Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper - Physics Secondary 3

Answer Key & Marking Scheme Paper: SA2 Practice Paper (Version 5)

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Section A: Multiple Choice & Short Structured Questions

1. B

• Reading = Main scale + (Thimble $\times$ Precision)
• Reading = $2.5 \text{ mm} + (32 \times 0.01 \text{ mm}) = 2.5 + 0.32 = 2.82 \text{ mm}$.

2. D

• Speed (scalar) / Velocity (vector).
• Mass (scalar) / Weight (vector).
• Distance (scalar) / Displacement (vector).
• All pairs contain one scalar and one vector.

3. A

• Let distance one way be $d$. Total distance = $2d$.
• Time there $t_1 = d/60$. Time back $t_2 = d/40$.
• Total time $T = d/60 + d/40 = (2d + 3d)/120 = 5d/120 = d/24$.
• Average Speed = Total Distance / Total Time = $2d / (d/24) = 48 \text{ km/h}$.

4. C

• Distance = Area under v-t graph.
• Area = Triangle (0-2s) + Rectangle (2-6s) + Triangle (6-8s).
• Area = $\frac{1}{2}(2)(4) + (4)(4) + \frac{1}{2}(2)(4) = 4 + 16 + 4 = 24 \text{ m}$.

5. B

• Block is stationary (equilibrium).
• Horizontal forces must balance.
• Friction = Applied Force = 20 N.

6. B

• Action-reaction pairs act on different objects and are of the same type of force.
• A: Act on same object (book). Different types (gravitational vs electromagnetic/contact).
• B: Earth pulls Moon, Moon pulls Earth. Correct.
• C: Thrust (gas on rocket) vs Air Resistance (air on rocket). Act on same object.
• D: Spring pull vs Weight. Act on same object.

7. B

• Forces: Weight ($mg$) down, Friction ($f$) up.
• Resultant force $F_{net} = ma$ (downwards).
• $mg - f = ma$
• $40(10) - f = 40(2)$
• $400 - f = 80$
• $f = 320 \text{ N}$.

8. C

• Pivot at 50 cm.
• 2 N weight at 20 cm: Distance from pivot = $50 - 20 = 30 \text{ cm}$.
• Moment = $2 \times 30 = 60 \text{ N cm}$ (Anticlockwise).
• 3 N weight must provide 60 N cm (Clockwise).
• $3 \times d = 60 \Rightarrow d = 20 \text{ cm}$.
• Position = $50 + 20 = 70 \text{ cm}$ mark.

9. B

• Stability depends on the position of the center of gravity relative to the base.
• A lower CG and wider base mean the object must be tilted further before the line of action of the weight falls outside the base, causing it to topple.

10. C

• Pascal's Principle: $P_1 = P_2 \Rightarrow F_1/A_1 = F_2/A_2$.
• $50 / 0.01 = F_2 / 0.5$.
• $5000 = F_2 / 0.5$.
• $F_2 = 2500 \text{ N}$.

11. Pressure due to seawater:

• Formula: $P = h \rho g$ [1]
• Substitution: $P = 20 \times 1030 \times 10$ [1]
• Answer: $206,000 \text{ Pa}$ (or $206 \text{ kPa}$) [1] (Note: Question asks for pressure due to seawater alone, so atmospheric pressure is not added.)

12. Definition of Moment:

• The turning effect of a force [1].
• (Accept: Product of force and perpendicular distance from the pivot).

13. Principle of Moments:

• For an object in rotational equilibrium, the sum of clockwise moments about any pivot is equal to the sum of anticlockwise moments about the same pivot. [1]

14. Gain in GPE:

• Formula: $\Delta GPE = mgh$ [1]
• Substitution: $10 \times 10 \times 3$ [1]
• Answer: $300 \text{ J}$ [1]

15. Work Done by Applied Force:

• Formula: $W = F \times d$ [1]
• Note: $d$ is the distance moved in the direction of the force (along the plane).
• Substitution: $80 \times 5$ [1]
• Answer: $400 \text{ J}$ [1]

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Section B: Structured Questions

16. Trolley Motion Investigation

(a) Description of motion:

• Uniform acceleration / Constant acceleration. [1] (Reason: Dots getting further apart at constant rate implies velocity increasing uniformly).

(b) (i) Gradient represents:

• Acceleration. [1]

(b) (ii) Displacement after 4s:

• Method 1 (Area under graph): Graph is triangle. Base=4, Height=$v=at=1.5 \times 4 = 6$. Area = $\frac{1}{2} \times 4 \times 6 = 12 \text{ m}$.
• Method 2 (Formula): $s = ut + \frac{1}{2}at^2$. $u=0, a=1.5, t=4$.
• $s = 0 + \frac{1}{2}(1.5)(4^2) = 0.75 \times 16 = 12 \text{ m}$. [2] (1 mark for correct formula/substitution, 1 mark for answer).

