Secondary 2 Science Practice Paper 1

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TuitionGoWhere Practice Paper - Science Secondary 2

TuitionGoWhere Practice Paper (AI) — Version 1

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper — Physical Sciences
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions.
3. Write your answers in the spaces provided on the question paper.
4. The number of marks is given in brackets [ ] at the end of each question or part question.
5. The total number of marks for this paper is 60.
6. You may use a calculator.
7. Where appropriate, take $g = 10 \text{ N/kg}$.
8. Show all working for calculation questions.

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Section A: Multiple Choice Questions [10 marks]

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

1. A ball of mass 0.2 kg is dropped from a height of 5 m. Ignoring air resistance, what is the kinetic energy of the ball just before it hits the ground? [1]

A. 1 J
B. 5 J
C. 10 J
D. 20 J

Answer: \fbox{\phantom{A}}

2. Which of the following energy conversions takes place when a candle burns? [1]

A. Chemical potential energy $\to$ Heat energy + Light energy
B. Heat energy $\to$ Chemical potential energy + Light energy
C. Light energy $\to$ Chemical potential energy + Heat energy
D. Chemical potential energy $\to$ Kinetic energy + Sound energy

Answer: \fbox{\phantom{A}}

3. A force of 15 N is used to push a box horizontally across a floor for a distance of 4 m. The work done on the box is: [1]

A. 3.75 J
B. 19 J
C. 60 J
D. 600 J

Answer: \fbox{\phantom{A}}

4. A student runs up a flight of stairs of vertical height 3 m in 6 seconds. If the student has a mass of 50 kg, what is the average power developed by the student against gravity? [1]

A. 25 W
B. 150 W
C. 250 W
D. 900 W

Answer: \fbox{\phantom{A}}

5. The diagram below shows a simple pendulum swinging from position P to Q to R. At which position(s) does the pendulum bob have maximum kinetic energy? [1]

<image_placeholder> id: Q5-fig1 type: diagram linked_question: Q5 description: Simple pendulum at three positions: P (highest left), Q (lowest centre), R (highest right). String length labelled L. Vertical height difference between P/Q and Q/R labelled h. labels: P, Q, R, L, h values: L = 1.0 m, h = 0.2 m must_show: Pendulum bob at three distinct positions with height difference clearly indicated </image_placeholder>

A. P only
B. Q only
C. R only
D. P and R

Answer: \fbox{\phantom{A}}

6. An electric kettle rated 2000 W is used to boil water for 3 minutes. How much electrical energy is consumed? [1]

A. 6000 J
B. 36000 J
C. 360000 J
D. 600000 J

Answer: \fbox{\phantom{A}}

7. A 2 kg object moves with a velocity of 5 m/s. Its kinetic energy is: [1]

A. 10 J
B. 25 J
C. 50 J
D. 100 J

Answer: \fbox{\phantom{A}}

8. Which of the following statements about energy is correct? [1]

A. Energy can be created but not destroyed.
B. Energy can be destroyed but not created.
C. Energy can neither be created nor destroyed.
D. Energy can be both created and destroyed.

Answer: \fbox{\phantom{A}}

9. A motor lifts a load of 200 N through a vertical height of 10 m in 20 s. The power output of the motor is: [1]

A. 10 W
B. 100 W
C. 200 W
D. 400 W

Answer: \fbox{\phantom{A}}

10. A roller coaster car starts from rest at the top of a hill 40 m high. Assuming no energy losses, what is its speed at the bottom of the hill? [1]

A. 20 m/s
B. 28 m/s
C. 40 m/s
D. 80 m/s

Answer: \fbox{\phantom{A}}

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Section B: Structured Questions [30 marks]

Answer all questions in the spaces provided.

11. A cyclist of total mass 70 kg (including bicycle) starts from rest at the top of a hill of vertical height 25 m. The cyclist freewheels down the hill without pedalling. At the bottom of the hill, the cyclist's speed is 20 m/s. [4]

(a) Calculate the loss in gravitational potential energy of the cyclist from the top to the bottom of the hill. [2]

Answer: _______________________________________________________________________

(b) Calculate the kinetic energy of the cyclist at the bottom of the hill. [1]

Answer: _______________________________________________________________________

(c) Explain why the kinetic energy at the bottom is less than the loss in gravitational potential energy. [1]

Answer: _______________________________________________________________________

12. A student investigates the bouncing of a tennis ball. The ball of mass 0.06 kg is dropped from a height of 2.0 m onto a hard floor. It rebounds to a height of 1.2 m. [5]

(a) Calculate the gravitational potential energy of the ball before it is dropped. [1]

