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

TuitionGoWhere Practice Paper - Combined Science Secondary 3

Version 3 of 5 - ANSWER KEY

SECTION A: Multiple Choice Questions [10 marks]

1. Answer: C

Explanation: At constant speed, kinetic energy is constant (not increasing, so A is wrong). The student does work against friction, and this work is converted to thermal energy (internal energy) of the floor and box. Not all work becomes kinetic energy—that would require acceleration. Gravitational potential energy stays constant on a horizontal floor. Common mistake: thinking "work done" automatically means kinetic energy gain. At constant velocity, net force is zero, so all applied work goes against friction.

2. Answer: B

Explanation: At maximum height, the ball momentarily stops before falling back down. Kinetic energy ($E_k = \frac{1}{2}mv^2$) is zero because $v = 0$. All the initial kinetic energy has been converted to gravitational potential energy ($E_p = mgh$), which is at its maximum. Total energy (kinetic + potential) remains constant if air resistance is ignored. Common mistake: confusing "maximum height" with having maximum speed—it's actually minimum (zero) speed.

3. Answer: B

Explanation: Using Pascal's principle for hydraulic systems: pressure is transmitted equally throughout the fluid.

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

$F_2 = F_1 \times \frac{A_2}{A_1} = 100 \times \frac{0.008}{0.002} = 100 \times 4 = 400 \text{ N}$

Common mistake: dividing by 4 instead of multiplying, or confusing which area goes on top.

4. Answer: C

Explanation: Using $Q = mc\Delta\theta$:

$Q = 2.0 \times 450 \times (120 - 20) = 2.0 \times 450 \times 100 = 90\,000 \text{ J}$

Common mistake: forgetting to calculate temperature change ($\Delta\theta = 100°C$, not 120°C) or mixing up mass and specific heat capacity values.

5. Answer: C

Explanation: Using $v = f\lambda$, so $f = \frac{v}{\lambda} = \frac{2.0}{0.40} = 5.0$ Hz.

SECTION B: Structured Questions [35 marks]

6. (a) [2 marks]

$E_p = mgh = 500 \times 10 \times 25 = 125\,000 \text{ J} = 1.25 \times 10^5 \text{ J}$

Marking: Formula [1], correct substitution and answer with unit [1]

6. (b) [3 marks]

By conservation of energy: $E_p$ at A = $E_k$ at B (assuming no energy loss)

$mgh_A = \frac{1}{2}mv^2$

$v = \sqrt{2gh_A} = \sqrt{2 \times 10 \times 25} = \sqrt{500} = 22.4 \text{ m/s}$

Marking: Statement of energy conservation or correct equating [1], correct formula/substitution [1], correct answer [1]

6. (c) [3 marks]

Actual kinetic energy at B: $E_k = \frac{1}{2} \times 500 \times 18^2 = 81\,000$ J

Expected kinetic energy (from part a): 125 000 J

Energy lost = $125\,000 - 81\,000 = 44\,000$ J = $4.4 \times 10^4$ J

Marking: Correct actual $E_k$ calculation [1], correct subtraction [1], correct answer with unit [1]

Common mistake: Using $v = 18$ in $mgh$ or forgetting to square the velocity.

6. (d) [2 marks]

The car can reach point C because:

• Point C is at 10 m, which requires $E_p = 500 \times 10 \times 10 = 50\,000$ J [1]
• The car still has 81 000 J of kinetic energy at B, which is greater than 50 000 J needed [1]
• Even after losses, sufficient energy remains to reach the lower height of point C

7. (a) [1 mark]

Principle of conservation of energy: electrical energy supplied by heater = thermal energy gained by block (assuming no heat losses)

Or: $VIt = mc\Delta\theta$

7. (b) [2 marks]

$E = VIt = 12 \times 4.0 \times 300 = 14\,400 \text{ J}$

Marking: Formula [1], correct answer [1]

7. (c) [3 marks]

Using $VIt = mc\Delta\theta$:

$14\,400 = 1.0 \times c \times (47 - 22)$

$14\,400 = c \times 25$

$c = \frac{14\,400}{25} = 576 \text{ J/(kg·°C)}$

Marking: Correct $\Delta\theta = 25°C$ [1], correct rearrangement [1], correct answer with unit [1]

Common mistake: Using time in minutes instead of seconds, or using wrong temperature change.