(c) Resultant Force:

• Formula: $F = ma$ [1]
• Substitution: $F = 0.8 \times 1.5$ [1]
• Answer: $1.2 \text{ N}$ [1]

(d) Frictional Force:

• Resultant Force = Pulling Force - Friction [1]
• $1.2 = 2.0 - f$
• $f = 2.0 - 1.2 = 0.8 \text{ N}$ [1]

17. Uniform Beam Equilibrium

(a) Forces Diagram:

• Weight of beam (50 N) acting downwards at center (1.0 m from A). [1]
• Load (100 N) acting downwards at 0.5 m from A. [1]
• Tension ($T$) acting upwards at B (2.0 m from A). [1]
• Reaction at A (vertical component upwards, horizontal component potentially, but usually just vertical shown in simple problems unless specified. Accept vertical reaction $R_A$ upwards). [1] (Award marks for correct direction and position. Max 2 marks).

(b) Calculate Tension $T$:

• Take moments about A. [1]
• Clockwise Moments = Anticlockwise Moments.
• $(100 \times 0.5) + (50 \times 1.0) = T \times 2.0$ [1]
• $50 + 50 = 2T$
• $100 = 2T$
• $T = 50 \text{ N}$ [1]

(c) Vertical Reaction at A ($R_A$):

• Upward Forces = Downward Forces (Vertical Equilibrium). [1]
• $R_A + T = 100 + 50$
• $R_A + 50 = 150$
• $R_A = 100 \text{ N}$ [1]

18. Car Dynamics

(a) Resistive Forces:

• Magnitude: 800 N. [1]
• Reason: The car is moving at constant speed, so acceleration is zero. By Newton's First Law, the resultant force is zero, meaning driving force equals resistive force. [1]

(b) (i) Resultant Force:

• $F_{net} = F_{drive} - F_{resist}$
• $F_{net} = 2000 - 800 = 1200 \text{ N}$ [1]

(b) (ii) Acceleration:

• $a = F_{net} / m$ [1]
• $a = 1200 / 1200 = 1 \text{ m/s}^2$ [1]

(c) Final Speed:

• $v = u + at$ [1]
• $v = 20 + (1)(10) = 30 \text{ m/s}$ [1]

(d) Distance Traveled:

• $s = ut + \frac{1}{2}at^2$ OR $s = \frac{u+v}{2} \times t$ [1]
• $s = 20(10) + \frac{1}{2}(1)(100) = 200 + 50 = 250 \text{ m}$ [1] (Or $s = \frac{20+30}{2} \times 10 = 25 \times 10 = 250 \text{ m}$).

19. Crane Power and Efficiency

(a) Work Done:

• Force required = Weight = $mg = 5000 \times 10 = 50,000 \text{ N}$. [1]
• $W = F \times d = 50,000 \times 15$ [1]
• $W = 750,000 \text{ J}$ (or $750 \text{ kJ}$) [1]

(b) Useful Power Output:

• $P = W / t$ [1]
• $P = 750,000 / 30$ [1]
• $P = 25,000 \text{ W}$ (or $25 \text{ kW}$) [1]

(c) Efficiency:

• Efficiency = (Useful Power Output / Total Power Input) $\times 100\%$ [1]
• Input Power = 30 kW = 30,000 W.
• Efficiency = $(25,000 / 30,000) \times 100\%$ [1]
• Efficiency = $83.3\%$ [1]

(d) Reason for Efficiency < 100%:

• Energy is lost/work is done against friction in the moving parts of the crane / air resistance / heating of the motor. [1]

20. Energy Conservation

(a) Speed at Bottom:

• Conservation of Energy: Loss in GPE = Gain in KE. [1]
• $mgh = \frac{1}{2}mv^2$
• $gh = \frac{1}{2}v^2 \Rightarrow v = \sqrt{2gh}$ [1]
• $v = \sqrt{2 \times 10 \times 5} = \sqrt{100}$ [1]
• $v = 10 \text{ m/s}$ [1]

(b) Average Frictional Force:

• Work Done by Friction = Loss in Kinetic Energy. [1]
• $KE_{initial} = \frac{1}{2}mv^2 = \frac{1}{2}(2)(10^2) = 100 \text{ J}$.
• $KE_{final} = 0 \text{ J}$.
• Work Done $W = F \times d$. [1]
• $100 = F \times 10$
• $F = 10 \text{ N}$ [1]

(c) Effect of Friction on Curved Track:

• The speed would be lower (less than 10 m/s). [1]
• Explanation: Some of the initial gravitational potential energy would be converted into heat/internal energy due to work done against friction on the curved track, leaving less energy to be converted into kinetic energy. [1]