Answer: _______________________________________________________________________

(b) Calculate the speed of the ball just before it hits the floor. [2]

Answer: _______________________________________________________________________

(c) Calculate the percentage of energy retained after the bounce. [2]

Answer: _______________________________________________________________________

13. An electric heater rated 1500 W is used to heat a room for 45 minutes. The cost of electricity is $0.28 per kWh. [4]

(a) Calculate the electrical energy consumed in kWh. [2]

Answer: _______________________________________________________________________

(b) Calculate the cost of using the heater for 45 minutes. [2]

Answer: _______________________________________________________________________

14. A crane lifts a concrete block of mass 500 kg vertically upwards at a constant speed of 0.5 m/s. [4]

(a) State the net force acting on the concrete block while it is moving at constant speed. Explain your answer. [2]

Answer: _______________________________________________________________________

(b) Calculate the power output of the crane motor. [2]

Answer: _______________________________________________________________________

15. The diagram below shows a roller coaster track. The car of mass 400 kg starts from rest at point A (height 30 m) and moves along the frictionless track. [5]

<image_placeholder> id: Q15-fig1 type: diagram linked_question: Q15 description: Roller coaster track profile with points A, B, C, D labelled. Point A at height 30 m, point B at height 10 m, point C at height 20 m, point D at ground level (0 m). Track shown as smooth curve connecting points. labels: A (30 m), B (10 m), C (20 m), D (0 m) values: h_A = 30 m, h_B = 10 m, h_C = 20 m, h_D = 0 m, mass = 400 kg must_show: Clear height labels at each point, track profile, car at starting position A </image_placeholder>

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

Answer: _______________________________________________________________________

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

Answer: _______________________________________________________________________

(c) Calculate the kinetic energy of the car at point C. [2]

Answer: _______________________________________________________________________

16. A spring-loaded toy gun fires a pellet of mass 0.01 kg vertically upwards. The spring is compressed by 0.05 m and the spring constant is 200 N/m. Assume all elastic potential energy is converted to gravitational potential energy at the maximum height. [4]

(a) Calculate the elastic potential energy stored in the compressed spring. [2]

Answer: _______________________________________________________________________

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

Answer: _______________________________________________________________________

17. A car of mass 1200 kg accelerates uniformly from rest to 25 m/s in 10 seconds along a horizontal road. [4]

(a) Calculate the acceleration of the car. [1]

Answer: _______________________________________________________________________

(b) Calculate the work done by the engine to accelerate the car (ignore friction and air resistance). [2]

Answer: _______________________________________________________________________

(c) Calculate the average power developed by the engine during this acceleration. [1]

Answer: _______________________________________________________________________

18. A hydroelectric power station uses water falling from a height of 50 m to generate electricity. Water flows at a rate of 200 kg/s. The overall efficiency of the system is 80%. [4]

(a) Calculate the gravitational potential energy lost by the water per second. [2]

Answer: _______________________________________________________________________

(b) Calculate the electrical power output of the power station. [2]

Answer: _______________________________________________________________________

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Section C: Longer Structured Questions [20 marks]

Answer all questions in the spaces provided.

19. A student sets up an experiment to investigate the relationship between the height of a ramp and the speed of a trolley at the bottom. The trolley of mass 0.5 kg is released from rest at different heights. The student records the following data: [6]

Height of ramp / m	Speed at bottom / m/s
0.10	1.3
0.20	1.9
0.30	2.3
0.40	2.7
0.50	3.0

(a) For the height of 0.30 m, calculate the gravitational potential energy lost by the trolley. [1]

Answer: _______________________________________________________________________

(b) For the height of 0.30 m, calculate the kinetic energy of the trolley at the bottom. [1]

Answer: _______________________________________________________________________

(c) Calculate the percentage of gravitational potential energy converted to kinetic energy at 0.30 m. [1]

Answer: _______________________________________________________________________

(d) Suggest one reason why the percentage calculated in (c) is less than 100%. [1]

Answer: _______________________________________________________________________

(e) The student plots a graph of $v^2$ against $h$. State the expected shape of the graph and explain why, using the principle of conservation of energy. [2]

Answer: _______________________________________________________________________

20. A solar panel of area 2.0 m$^2$ receives sunlight with an intensity of 800 W/m$^2$. The panel has an efficiency of 18%. The electrical energy generated is used to charge a battery. [6]

(a) Calculate the power input from sunlight to the solar panel. [1]

Answer: _______________________________________________________________________

(b) Calculate the electrical power output of the solar panel. [1]

Answer: _______________________________________________________________________

(c) Calculate the electrical energy generated in 3 hours, in kWh. [2]

Answer: _______________________________________________________________________

(d) The battery stores energy with an efficiency of 90%. Calculate the energy actually stored in the battery after 3 hours of charging, in kWh. [2]

Answer: _______________________________________________________________________

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End of Paper

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TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper — Physical Sciences (Version 1)
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

1. Answer: C [1]

Working:
Loss in GPE = Gain in KE (conservation of energy, no air resistance)
$\text{GPE} = mgh = 0.2 \times 10 \times 5 = 10 \text{ J}$
$\therefore \text{KE} = 10 \text{ J}$

Key concept: When an object falls without air resistance, all gravitational potential energy converts to kinetic energy.

