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

TuitionGoWhere Practice Paper - Combined Science Secondary 3 SA2 Version 1 - Answer Key

Total Marks: 60

Section A: Multiple Choice Questions [15 marks]

1. Answer: A [1]

Explanation: As the ball rises, kinetic energy decreases while gravitational potential energy increases. Some energy is also converted to thermal energy due to air resistance (though often neglected in idealised problems, the most complete correct statement includes thermal energy). The principle of conservation of energy states total energy remains constant, but the conversion involves KE → GPE + thermal.

Marking note: Award 1 mark for A. Option C is incomplete as it ignores air resistance/thermal energy.

2. Answer: A [1]

Working:

• Work done by applied force = $F \times s = 20 \times 5 = 100 \text{ J}$
• Work done against friction = 30 J (energy lost)
• Net work done on box = Work by applied force - Work against friction = $100 - 30 = 70 \text{ J}$

Alternative: Net work = Change in kinetic energy = 70 J.

3. Answer: D [1]

Explanation: Force is a vector quantity (has magnitude and direction). Energy, power, and work are scalar quantities (magnitude only).

4. Answer: Answer: A [1]

Working:

• Final kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = 240,000 \text{ J}$
• Time = 10 s
• Average power = $\frac{\text{Work done}}{\text{Time}} = \frac{240,000}{10} = 24,000 \text{ W} = 24 \text{ kW}$

5. Answer: B [1]

Explanation: At the lowest point Q, gravitational potential energy is minimum, so kinetic energy is maximum (by conservation of energy). At P and R (highest points), speed is zero, so KE = 0.

6. Answer: B [1]

Working:

• Loss in GPE = Gain in KE (conservation of energy)
• $mgh = 0.5 \times 10 \times 10 = 50 \text{ J}$
• KE just before impact = 50 J

7. Answer: B [1]

Explanation: Water at height has gravitational potential energy → flows down gaining kinetic energy → turns turbines → generates electrical energy.

8. Answer: C [1]

Working:

• Useful work output = Load × distance lifted = $200 \times 3 = 600 \text{ J}$
• Work input = Effort × distance moved by effort = $80 \times 10 = 800 \text{ J}$
• Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{600}{800} \times 100\% = 75\%$

9. Answer: B [1]

Working:

• Work done = Gain in GPE = $mgh = 60 \times 10 \times 4 = 2400 \text{ J}$
• Power = $\frac{\text{Work}}{\text{Time}} = \frac{2400}{5} = 480 \text{ W}$

10. Answer: A [1]

Working:

• For a spring, average force during compression = $\frac{0 + 10}{2} = 5 \text{ N}$ (if linear)
• Elastic PE = Average force × compression = $5 \times 0.05 = 0.25 \text{ J}$
• Or using formula: $EPE = \frac{1}{2}Fx = \frac{1}{2} \times 10 \times 0.05 = 0.25 \text{ J}$

11. Answer: C [1]

Explanation: Work done = $F \cdot s \cdot \cos\theta$. If force is perpendicular to displacement ($\theta = 90^\circ$), $\cos 90^\circ = 0$, so no work is done.

12. Answer: B [1]

Working:

• Loss in GPE = Gain in KE
• $mgh = \frac{1}{2}mv^2$
• $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 3} = \sqrt{60} \approx 7.7 \text{ m/s}$

13. Answer: C [1]

Explanation: Power is defined as the rate of doing work, or rate of energy transfer. $P = \frac{W}{t}$.

14. Answer: A [1]

Explanation: Work done = Force × displacement in direction of force. The barbell is stationary (displacement = 0), so work done = 0 J. The weightlifter exerts a force but does no work on the barbell while holding it stationary.

