Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 4

TuitionGoWhere Practice Paper - Physics Secondary 3 (Version 4)

Answer Key & Marking Scheme

Section A

1. C
Reasoning: Weight is a force ($W=mg$) and has direction (downwards), making it a vector. Mass, speed, and energy are scalars.
[1]

2. A
Reasoning:
Displacement = $120 \text{ km North} - 60 \text{ km South} = 60 \text{ km North}$.
Total time = $2 \text{ h} + 1 \text{ h} = 3 \text{ h}$.
Average Velocity = $\frac{\text{Displacement}}{\text{Time}} = \frac{60}{3} = 20 \text{ km/h North}$.
[1]

3. C
Reasoning: A horizontal line on a v-t graph indicates constant velocity. In the context of falling with air resistance, this constant velocity is terminal velocity where drag equals weight.
[1]

4. B
Reasoning: Since the block does not move, it is in equilibrium. The applied force (20 N) is balanced by the static frictional force. Therefore, friction = 20 N.
[1]

5. B
Reasoning: Resultant $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ N}$.
[1]

6. 500 J
Working:
$W = F \times d$
$W = 50 \times 10 = 500 \text{ J}$
[1]

7. Inertia is the resistance of an object to change its state of motion (or rest).
Accept: "Tendency of an object to remain at rest or in uniform motion unless acted upon by an external force."
[1]

8. 1000 N
Working:
Pressure is transmitted equally: $P_1 = P_2$
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
$\frac{100}{0.01} = \frac{F_2}{0.1}$
$10,000 = \frac{F_2}{0.1}$
$F_2 = 10,000 \times 0.1 = 1000 \text{ N}$
[2] (1 for formula/substitution, 1 for answer)

9. A sharp knife has a very small surface area (contact area).
Since $P = \frac{F}{A}$, for the same force, a smaller area results in a higher pressure.
Higher pressure allows the knife to penetrate the object more easily.
[2] (1 for area/pressure relationship, 1 for explanation)

10. Kinetic energy decreases as the ball slows down.
Gravitational potential energy increases as the ball gains height.
(Kinetic energy is converted to gravitational potential energy).
[2] (1 for KE decrease, 1 for GPE increase/conversion)

Section B

11. (a) Acceleration $a = \frac{v - u}{t}$
$a = \frac{12 - 0}{4} = 3 \text{ m/s}^2$
[2]

(b) Graph Sketch:

• Y-axis: Velocity (m/s), X-axis: Time (s).
• Line from (0,0) to (4,12) [Straight diagonal].
• Line from (4,12) to (14,12) [Horizontal].
• Line from (14,12) to (16,0) [Straight diagonal down].
  [2] (1 for shape, 1 for correct coordinates)

(c) Distance = Area under graph.
Area 1 (Triangle): $\frac{1}{2} \times 4 \times 12 = 24 \text{ m}$
Area 2 (Rectangle): $10 \times 12 = 120 \text{ m}$
Area 3 (Triangle): $\frac{1}{2} \times 2 \times 12 = 12 \text{ m}$
Total Distance = $24 + 120 + 12 = 156 \text{ m}$
[2] (1 for method, 1 for answer)

12. (a) Free Body Diagram:

• Weight ($W$ or $mg$) acting vertically downwards from center.
• Normal Contact Force ($N$ or $R$) acting perpendicular to the slope.
• Friction ($f$) acting down the slope (opposing motion up).
• Tension ($T$) acting up the slope.
  [2] (0.5 per correct force vector)

(b) Component of weight down slope = $mg \sin \theta$
$= 20 \times 10 \times \sin(30^\circ)$
$= 200 \times 0.5 = 100 \text{ N}$
[2]

(c) Since speed is constant, forces are balanced.
$T = \text{Friction} + \text{Weight Component down slope}$
$T = 40 + 100 = 140 \text{ N}$
[2]

13. (a) For an object in equilibrium, the sum of clockwise moments about any pivot is equal to the sum of anticlockwise moments about the same pivot.
[1]

(b) Pivot at 50 cm.
2 N weight at 10 cm: Distance = $50 - 10 = 40 \text{ cm} = 0.4 \text{ m}$.
Moment = $2 \times 0.4 = 0.8 \text{ Nm}$ (Anticlockwise).
Weight $W$ at 80 cm: Distance = $80 - 50 = 30 \text{ cm} = 0.3 \text{ m}$.
Moment = $W \times 0.3$ (Clockwise).
$0.8 = 0.3 W$
$W = \frac{0.8}{0.3} = 2.67 \text{ N}$
[2]