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2. Answer: A [1]

Explanation:
A burning candle converts the chemical potential energy stored in the wax into heat energy and light energy. This is a chemical reaction (combustion) releasing energy.

Common mistake: Confusing the direction of energy conversion or including kinetic/sound energy which are negligible.

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3. Answer: C [1]

Working:
$\text{Work done} = \text{Force} \times \text{Distance moved in direction of force}$
$W = 15 \times 4 = 60 \text{ J}$

Key concept: Work done = $F \times s$ when force and displacement are in the same direction.

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4. Answer: C [1]

Working:
$\text{Work done against gravity} = mgh = 50 \times 10 \times 3 = 1500 \text{ J}$
$\text{Power} = \frac{\text{Work done}}{\text{Time}} = \frac{1500}{6} = 250 \text{ W}$

Key concept: Power = rate of doing work = $\frac{E}{t}$.

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5. Answer: B [1]

Explanation:
At position Q (lowest point), the pendulum bob has minimum gravitational potential energy and maximum kinetic energy. At P and R (highest points), speed is momentarily zero so KE = 0.

Key concept: In a swinging pendulum, energy continuously converts between GPE (max at extremes) and KE (max at centre).

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6. Answer: C [1]

Working:
$\text{Energy} = \text{Power} \times \text{Time}$
$E = 2000 \text{ W} \times (3 \times 60) \text{ s} = 2000 \times 180 = 360\,000 \text{ J}$

Key concept: Convert time to seconds when power is in watts (J/s). 3 minutes = 180 seconds.

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7. Answer: B [1]

Working:
$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 5^2 = 1 \times 25 = 25 \text{ J}$

Key concept: Kinetic energy formula: $\text{KE} = \frac{1}{2}mv^2$.

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8. Answer: C [1]

Explanation:
The Principle of Conservation of Energy states that energy cannot be created or destroyed, only converted from one form to another. The total energy in a closed system remains constant.

Key concept: This is the fundamental principle of energy conservation.

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9. Answer: B [1]

Working:
$\text{Work done} = \text{Force} \times \text{Distance} = 200 \times 10 = 2000 \text{ J}$
$\text{Power} = \frac{\text{Work done}}{\text{Time}} = \frac{2000}{20} = 100 \text{ W}$

Key concept: Power = Work/Time. Force is given directly as 200 N (weight of load).

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10. Answer: B [1]

Working:
$\text{Loss in GPE} = \text{Gain in KE}$
$mgh = \frac{1}{2}mv^2$
$gh = \frac{1}{2}v^2$
$v^2 = 2gh = 2 \times 10 \times 40 = 800$
$v = \sqrt{800} \approx 28.3 \text{ m/s} \approx 28 \text{ m/s}$

Key concept: Using $v = \sqrt{2gh}$ for free fall from rest (or frictionless slope).

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Section B: Structured Questions [30 marks]

11. (a) Answer: 17 500 J [2]

Working:
$\text{Loss in GPE} = mgh = 70 \times 10 \times 25 = 17\,500 \text{ J}$

Mark breakdown:

• Correct formula/substitution: 1 mark
• Correct answer with unit: 1 mark

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11. (b) Answer: 14 000 J [1]

Working:
$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 70 \times 20^2 = 35 \times 400 = 14\,000 \text{ J}$

Mark breakdown:

• Correct answer with unit: 1 mark

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11. (c) Answer: Some gravitational potential energy is converted to heat and sound energy due to friction (air resistance and friction between wheels/ground). [1]

Explanation:
The "missing" energy (17 500 - 14 000 = 3 500 J) is dissipated as thermal energy (heat) and sound due to non-conservative forces (friction, air resistance). This is why the KE at the bottom is less than the GPE lost.

Marking note: Must mention energy conversion to heat/sound or work done against friction. "Energy is lost" alone is insufficient — energy is not lost, it is converted to other forms.