15. Answer: C [1]

Working:

• Loss in GPE = Gain in KE
• $mg(h_1 - h_2) = \frac{1}{2}mv^2$
• $v = \sqrt{2g(h_1 - h_2)} = \sqrt{2 \times 10 \times (50 - 20)} = \sqrt{600} \approx 24.5 \text{ m/s} \approx 24 \text{ m/s}$

Section B: Structured Questions [30 marks]

16. [5 marks]

(a) GPE at A = $mgh_A = 0.2 \times 10 \times 1.0 = 2.0 \text{ J}$ [1]

(b) At point B (top of loop), height = 0.6 m

• Total energy at A = Total energy at B (conservation)
• $mgh_A = mgh_B + \frac{1}{2}mv_B^2$
• $2.0 = (0.2 \times 10 \times 0.6) + \frac{1}{2}(0.2)v_B^2$
• $2.0 = 1.2 + 0.1v_B^2$
• $0.1v_B^2 = 0.8$
• $v_B^2 = 8$
• $v_B = \sqrt{8} = 2.83 \text{ m/s}$ [2]

Mark breakdown: 1 mark for correct energy equation, 1 mark for correct final answer with unit.

(c) Centripetal force at B = $\frac{mv_B^2}{R} = \frac{0.2 \times 8}{0.3} = \frac{1.6}{0.3} = 5.33 \text{ N}$ [1]

(d) Weight of car = $mg = 0.2 \times 10 = 2 \text{ N}$

• At top of loop, centripetal force required = 5.33 N downwards
• Forces providing centripetal force: Weight (2 N down) + Normal reaction (N down)
• $N + 2 = 5.33 \Rightarrow N = 3.33 \text{ N}$
• Since $N > 0$, the track pushes down on the car, so the car maintains contact with the track. [1]

Marking note: Must state N > 0 or normal reaction is positive to show contact maintained.

17. [5 marks]

(a) Net force = 0 N [1] Explanation: Constant speed means zero acceleration. By Newton's First Law, net force = 0.

(b) Tension = Weight = $mg = 500 \times 10 = 5000 \text{ N}$ [1] Explanation: Net force = 0, so Tension - Weight = 0 → Tension = Weight.

(c) Power = Force × velocity = Tension × speed = $5000 \times 2 = 10,000 \text{ W} = 10 \text{ kW}$ [2] Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(d) Efficiency = $\frac{\text{Output power}}{\text{Input power}} \times 100\%$

• $0.80 = \frac{10,000}{\text{Input power}}$
• Input power = $\frac{10,000}{0.80} = 12,500 \text{ W} = 12.5 \text{ kW}$ [1]

18. [6 marks]

(a) Elastic PE = $\frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 100 \times 0.01 = 1.0 \text{ J}$ [1]

(b) Friction force = $\mu_k N = \mu_k mg = 0.2 \times 0.5 \times 10 = 1.0 \text{ N}$

• Work against friction = friction force × distance = $1.0 \times 0.5 = 0.5 \text{ J}$ [2] Mark breakdown: 1 mark for friction force calculation, 1 mark for work done.

(c) Initial EPE = 1.0 J

• Work done against friction in 0.5 m = 0.5 J
• KE after 0.5 m = Initial EPE - Work against friction = $1.0 - 0.5 = 0.5 \text{ J}$ [2] Mark breakdown: 1 mark for correct energy balance equation, 1 mark for correct answer.

(d) Total distance before rest: All initial EPE converted to work against friction

• $EPE_{initial} = f \times d_{total}$
• $1.0 = 1.0 \times d_{total}$
• $d_{total} = 1.0 \text{ m}$ [1]

19. [5 marks]

(a) GPE lost per second = $mgh$ per second = (mass flow rate) × $g$ × $h$

• $= 500 \times 10 \times 80 = 400,000 \text{ J/s} = 400 \text{ kW}$ [2] Mark breakdown: 1 mark for correct formula/approach, 1 mark for correct answer with unit.

(b) Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{300}{400} \times 100\% = 75\%$ [2] Mark breakdown: 1 mark for correct formula, 1 mark for correct answer.

(c) Any one valid reason, e.g.:

• Energy lost as heat due to friction in turbines/generators
• Sound energy produced
• Kinetic energy of water leaving turbines not fully recovered
• Electrical resistance losses in generators/transmission [1]

20. [4 marks]

(a) Gravitational potential energy → Kinetic energy (+ some thermal energy due to friction) [1]

(b) Graph of $v^2$ against $h$ is a straight line passing through the origin. [1] Explanation: From conservation of energy: $mgh = \frac{1}{2}mv^2 \Rightarrow v^2 = 2gh$. This is of the form $y = mx$ where $y = v^2$, $x = h$, gradient = $2g$. Since $g$ is constant, the graph is linear through origin. [1]