7. (d) [2 marks]

The insulated jacket reduces heat loss to the surroundings [1], ensuring that electrical energy ≈ thermal energy gained by block.

Other modification: Stir the metal block gently to ensure uniform temperature distribution; or use a more sensitive thermometer; or allow more time for temperature to stabilize before recording. [1]

8. (a) [2 marks]

Snell's Law: $n_1 \sin i = n_2 \sin r$

Or in words: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media / equals the refractive index.

Marking: Correct formula or statement [1], correct definition of terms [1]

8. (b) [3 marks]

$n_{air} \sin i = n_{glass} \sin r$

$1.0 \times \sin 45° = 1.52 \times \sin r$

$\sin r = \frac{\sin 45°}{1.52} = \frac{0.707}{1.52} = 0.465$

$r = \sin^{-1}(0.465) = 27.7° ≈ 28°$

Marking: Correct formula/substitution [1], correct calculation of sin r [1], correct answer [1]

8. (c) [2 marks]

As light enters glass: speed decreases [1] because glass is optically denser than air; wavelength decreases [1] while frequency stays constant (since $v = f\lambda$ and $f$ is determined by the source).

8. (d) [2 marks]

$\sin r = \frac{\sin 60°}{1.52} = \frac{0.866}{1.52} = 0.570$

$r = \sin^{-1}(0.570) = 34.7° ≈ 35°$

Marking: Correct substitution [1], correct answer [1]

9. (a) [2 marks]

For 2 complete cycles: 4 horizontal divisions
Time per division = 2 ms

Period for 2 cycles = $4 \times 2 = 8$ ms
Period for 1 cycle = $\frac{8}{2} = 4$ ms = $4 \times 10^{-3}$ s

Marking: Correct method for finding time for cycles [1], correct period [1]

9. (b) [2 marks]

$f = \frac{1}{T} = \frac{1}{4 \times 10^{-3}} = 250 \text{ Hz}$

Marking: Formula [1], correct answer with unit [1]

9. (c) [2 marks]

Peak-to-peak = 3 divisions
Peak-to-peak voltage = $3 \times 0.5 = 1.5$ V

Amplitude = $\frac{1.5}{2} = 0.75$ V

Marking: Correct peak-to-peak voltage [1], correct amplitude [1]

Common mistake: Forgetting to divide by 2 for amplitude (giving 1.5 V instead of 0.75 V).

10. (a) [1 mark]

The substance is melting / undergoing phase change from solid to liquid.

10. (b) [2 marks]

During melting, energy supplied is used to break intermolecular bonds [1] and increase potential energy of molecules, not to increase kinetic energy (which determines temperature) [1]. The temperature remains constant at the melting point during this phase change.

10. (c) [2 marks]

Time = 4 minutes = $4 \times 60 = 240$ s

$E = Pt = 200 \times 240 = 48\,000 \text{ J}$

Marking: Correct time conversion [1], correct energy calculation [1]

10. (d) [3 marks]

Energy supplied = heat for melting = $mL_f$

$48\,000 = 0.20 \times L_f$

$L_f = \frac{48\,000}{0.20} = 240\,000 \text{ J/kg} = 2.4 \times 10^5 \text{ J/kg}$

Marking: Correct equating [1], correct rearrangement [1], correct answer with unit [1]

Common mistake: Using mass in grams or wrong power/time values.

11. (a) [1 mark]

Electrical energy → kinetic energy (rotational) / mechanical energy

11. (b) [3 marks]

• Current flows in coil sides AB and CD which are in a magnetic field [1]
• A current-carrying conductor in a magnetic field experiences a force (motor effect) [1]
• Forces on AB and CD are in opposite directions (Fleming's left-hand rule: AB force up, CD force down, or vice versa depending on field and current directions) [1]
• These equal and opposite forces on opposite sides create a couple/torque, causing rotation

11. (c) [2 marks]

The split-ring commutator reverses the direction of current in the coil every half rotation [1]. This ensures that the forces on AB and CD always produce torque in the same direction, maintaining continuous rotation in one direction [1].