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12. (a) Answer: 1.2 J [1]

Working:
$\text{GPE} = mgh = 0.06 \times 10 \times 2.0 = 1.2 \text{ J}$

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12. (b) Answer: 6.3 m/s (or 6.32 m/s) [2]

Working:
$\text{GPE at top} = \text{KE just before impact}$ (ignoring air resistance)
$mgh = \frac{1}{2}mv^2$
$v^2 = 2gh = 2 \times 10 \times 2.0 = 40$
$v = \sqrt{40} \approx 6.32 \text{ m/s}$

Mark breakdown:

• Correct use of $v = \sqrt{2gh}$ or energy conservation: 1 mark
• Correct calculation and unit: 1 mark

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12. (c) Answer: 60% [2]

Working:
$\text{GPE after bounce} = mgh_{\text{rebound}} = 0.06 \times 10 \times 1.2 = 0.72 \text{ J}$
$\text{Initial GPE} = 1.2 \text{ J}$
$\text{Percentage retained} = \frac{0.72}{1.2} \times 100\% = 60\%$

Alternative method: $\frac{h_{\text{rebound}}}{h_{\text{drop}}} \times 100\% = \frac{1.2}{2.0} \times 100\% = 60\%$

Mark breakdown:

• Correct rebound GPE or ratio method: 1 mark
• Correct percentage: 1 mark

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13. (a) Answer: 1.125 kWh [2]

Working:
$\text{Power} = 1500 \text{ W} = 1.5 \text{ kW}$
$\text{Time} = 45 \text{ min} = 0.75 \text{ h}$
$\text{Energy} = \text{Power} \times \text{Time} = 1.5 \times 0.75 = 1.125 \text{ kWh}$

Mark breakdown:

• Correct conversion to kW and hours: 1 mark
• Correct answer with unit: 1 mark

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13. (b) Answer: $0.315 (or 31.5 cents) [2]

Working:
$\text{Cost} = \text{Energy} \times \text{Rate} = 1.125 \times 0.28 = 0.315$
= \0.315 \text{ or } 31.5 \text{ cents}$

Mark breakdown:

• Correct multiplication: 1 mark
• Correct answer with $ sign: 1 mark

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14. (a) Answer: Net force = 0 N. The block moves at constant velocity, so by Newton's First Law, the net force acting on it is zero. The upward tension force equals the downward weight. [2]

Mark breakdown:

• Net force = 0 N: 1 mark
• Correct explanation (constant velocity $\to$ balanced forces / Newton's First Law): 1 mark

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14. (b) Answer: 2500 W [2]

Working:
$\text{Force needed} = \text{Weight} = mg = 500 \times 10 = 5000 \text{ N}$
$\text{Power} = \text{Force} \times \text{Velocity} = 5000 \times 0.5 = 2500 \text{ W}$

Alternative method:
$\text{Work done per second} = mgh/t = 500 \times 10 \times 0.5 = 2500 \text{ J/s} = 2500 \text{ W}$

Mark breakdown:

• Correct force/weight calculation: 1 mark
• Correct power calculation with unit: 1 mark

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15. (a) Answer: 120 000 J [1]

Working:
At point A, car is at rest so KE = 0.
$\text{Total mechanical energy} = \text{GPE} = mgh = 400 \times 10 \times 30 = 120\,000 \text{ J}$

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15. (b) Answer: 20 m/s [2]

Working:
At point B (height 10 m):
$\text{GPE}_B = mgh_B = 400 \times 10 \times 10 = 40\,000 \text{ J}$
$\text{Total energy} = 120\,000 \text{ J}$ (conserved, frictionless)
$\text{KE}_B = 120\,000 - 40\,000 = 80\,000 \text{ J}$
$\frac{1}{2}mv^2 = 80\,000$
$v^2 = \frac{160\,000}{400} = 400$
$v = 20 \text{ m/s}$

Alternative (shorter):
Loss in GPE from A to B = $mg(h_A - h_B) = 400 \times 10 \times 20 = 80\,000 \text{ J} = \text{KE}_B$
$v = \sqrt{2g(h_A - h_B)} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}$

Mark breakdown:

• Correct energy conservation approach: 1 mark
• Correct answer with unit: 1 mark

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15. (c) Answer: 40 000 J [2]

Working:
At point C (height 20 m):
$\text{GPE}_C = mgh_C = 400 \times 10 \times 20 = 80\,000 \text{ J}$
$\text{KE}_C = \text{Total energy} - \text{GPE}_C = 120\,000 - 80\,000 = 40\,000 \text{ J}$

Mark breakdown:

• Correct GPE at C: 1 mark
• Correct KE by subtraction: 1 mark

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16. (a) Answer: 0.25 J [2]

Working:
$\text{Elastic PE} = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.05)^2 = 100 \times 0.0025 = 0.25 \text{ J}$