(c) The gradient (18 m/s²) is less than $2g$ (20 m/s²) because energy is lost to friction/air resistance (or work done against friction), so not all GPE is converted to KE. The actual KE gained is less than the GPE lost. [1]

Section C: Longer Structured Questions [15 marks]

21. [8 marks]

(a) Total mechanical energy at A = GPE at A (since KE = 0 at rest)

• $E = mgh_A = 800 \times 10 \times 50 = 400,000 \text{ J} = 400 \text{ kJ}$ [1]

(b) At point B (h = 0), all energy is KE:

• $400,000 = \frac{1}{2} \times 800 \times v_B^2$
• $v_B^2 = \frac{400,000 \times 2}{800} = 1000$
• $v_B = \sqrt{1000} = 31.6 \text{ m/s}$ [2] Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct answer.

(c) At point C (top of loop, h = 30 m):

• $E_{total} = GPE_C + KE_C$
• $400,000 = (800 \times 10 \times 30) + \frac{1}{2}(800)v_C^2$
• $400,000 = 240,000 + 400v_C^2$
• $400v_C^2 = 160,000$
• $v_C^2 = 400$
• $v_C = 20 \text{ m/s}$ [2] Mark breakdown: 1 mark for correct energy equation, 1 mark for correct answer.

(d) At point C (top of loop):

• Centripetal force required = $\frac{mv_C^2}{R} = \frac{800 \times 400}{15} = \frac{320,000}{15} = 21,333 \text{ N}$
• Forces downwards: Weight ($mg = 8000 \text{ N}$) + Normal reaction ($N$)
• $N + 8000 = 21,333$
• $N = 13,333 \text{ N}$ [2] Mark breakdown: 1 mark for centripetal force calculation, 1 mark for normal reaction calculation.

(e) Minimum speed at C for contact: $N = 0$, so $mg = \frac{mv_{min}^2}{R}$

• $v_{min}^2 = gR = 10 \times 15 = 150$
• $v_{min} = \sqrt{150} = 12.2 \text{ m/s}$
• Minimum height $h_{min}$: $mgh_{min} = mg(2R) + \frac{1}{2}mv_{min}^2$
• $h_{min} = 2R + \frac{v_{min}^2}{2g} = 30 + \frac{150}{20} = 30 + 7.5 = 37.5 \text{ m}$ [1] Marking note: 1 mark for both minimum speed and minimum height correctly stated.

22. [7 marks]

(a) Forces on diagram [1]:

• Weight $mg$ vertically downwards
• Normal reaction $N$ perpendicular to plane (upwards)
• Applied force $F = 40 \text{ N}$ up the plane
• Friction $f = \mu_k N$ down the plane (opposing motion)

(b) Work done by applied force = $F \times s = 40 \times 5 = 200 \text{ J}$ [1]

(c) Normal reaction $N = mg\cos\theta = 3 \times 10 \times \cos 30^\circ = 30 \times 0.866 = 25.98 \text{ N}$

• Friction $f = \mu_k N = 0.15 \times 25.98 = 3.897 \text{ N}$
• Work against friction = $f \times s = 3.897 \times 5 = 19.485 \text{ J} \approx 19.5 \text{ J}$ [2] Mark breakdown: 1 mark for normal reaction/friction calculation, 1 mark for work done.

(d) Vertical height gained = $s \sin\theta = 5 \times \sin 30^\circ = 5 \times 0.5 = 2.5 \text{ m}$

• Gain in GPE = $mgh = 3 \times 10 \times 2.5 = 75 \text{ J}$ [1]

(e) Work-energy theorem: Net work done = Change in KE

• Net work = Work by applied force - Work against friction - Gain in GPE
• $= 200 - 19.5 - 75 = 105.5 \text{ J}$
• $\frac{1}{2}mv^2 = 105.5$
• $\frac{1}{2} \times 3 \times v^2 = 105.5$
• $1.5v^2 = 105.5$
• $v^2 = 70.33$
• $v = 8.39 \text{ m/s}$ [2] Mark breakdown: 1 mark for correct work-energy equation setup, 1 mark for correct final answer.

Alternative method: Find net force, then acceleration, then $v^2 = u^2 + 2as$. Both methods accepted.

End of Answer Key

Total Marks: 60