SECTION C: Data Analysis and Application [15 marks]

12. (a) [2 marks]

Expected plot: Six points (0,0), (2, 0.8), (4, 1.6), (6, 2.4), (8, 3.2), (10, 4.0) lie on a straight line; point (12, 5.6) deviates above this line.

Marking: Correct plotting of points [1], suitable line of best fit with last point treated as anomaly or clear indication of deviation [1]

12. (b) [2 marks]

Using linear region: from graph or calculation, gradient = $\frac{4.0}{10.0} = 0.4$ cm/N

For e = 3.6 cm: $F = \frac{3.6}{0.4} = 9.0$ N

Or by reading directly from best-fit line: approximately 9.0 N

Marking: Method shown [1], correct answer [1]

12. (c) [2 marks]

From linear portion (Hooke's Law region):

$k = \frac{F}{e} = \frac{10.0}{4.0} = 2.5 \text{ N/cm}$

Or: $k = \frac{4.0}{1.6} = \frac{6.0}{2.4} = 2.5$ N/cm

Marking: Correct method using linear data [1], correct answer with unit [1]

12. (d) [2 marks]

The 12.0 N load exceeds the elastic limit / limit of proportionality of the spring [1]. Beyond this point, Hooke's Law no longer applies; the spring undergoes plastic deformation and does not return to original length, causing greater extension per unit load [1].

13. (a) [3 marks]

Efficiency = $\frac{\text{useful power output}}{\text{total power input}} \times 100\%$

$0.35 = \frac{600}{\text{power input}}$

$\text{Power input} = \frac{600}{0.35} = 1714.3 \text{ MW}$

Energy input per hour = $1714.3 \times 10^6 \times 3600 = 6.17 \times 10^{12}$ J

Or: 1714.3 MWh

Marking: Correct formula/rearrangement [1], correct power input [1], correct energy in suitable unit [1]

13. (b) [3 marks]

For 200 MW demand:

Power Station B (natural gas): Need $\frac{200}{0.50} = 400$ MW thermal input. CO₂ = $400 \times 0.35 = 140$ kg/hr. Smaller land footprint, moderate emissions. [1]

Power Station D (solar): Need $\frac{200}{0.15} = 1333$ MW installed capacity (area-dependent). Zero direct CO₂ but requires large land area for panels, intermittent supply needs storage. [1]

Comparison: B is more land-efficient and reliable but emits CO₂; D is renewable with no direct emissions but needs much more space and backup systems for continuous supply. [1]

13. (c) [2 marks]

Nuclear fission does not produce CO₂ as no fossil fuel combustion occurs [1]. However, the process generates large amounts of waste heat from the reactor that must be dissipated, typically into cooling water or air, causing thermal pollution [1].

14. (a) [2 marks]

$E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 4000 \times 30^2 = \frac{1}{2} \times 4000 \times 900 = 1\,800\,000 \text{ J} = 1.8 \times 10^6 \text{ J}$

Marking: Formula [1], correct answer [1]

14. (b) [2 marks]

$\text{Energy supplied} = \frac{E_k}{\text{efficiency}} = \frac{1.8 \times 10^6}{0.75} = 2.4 \times 10^6 \text{ J}$

Marking: Correct division by efficiency [1], correct answer [1]

14. (c) [3 marks]

Energy after 10% loss during climb: $0.90 \times 1.8 \times 10^6 = 1.62 \times 10^6$ J

Energy needed to reach 45 m: $E_p = mgh = 4000 \times 10 \times 45 = 1\,800\,000 = 1.8 \times 10^6$ J

Comparison: $1.62 \times 10^6$ J < $1.8 \times 10^6$ J

Conclusion: The train will not successfully reach the top of the 45 m hill [1]. The kinetic energy remaining after launch losses is insufficient to overcome the gravitational potential energy required, falling short by $1.8 \times 10^6 - 1.62 \times 10^6 = 0.18 \times 10^6$ J or 180 kJ.

Marking: Correct calculation of remaining energy [1], correct required potential energy [1], correct conclusion with justification [1]

TOTAL MARKS: 60