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

TuitionGoWhere Practice Paper - Science Secondary 2

SA2 Version 3 - Answer Key and Marking Scheme

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

Question 1 [1]

Answer: B (100 J)

Working:

• Gravitational potential energy at start = $mgh = 0.5 \times 10 \times 20 = 100 \text{ J}$
• By conservation of energy, this is converted entirely to kinetic energy just before impact (ignoring air resistance).
• Kinetic energy = 100 J

Key concept: Conservation of mechanical energy for a falling object.

Question 2 [1]

Answer: A (Chemical potential energy $\rightarrow$ Heat energy + Light energy)

Explanation: A candle burns wax (chemical potential energy) releasing heat and light. This is a chemical reaction (combustion) converting stored chemical energy to thermal and radiant energy.

Question 3 [1]

Answer: C (100 J)

Working:

• Work done = Force $\times$ Distance moved in direction of force
• $W = 20 \text{ N} \times 5 \text{ m} = 100 \text{ J}$

Key concept: $W = F \times s$ (only when force and displacement are in the same direction).

Question 4 [1]

Answer: B (360 000 J)

Working:

• Power = 2000 W, Time = 3 minutes = 180 seconds
• Energy = Power $\times$ Time = $2000 \times 180 = 360\,000 \text{ J}$

Common mistake: Forgetting to convert minutes to seconds (would give 6000 J, option A).

Question 5 [1]

Answer: B (Position B only)

Explanation: At the lowest point (B), gravitational potential energy is minimum and kinetic energy is maximum.kinetic energy is maximum. At highest points (A and C), speed is zero so kinetic energy is zero.

Question 6 [1]

Answer: B (16 J)

Working:

• Kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 4^2 = 1 \times 16 = 16 \text{ J}$

Question 7 [1]

Answer: A (Power is the rate of doing work.)

Explanation: Power = Work done / Time taken. Unit is watt (W) = joule per second (J/s). It is a rate, not a total amount.

Question 8 [1]

Answer: A (375 W)

Working:

• Work done against gravity = $mgh = 50 \times 10 \times 3 = 1500 \text{ J}$
• Power = Work / Time = $1500 / 4 = 375 \text{ W}$

Question 9 [1]

Answer: B (Gravitational potential energy $\rightarrow$ Kinetic energy $\rightarrow$ Electrical energy)

Explanation: Water at height has GPE $\rightarrow$ flows down gaining KE $\rightarrow$ turns turbine/generator producing electrical energy.

Question 10 [1]

Answer: C (Elastic potential energy)

Explanation: A compressed (or stretched) spring stores elastic potential energy due to its deformation.

Section B: Structured Questions [30 marks]

Question 11 [4]

(a) [1] GPE at A = $mgh = 500 \times 10 \times 40 = 200\,000 \text{ J}$ (or $2.0 \times 10^5 \text{ J}$)

(b) [1] Total mechanical energy at A = GPE + KE = $200\,000 + 0 = 200\,000 \text{ J}$ (since car is at rest, KE = 0)

(c) [2]

• Total mechanical energy conserved (frictionless track) = $200\,000 \text{ J}$
• GPE at B = $mgh = 500 \times 10 \times 10 = 50\,000 \text{ J}$
• KE at B = Total energy - GPE at B = $200\,000 - 50\,000 = 150\,000 \text{ J}$

Mark breakdown: 1 mark for correct GPE at B, 1 mark for correct KE using conservation.

Question 12 [5]

(a) [1] $a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m/s}^2$

(b) [1] $F = ma = 1200 \times 2.5 = 3000 \text{ N}$

(c) [2] Method 1 (Work-Energy Theorem): Work done = Change in KE = $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2} \times 1200 \times 25^2 - 0 = 375\,000 \text{ J}$

Method 2 (Force $\times$ Distance): Distance $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2.5 \times 10^2 = 125 \text{ m}$ Work done = $F \times s = 3000 \times 125 = 375\,000 \text{ J}$

Mark breakdown: 1 mark for correct method/approach, 1 mark for correct answer with unit.

(d) [1] Average power = Work done / Time = $375\,000 / 10 = 37\,500 \text{ W}$ (or $37.5 \text{ kW}$)

Question 13 [4]

(a) [1] Work done by applied force = $F \times s = 25 \times 4 = 100 \text{ J}$

(b) [1] Gain in GPE = $mgh = 3 \times 10 \times 2 = 60 \text{ J}$

(c) [1] Work done against friction = Work input - Gain in GPE = $100 - 60 = 40 \text{ J}$ (Alternatively: Net work = 0 since constant speed $\rightarrow$ Work by applied force = Work against friction + Gain in GPE)

(d) [1] Work against friction = Friction force $\times$ distance $40 = f \times 4$ $f = 10 \text{ N}$

Question 14 [5]

(a) [1] Work done = Gain in GPE = $mgh = 80 \times 10 \times 15 = 12\,000 \text{ J}$

(b) [1] Useful power output = Work done / Time = $12\,000 / 20 = 600 \text{ W}$

(c) [1] Electrical power input = $VI = 240 \times 5 = 1200 \text{ W}$

(d) [2] Efficiency = $\frac{\text{Useful power output}}{\text{Power input}} \times 100\% = \frac{600}{1200} \times 100\% = 50\%$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with % sign.

