000 J, which is less than the initial 300,000 J. For conservation of energy to hold, the car would have 150,000 J of KE at point E, meaning it would still be moving.

Mark breakdown:

1 mark for "No / not consistent"
1 mark for correct explanation using energy values

(e) [2] Answer: In reality, friction between wheels/track and air resistance do negative work, converting mechanical energy to heat and sound. The "missing" mechanical energy is dissipated as thermal energy (heat) and sound energy.

Mark breakdown:

1 mark for identifying friction/air resistance as cause
1 mark for stating energy converted to heat/sound

17

(a) [1] Answer: 0.5 J

Working:

Elastic PE = $\frac{1}{2}kx^2 = \frac{1}{2} \times 400 \times (0.05)^2 = 200 \times 0.0025 = 0.5 \text{ J}$

(b) [2] Answer: 5.0 m

Working:

Elastic PE → GPE at max height
$\frac{1}{2}kx^2 = mgh$
$0.5 = 0.01 \times 10 \times h$
$0.5 = 0.1h$
$h = 5.0 \text{ m}$

Mark breakdown:

1 mark for correct energy equation
1 mark for correct calculation with unit

(c) [2] Answer: Some energy is lost to air resistance during the up and down motion, and possibly to heat/sound when the pellet hits the spring. The pellet has less kinetic energy when it returns to the spring, so it compresses the spring less (0.04 m vs 0.05 m).

Mark breakdown:

1 mark for identifying energy loss (air resistance / impact)
1 mark for linking less energy to less compression

18

(a) [1] Answer: 30,000 J (or $3.0 \times 10^4 \text{ J}$ )

Working:

GPE gain = $mgh = 200 \times 10 \times 15 = 30,000 \text{ J}$

(b) [1] Answer: 6,400 J (or $6.4 \times 10^3 \text{ J}$ )

Working:

KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 200 \times 8^2 = 100 \times 64 = 6,400 \text{ J}$

(c) [2] Answer: 606.7 W (or 607 W)

Working:

Total energy per minute = GPE + KE = 30,000 + 6,400 = 36,400 J
Power = Energy / Time = 36,6,6400 J / 60 s = 606.67 W ≈ 607 W

Mark breakdown:

1 mark for total energy per minute (36,400 J)
1 mark for power calculation with unit (W)

(d) [1] Answer: 50.6% (or 51%)

Working:

Efficiency = $\frac{\text{Output power}}{\text{Input power}} \times 100\% = \frac{606.7}{1200} \times 100\% = 50.6\%$

19

(a) [2] Answer: 0.134 m (or 13.4 cm)

Working:

$h = L - L\cos\theta = L(1 - \cos\theta)$
$h = 1.0 \times (1 - \cos 30°) = 1.0 \times (1 - 0.866) = 0.134 \text{ m}$

Mark breakdown:

1 mark for correct method ( $h = L(1 - \cos\theta)$ or equivalent)
1 mark for correct calculation with unit

(b) [2] Answer: 1.64 m/s (or 1.6 m/s)

Working:

GPE at max displacement = KE at bottom
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} \approx 1.64 \text{ m/s}$

Mark breakdown:

1 mark for correct energy equation
1 mark for correct calculation with unit

(c) [2] Answer:

At release: GPE (maximum) → converts to KE as bob swings down
At bottom: KE (maximum) → converts to GPE as bob swings up
At each swing: Some energy lost to air resistance (and friction at pivot) → heat/sound
Eventually: All mechanical energy dissipated as heat and sound; bob comes to rest at bottom with zero KE and zero GPE (relative to bottom)

Mark breakdown:

1 mark for describing GPE ↔ KE conversions during oscillation
1 mark for mentioning energy loss to air resistance/friction and final dissipation

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Science Secondary 2

TuitionGoWhere Secondary School (AI)

Subject: Science
Level: Secondary 2 (G3)
Paper: SA2 Version 5
Duration: 1 hour 30 minutes
Total Marks: 60

Marking Scheme and Model Answers

Section A: Multiple Choice Questions [10 marks]

Question	Answer	Explanation
1	B	GPE at top = mgh = 0.5 × 10 × 20 = 100 J. By conservation of energy, KE at bottom = 100 J.
2	B	Water at height has GPE → flows down gaining KE → turns turbine generating electrical energy.
3	C	Work done = Force × Distance = 25 N × 4 m = 100 J.
4	C	GPE gained = mgh = 3 × 10 × 1.5 = 45 J.
5	B	Work done = Force × Distance moved in direction of force. No displacement while holding stationary.
6	B	Maximum KE at lowest point (Y) where GPE is minimum and speed is maximum.
7	B	Energy = Power × Time = 2000 W × (10 × 60) s = 2000 × 600 = 1,200,000 J.
8	B	KE gained = ½mv² = 0.5 × 1000 × 20² = 200,000 J. Average Power = Work/Time = 200,000/10 = 20,000 W.
9	C	Principle of conservation of energy: Total energy of an isolated system remains constant.
10	C	mgh = ½mv² → v = √(2gh) = √(2 × 10 × 50) = √1000 ≈ 31.6 m/s ≈ 32 m/s.

