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

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Paper: SA2 Practice Paper (Version 1 of 5)
Total Marks: 50

Section A: Multiple Choice & Short Structured Questions

1. B [1]
Reasoning: Reading = Main Scale + (Thimble $\times$ Precision) = $2.5 + (32 \times 0.01) = 2.5 + 0.32 = 2.82$ mm.

2. D [1]
Reasoning: Speed (scalar) vs Velocity (vector); Mass (scalar) vs Weight (vector); Distance (scalar) vs Displacement (vector). All pairs fit the criteria.

3. B [2]
Reasoning: Let distance one way be $d$. Total distance = $2d$.
Time$_1$ = $d/60$, Time$_2$ = $d/40$.
Total Time = $d/60 + d/40 = (2d + 3d)/120 = 5d/120 = d/24$.
Average Speed = Total Distance / Total Time = $2d / (d/24) = 48$ km/h.

4. 26 m [2]
Reasoning: Distance = Area under v-t graph.
Area = Area of triangle (acceleration) + Area of rectangle (constant) + Area of triangle (deceleration).
Area 1 = $\frac{1}{2} \times 2 \times 4 = 4$ m.
Area 2 = $5 \times 4 = 20$ m.
Area 3 = $\frac{1}{2} \times 2 \times 4 = 4$ m.
Total = $4 + 20 + 4 = 28$ m.
(Correction: Wait, let's re-read the graph description in Q4. "increases... in 2s, constant... for 5s, decreases... in 2s". Total time = 9s. Area = $0.5(2)(4) + 5(4) + 0.5(2)(4) = 4 + 20 + 4 = 28$ m.)
Self-Correction for Answer Key: The calculated answer is 28 m.
Answer: 28 m

5. B [1]
Reasoning: Since the block does not move, it is in equilibrium. The applied force is balanced by static friction. $f_s = F_{applied} = 20$ N.

6. Newton's First Law: [2]
An object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force.
(1 mark for "rest or uniform motion/constant velocity", 1 mark for "unless acted on by resultant force")

7. (a) Explanation: [2]
Initially, weight is greater than air resistance, so there is a resultant downward force causing acceleration. As speed increases, air resistance increases. Eventually, air resistance equals weight. The resultant force becomes zero, so acceleration becomes zero and velocity becomes constant (terminal velocity).

(b) Air Resistance: [1]
At terminal velocity, forces are balanced.
Air Resistance = Weight = $mg = 80 \times 10 = 800$ N.

8. 40 cm mark [2]
Reasoning: Principle of Moments: Clockwise Moment = Anticlockwise Moment.
Pivot at 50 cm.
Weight 2 N at 20 cm: Distance from pivot = $50 - 20 = 30$ cm.
Anticlockwise Moment = $2 \text{ N} \times 30 \text{ cm} = 60 \text{ N cm}$.
Let $x$ be the distance of the 3 N weight from the pivot on the other side.
Clockwise Moment = $3 \text{ N} \times x$.
$3x = 60 \Rightarrow x = 20$ cm.
Position = $50 \text{ cm} + 20 \text{ cm} = 70$ cm mark?
Wait, let's check sides. 20 cm is to the left of 50 cm. So 3 N must be to the right.
Position = $50 + 20 = 70$ cm.
Re-reading question: "Where must a weight... be hung".
Answer: At the 70 cm mark.

9. Definition: [1]
The product of the force and the perpendicular distance from the pivot to the line of action of the force.

10. (a) Pressure: [1]
$P = F/A = 100 / 0.01 = 10,000$ Pa (or N/m²).

(b) Output Force: [1]
$F = P \times A = 10,000 \times 0.1 = 1,000$ N.

Section B: Structured Questions

11. (a) Graph: [3]

• Axes labeled correctly with units ($s$/m and $t^2$/s²). [1]
• Points plotted correctly: (0,0), (0.25, 0.2), (1.0, 0.8), (2.25, 1.8), (4.0, 3.2). [1]
• Best-fit straight line drawn through the origin. [1]

(b) Gradient: [2]
Using points from the line, e.g., (4.0, 3.2) and (0,0).
Gradient = $\frac{\Delta y}{\Delta x} = \frac{3.2 - 0}{4.0 - 0} = 0.8$.
Answer: 0.8 m/s² (Note: Unit of gradient is m/s² because y is m and x is s²).

(c) Acceleration: [2]
Equation: $s = \frac{1}{2}at^2$. This is in the form $y = mx$ where $m = \frac{1}{2}a$.
Gradient = $\frac{1}{2}a$.
$0.8 = \frac{1}{2}a$.
$a = 1.6$ m/s².

12. (a) Free-Body Diagram: [3]

• Weight ($W$ or $mg$) acting vertically downwards from center. [1]
• Normal Reaction ($N$ or $R$) acting vertically upwards from contact point. [1]
• Tension ($T$) acting at $30^\circ$ above horizontal to the right. [1]
• Friction ($f$) acting horizontally to the left. [1]
  (Max 3 marks, deduct for missing labels or wrong directions)

(b) Horizontal Tension: [1]
$T_x = T \cos(30^\circ) = 50 \cos(30^\circ) = 50 \times 0.866 = 43.3$ N.

(c) Frictional Force: [1]
Since velocity is constant, horizontal forces are balanced.
$f = T_x = 43.3$ N.

(d) Vertical Tension: [1]
$T_y = T \sin(30^\circ) = 50 \sin(30^\circ) = 50 \times 0.5 = 25$ N.

(e) Normal Reaction: [2]
Vertical forces are balanced.
Upward forces = Downward forces.
$N + T_y = W$.
$N + 25 = mg = 12 \times 10 = 120$.
$N = 120 - 25 = 95$ N.

13. (a) Work Done: [2]
Force required to lift = Weight = $mg = 500 \times 10 = 5000$ N.
Work = Force $\times$ Distance = $5000 \times 20 = 100,000$ J.

(b) Power: [2]
Power = Work / Time = $100,000 / 10 = 10,000$ W (or 10 kW).

(c) Efficiency: [2]
Efficiency = (Useful Energy Output / Total Energy Input) $\times$ 100%.
Efficiency = $(100,000 / 150,000) \times 100\% = 66.7\%$.

14. (a) Principle: [1]
Energy cannot be created or destroyed, only converted from one form to another.

(b) Speed: [3]
Loss in GPE = Gain in KE.
$mgh = \frac{1}{2}mv^2$.
$gh = \frac{1}{2}v^2$.
$v^2 = 2gh = 2 \times 10 \times 10 = 200$.
$v = \sqrt{200} = 14.14$ m/s.

(c) Explanation: [1]
Work is done against air resistance, so some gravitational potential energy is converted to heat/internal energy instead of kinetic energy.

15. (a) Acceleration: [2]
Total Mass = $2 + 3 = 5$ kg.
Resultant Force = 20 N (smooth surface).
$a = F/m = 20 / 5 = 4$ m/s².

(b) Tension: [2]
Consider Block A (mass 2 kg). The only horizontal force acting on it is Tension $T$.
$T = m_A \times a = 2 \times 4 = 8$ N.
(Alternatively, consider Block B: $20 - T = m_B a \Rightarrow 20 - T = 3(4) \Rightarrow T = 8$ N).

(c) Effect of Friction: [2]
Decrease.
New Resultant Force = Applied Force - Total Friction = $20 - (5 + 5) = 10$ N.
New Acceleration = $10 / 5 = 2$ m/s².
Since the net force decreases while mass remains constant, acceleration decreases.