Question 15 [4]

(a) [2] Elastic potential energy = $\frac{1}{2}kx^2 = \frac{1}{2} \times 3200 \times (0.5)^2 = 1600 \times 0.25 = 400 \text{ J}$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(b) [2] By conservation of energy: Elastic PE at bottom = GPE at max height $400 = mgh = 40 \times 10 \times h$ $h = \frac{400}{400} = 1.0 \text{ m}$

Mark breakdown: 1 mark for equating energies, 1 mark for correct answer with unit.

Question 16 [4]

(a) [2] Loss in GPE = Gain in KE (ideal, no air resistance) $mgh = \frac{1}{2}mv^2$ $v^2 = 2gh = 2 \times 10 \times 0.3 = 6$ $v = \sqrt{6} \approx 2.45 \text{ m/s}$

Mark breakdown: 1 mark for correct use of $v^2 = 2gh$ or energy conservation, 1 mark for correct answer with unit.

(b) [2] Initial GPE = $mgh = 0.2 \times 10 \times 0.3 = 0.6 \text{ J}$ Actual KE at bottom = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.2 \times (2.2)^2 = 0.1 \times 4.84 = 0.484 \text{ J}$ Energy lost = Initial GPE - Actual KE = $0.6 - 0.484 = 0.116 \text{ J}$

Mark breakdown: 1 mark for calculating actual KE, 1 mark for correct energy lost with unit.

Question 17 [4]

(a) [1] Constant velocity $\rightarrow$ net force = 0 $\rightarrow$ Tension = Weight $T = mg = 2000 \times 10 = 20\,000 \text{ N}$

(b) [1] Useful power output = Force $\times$ velocity = $T \times v = 20\,000 \times 0.5 = 10\,000 \text{ W}$ (or $10 \text{ kW}$)

(c) [2] Efficiency = $\frac{\text{Useful power output}}{\text{Power input}} \times 100\%$ $80\% = \frac{10\,000}{\text{Power input}} \times 100\%$ Power input = $\frac{10\,000}{0.8} = 12\,500 \text{ W}$ (or $12.5 \text{ kW}$)

Mark breakdown: 1 mark for correct rearrangement of efficiency formula, 1 mark for correct answer with unit.

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

Question 18 [6]

(a) [2] Plotting and best-fit line:

• Calculate $v^2$ values: $1.4^2 = 1.96$, $2.0^2 = 4.00$, $2.4^2 = 5.76$, $2.8^2 = 7.84$, $3.1^2 = 9.61$
• Plot points $(h, v^2)$ on grid
• Draw best-fit straight line through origin

Mark breakdown: 1 mark for correct plotting of all 5 points, 1 mark for best-fit straight line through origin.

(b) [1] Gradient = $\frac{\Delta(v^2)}{\Delta h}$ = using points on best-fit line, e.g., $(0.5, 9.61)$ and $(0, 0)$ Gradient = $\frac{9.61 - 0}{0.5 - 0} = 19.22 \text{ m/s}^2$ (accept $19.2 \text{ m/s}^2$ or similar based on drawn line)

(c) [2] Theory: $v^2 = 2gh$, so gradient = $2g$ $g = \frac{\text{gradient}}{2} = \frac{19.22}{2} = 9.61 \text{ m/s}^2$

Mark breakdown: 1 mark for identifying gradient = $2g$, 1 mark for correct calculation of $g$.

(d) [1] Any one valid reason:

• Friction between car and ramp / air resistance
• Rotational kinetic energy of wheels not accounted for (translational KE only measured)
• Measurement errors in height or speed
• Ramp not perfectly rigid

Question 19 [7]

(a) [1] Volume per minute = $0.02 \text{ m}^3$ Mass per minute = Density $\times$ Volume = $1000 \times 0.02 = 20 \text{ kg/min}$

(b) [2] Gain in GPE per minute = $mgh = 20 \times 10 \times 12 = 2400 \text{ J/min}$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(c) [1] Useful power output = Energy per minute / 60 s = $2400 / 60 = 40 \text{ W}$

(d) [2] Efficiency = $\frac{\text{Useful power output}}{\text{Power input}} \times 100\% = \frac{40}{200} \times 100\% = 20\%$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with % sign.

(e) [1] Any two valid suggestions (1 mark for any two):

• Reduce friction in pump mechanism / use better bearings
• Use pipes with larger diameter / smoother interior to reduce fluid friction
• Improve motor efficiency (better electrical motor)
• Reduce leaks in the system
• Optimise pump speed for maximum efficiency point

Question 20 [7]

(a) [1] Initial KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1500 \times 30^2 = 750 \times 900 = 675\,000 \text{ J}$

(b) [1] Final KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1500 \times 10^2 = 750 \times 100 = 75\,000 \text{ J}$

(c) [1] Loss in KE = Initial KE - Final KE = $675\,000 - 75\,000 = 600\,000 \text{ J}$

(d) [2] Electrical energy stored = Efficiency $\times$ Loss in KE = $0.60 \times 600\,000 = 360\,000 \text{ J}$

Mark breakdown: 1 mark for correct use of efficiency, 1 mark for correct answer with unit.

(e) [1] The remaining 40% of the kinetic energy loss (240 000 J) is converted to:

• Heat energy due to friction in brake pads/discs
• Sound energy
• Some heat in the electrical system (wires, motor/generator)

Acceptable answer: "Converted to heat and sound energy due to friction in the braking system."

(f) [1] Any one valid advantage:

• Improves overall fuel efficiency / reduces petrol consumption
• Extends driving range of the car
• Reduces wear on conventional friction brakes
• Recovers energy that would otherwise be wasted as heat

End of Answer Key