Section B: Structured Questions [30 marks]

11

(a) GPE at A = mgh = 0.2 × 10 × 0.5 = 1.0 J [1]

(b) KE at B = 1.0 J [1]
Explanation: By conservation of energy on a frictionless ramp, all GPE at A is converted to KE at B. [1]

(c) At maximum compression, KE at B = Elastic PE in spring
1.0 = ½kx² = ½ × 200 × x²
x² = 1.0 / 100 = 0.01
x = 0.10 m (or 10 cm) [2]

(d) Friction does negative work / dissipates energy as heat/sound. [½]
Less energy reaches the spring, so maximum compression is smaller. [½]

12

(a) Work done = Force × Distance = Weight × Height = mg × h = 120 × 10 × 2.0 = 2400 J [2]

(b) Average Power = Work / Time = 2400 / 1.5 = 1600 W [1]

(c) Work done = 0 J [½]
Explanation: No displacement while holding stationary, so no work done (Work = Force × Distance). [½]

(d) Work done is negative. [1]
Explanation: The weightlifter exerts an upward force, but the displacement is downward. Force and displacement are in opposite directions, so work done is negative. [1]

13

(a) GPE lost = mgh = 500 × 10 × 80 = 400,000 J (or 400 kJ) [1]

(b) Input power = Rate of GPE loss = 400,000 J/s = 400 kW
Efficiency = (Output Power / Input Power) × 100% = (300 / 400) × 100% = 75% [2]

(c) Heat and sound (also accept: kinetic energy of water turbulence, internal energy) [1]

14

(a) For h = 0.30 m, v = 2.4 m/s → v² = 2.4² = 5.76 m²/s² [1] (Already filled in table)

(b) Graph plotting: [2]

Axes labelled with units: x-axis "Height h (m)", y-axis "v² (m²/s²)" [½]
Appropriate scales covering data range [½]
All 5 points plotted correctly [½]
Best-fit straight line through origin [½]

(c) Gradient = Δv² / Δh = (10.24 - 1.96) / (0.50 - 0.10) = 8.28 / 0.40 = 20.7 m/s² (accept 20–21) [1]

(d) Theoretical: v² = 2gh → Gradient = 2g
g = Gradient / 2 = 20.7 / 2 = 10.35 m/s² (accept 10–10.5) [1]

(e) Any one: Friction between trolley and ramp / air resistance / light gate measurement error / ramp not perfectly rigid / trolley wobble / rotational KE of wheels not accounted for. [1]

15

(a) P = VI → I = P/V = 60 / 240 = 0.25 A [1]

(b) Energy = Power × Time = 60 W × 5 h = 300 Wh = 0.3 kWh [1]

(c) Cost = 0.3 kWh × $0.28/kWh = **$ 0.084** (or 8.4 cents) [1]

(d) Total electrical energy = 0.3 kWh = 0.3 × 3.6 × 10⁶ = 1,080,000 J
Useful light energy = 10% × 1,080,000 = 108,000 J [2]

Section C: Longer Structured and Data-Based Questions [20 marks]

16

(a) Total mechanical energy at A = GPE at A (KE = 0) = mgh = 500 × 10 × 60 = 300,000 J (or 300 kJ) [1]

(b) At B (h = 0), all GPE → KE
½mv² = 300,000
v² = 600,000 / 500 = 1200
v = √1200 = 34.6 m/s (accept 34.6–35) [2]

(c) At C (h = 40 m), GPE = mgh = 500 × 10 × 40 = 200,000 J
KE = Total Energy - GPE = 300,000 - 200,000 = 100,000 J [2]

(d) Yes, it is consistent. [1]
At point E (h = 30 m), GPE = 500 × 10 × 30 = 150,000 J.
If speed is zero, KE = 0, so total mechanical energy = 150,000 J.
This is less than the initial 300,000 J, so energy is not conserved if the track is frictionless.
Correction: The car cannot reach point E with zero speed on a frictionless track because total mechanical energy must remain 300,000 J. At h = 30 m, it would still have KE = 150,000 J and would continue past point E. [1]

(e) In reality, friction (between wheels/track) and air resistance do negative work. [1]
The "missing" mechanical energy is converted to heat and sound (internal energy of track/air). [1]

17

(a) Elastic PE = ½kx² = ½ × 400 × (0.05)² = 200 × 0.0025 = 0.5 J [1]

(b) Elastic PE → GPE at max height
0.5 = mgh = 0.01 × 10 × h
h = 0.5 / 0.1 = 5.0 m [2]

(c) During flight, air resistance does negative work on the pellet. [1]
Some mechanical energy is lost as heat, so the pellet returns with less KE than it left with. [1]
Less KE → less elastic PE stored at maximum compression → smaller compression (0.04 m < 0.05 m). [Implied in explanation]

18

(a) Gain in GPE = mgh = 200 × 10 × 15 = 30,000 J [1]

(b) KE = ½mv² = ½ × 200 × 8² = 100 × 64 = 6,400 J [1]

(c) Total energy per minute = GPE + KE = 30,000 + 6,400 = 36,400 J
Power = Energy / Time = 36,400 / 60 = 606.7 W (accept 607 W) [2]

(d) Efficiency = (Output Power / Input Power) × 100% = (606.7 / 1200) × 100% = 50.6% (accept 50–51%) [1]

19

(a) h = L - L cosθ = L(1 - cosθ) = 1.0 × (1 - cos30°) = 1.0 × (1 - 0.866) = 0.134 m [2]

(b) GPE at top = KE at bottom
mgh = ½mv²
v = √(2gh) = √(2 × 10 × 0.134) = √2.68 = 1.64 m/s [2]

(c) The bob eventually comes to rest due to air resistance. Describe the energy conversions from the moment it is released until it comes to rest. [2]
Answer: GPE → KE (and back to GPE) repeatedly. With each swing, mechanical energy is dissipated as heat due to air resistance (and friction at pivot). Eventually, all initial GPE is converted to heat/internal energy of the surrounding air and support. [2]

Summary of Marks Allocation

Section	Questions	Total Marks
A	1–10	10
B	11–15	30
C	16–19	20
Total		60

End of Marking Scheme

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