Mark breakdown:

• Correct formula $\frac{1}{2}kx^2$: 1 mark
• Correct calculation with unit: 1 mark

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16. (b) Answer: 2.5 m [2]

Working:
$\text{Elastic PE} = \text{GPE at max height}$
$0.25 = mgh = 0.01 \times 10 \times h$
$h = \frac{0.25}{0.1} = 2.5 \text{ m}$

Mark breakdown:

• Correct energy conservation equation: 1 mark
• Correct answer with unit: 1 mark

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17. (a) Answer: 2.5 m/s$^2$ [1]

Working:
$a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m/s}^2$

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17. (b) Answer: 375 000 J [2]

Working:
Work done = Gain in KE (no friction/air resistance)
$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = 375\,000 \text{ J}$

Mark breakdown:

• Correct use of work-energy theorem: 1 mark
• Correct calculation with unit: 1 mark

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17. (c) Answer: 37 500 W [1]

Working:
$\text{Average Power} = \frac{\text{Work done}}{\text{Time}} = \frac{375\,000}{10} = 37\,500 \text{ W}$

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18. (a) Answer: 100 000 J/s (or 100 000 W) [2]

Working:
$\text{GPE lost per second} = \text{mass per second} \times g \times h$
$= 200 \times 10 \times 50 = 100\,000 \text{ J/s} = 100\,000 \text{ W}$

Mark breakdown:

• Correct formula (rate of GPE loss): 1 mark
• Correct calculation with unit: 1 mark

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18. (b) Answer: 80 000 W (or 80 kW) [2]

Working:
$\text{Electrical power output} = \text{Efficiency} \times \text{Input power}$
$= 0.80 \times 100\,000 = 80\,000 \text{ W} = 80 \text{ kW}$

Mark breakdown:

• Correct use of efficiency: 1 mark
• Correct answer with unit: 1 mark

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Section C: Longer Structured Questions [20 marks]

19. (a) Answer: 1.5 J [1]

Working:
$\text{GPE lost} = mgh = 0.5 \times 10 \times 0.30 = 1.5 \text{ J}$

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19. (b) Answer: 1.3225 J (or 1.32 J) [1]

Working:
$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (2.3)^2 = 0.25 \times 5.29 = 1.3225 \text{ J}$

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19. (c) Answer: 88.2% (or 88%) [1]

Working:
$\text{Percentage} = \frac{\text{KE}}{\text{GPE lost}} \times 100\% = \frac{1.3225}{1.5} \times 100\% = 88.17\% \approx 88.2\%$

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19. (d) Answer: Some gravitational potential energy is converted to heat and sound due to friction between the trolley wheels/axle and air resistance. [1]

Marking note: Must identify a dissipative mechanism (friction, air resistance) and the resulting energy forms (heat, sound). "Energy is lost" is not accepted.

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19. (e) Answer: The graph of $v^2$ against $h$ is a straight line passing through the origin. [1]

Explanation: By conservation of energy, $mgh = \frac{1}{2}mv^2$ (assuming no energy losses), so $v^2 = 2gh$. This is of the form $y = mx$ where $y = v^2$, $x = h$, and gradient $m = 2g$. Since the line passes through the origin (when $h=0$, $v=0$), it is a straight line through the origin. [1]

Mark breakdown:

• Correct shape (straight line through origin): 1 mark
• Correct explanation using $v^2 = 2gh$ / conservation of energy: 1 mark

Key concept: $v^2 \propto h$ when energy is conserved. The gradient gives $2g$.

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20. (a) Answer: 1600 W [1]

Working:
$\text{Power input} = \text{Intensity} \times \text{Area} = 800 \times 2.0 = 1600 \text{ W}$

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20. (b) Answer: 288 W [1]

Working:
$\text{Power output} = \text{Efficiency} \times \text{Power input} = 0.18 \times 1600 = 288 \text{ W}$

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20. (c) Answer: 0.864 kWh [2]

Working:
$\text{Energy} = \text{Power} \times \text{Time} = 288 \text{ W} \times 3 \text{ h} = 864 \text{ Wh} = 0.864 \text{ kWh}$

Mark breakdown:

• Correct power × time in Wh: 1 mark
• Correct conversion to kWh: 1 mark

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20. (d) Answer: 0.7776 kWh [2]

Working:
$\text{Energy stored} = \text{Battery efficiency} \times \text{Energy generated}$
$= 0.90 \times 0.864 = 0.7776 \text{ kWh}$

Mark breakdown:

• Correct use of battery efficiency: 1 mark
• Correct calculation with unit: 1 mark

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End of Answer Key