(c) Pivot at 30 cm.
2 N weight at 10 cm: Distance = $30 - 10 = 20 \text{ cm} = 0.2 \text{ m}$.
Moment = $2 \times 0.2 = 0.4 \text{ Nm}$ (Anticlockwise).
Weight $W$ (2.67 N) must create Clockwise moment.
Let distance from pivot be $d$.
$0.4 = 2.67 \times d$
$d = \frac{0.4}{2.67} \approx 0.15 \text{ m} = 15 \text{ cm}$.
Position on rule = Pivot + $d$ = $30 + 15 = 45 \text{ cm}$ mark.
[3] (1 for moment balance, 1 for distance calc, 1 for position)

14. (a) $GPE = mgh$
$GPE = 60 \times 10 \times 10 = 6000 \text{ J}$
[2]

(b) Conservation of Energy: Loss in GPE = Gain in KE
$6000 = \frac{1}{2} mv^2$
$6000 = \frac{1}{2} \times 60 \times v^2$
$6000 = 30 v^2$
$v^2 = 200$
$v = \sqrt{200} \approx 14.14 \text{ m/s}$
[3] (1 for principle, 1 for substitution, 1 for answer)

(c) Some energy is lost to air resistance (work done against air resistance).
This energy is converted to heat/internal energy, so less GPE is converted to KE.
[2]

15. (a) Boyle's Law: $P_1 V_1 = P_2 V_2$ (since T is constant)
$100,000 \times 0.002 = P_2 \times 0.001$
$200 = 0.001 P_2$
$P_2 = \frac{200}{0.001} = 200,000 \text{ Pa}$
[2]

(b) When volume decreases, gas particles are confined to a smaller space.
The frequency of collisions with the walls of the container increases.
Since pressure is force per unit area caused by these collisions, the pressure increases.
[3] (1 for particles closer/freq collisions, 1 for collision with walls, 1 for link to pressure)

Section C

16. (a) $a = \frac{v - u}{t}$
$a = \frac{0 - 20}{5} = -4 \text{ m/s}^2$
Deceleration = $4 \text{ m/s}^2$
[2]

(b) $F = ma$
$F = 1000 \times 4 = 4000 \text{ N}$
[2]

(c) $s = \frac{u + v}{2} \times t$
$s = \frac{20 + 0}{2} \times 5 = 10 \times 5 = 50 \text{ m}$
Alternatively: $v^2 = u^2 + 2as \rightarrow 0 = 400 + 2(-4)s \rightarrow 8s = 400 \rightarrow s = 50$.
[2]

17. (a) Work Done = Force $\times$ Distance
Force = Weight = $mg = 500 \times 10 = 5000 \text{ N}$
$W = 5000 \times 20 = 100,000 \text{ J}$
[2]

(b) Power = $\frac{\text{Work Done}}{\text{Time}}$
$P = \frac{100,000}{10} = 10,000 \text{ W}$ (or 10 kW)
[2]

(c) Efficiency = $\frac{\text{Useful Output Energy}}{\text{Total Input Energy}} \times 100\%$
Efficiency = $\frac{100,000}{120,000} \times 100\%$
Efficiency = $83.3\%$
[2]

18. (a) Beng is correct.
Velocity is a vector quantity (has magnitude and direction).
Although speed (magnitude) is constant, the direction of the satellite is constantly changing.
Therefore, velocity is changing, which means there is acceleration.
[2]

(b) Gravitational force (or Gravity).
[1]

(c) Decrease.
In a higher orbit, the gravitational force is weaker.
To maintain orbit, the required centripetal force is lower, which corresponds to a lower orbital speed ($v = \sqrt{\frac{GM}{r}}$).
[2]

19. (a) Pressure Difference $\Delta P = h \rho g$
$\Delta P = 0.2 \times 13,600 \times 10$
$\Delta P = 27,200 \text{ Pa}$
[2]

(b) If the open arm level is higher, the gas pressure is lower than atmospheric pressure.
$P_{\text{gas}} = P_{\text{atm}} - \Delta P$
$P_{\text{gas}} = 100,000 - 27,200 = 72,800 \text{ Pa}$
[2]

(c) Larger.
Water has a much lower density ($1000 \text{ kg/m}^3$) compared to the original liquid ($13,600 \text{ kg/m}^3$).
Since $\Delta P = h \rho g$, for the same pressure difference, if $\rho$ decreases, $h$ must increase.
[2]

20. (a) The temperature remains constant at $0^\circ\text{C}$.
[1]

(b) The heat energy absorbed is used to overcome the strong forces of attraction between the ice particles (breaking the lattice structure).
The energy increases the potential energy of the particles, not their kinetic energy.
Since temperature is a measure of average kinetic energy, the temperature does not rise.
[3] (1 for overcoming forces, 1 for PE vs KE, 1 for link to temp)

(c) The particles gain kinetic energy.
They move/vibrate faster.
The average speed of the particles increases.
[2]