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

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Questions

TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Secondary School (AI)
Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper (Version 1 of 5)
Duration: 1 hour 15 minutes
Total Marks: 50

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates:

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
Take the acceleration due to gravity, $g = 10 \text{ m/s}^2$ .

Section A: Multiple Choice & Short Structured Questions [20 marks]

1. A student measures the diameter of a wire using a micrometer screw gauge. The main scale reads 2.5 mm and the thimble scale reads 32 divisions. If one division on the thimble scale represents 0.01 mm, what is the diameter of the wire?
[1]

A. 2.532 mm
B. 2.82 mm
C. 2.5032 mm
D. 2.32 mm

Answer: _______________

2. Which of the following pairs contains one scalar quantity and one vector quantity?
[1]

A. Speed and Velocity
B. Mass and Weight
C. Distance and Displacement
D. All of the above

Answer: _______________

3. A car travels from Town A to Town B at an average speed of 60 km/h and returns from Town B to Town A at an average speed of 40 km/h. What is the average speed for the entire journey?
[2]

A. 50 km/h
B. 48 km/h
C. 55 km/h
D. 100 km/h

Answer: _______________

4. The velocity-time graph below shows the motion of a lift moving between floors.

(Imagine a graph: Velocity starts at 0, increases linearly to 4 m/s in 2s, stays constant at 4 m/s for 5s, then decreases linearly to 0 in 2s.)

Calculate the total distance traveled by the lift.
[2]

5. A block of mass 5 kg rests on a rough horizontal surface. A horizontal force of 20 N is applied to the block, but it does not move. What is the magnitude of the frictional force acting on the block?
[1]

A. 0 N
B. 20 N
C. 50 N
D. Greater than 20 N

Answer: _______________

6. State Newton’s First Law of Motion.
[2]

7. A skydiver of mass 80 kg falls through the air. Initially, he accelerates downwards. Eventually, he reaches a constant terminal velocity.
(a) Explain, in terms of forces, why the skydiver reaches terminal velocity.
[2]

(b) Calculate the air resistance acting on the skydiver when he is at terminal velocity.
[1]

8. A uniform metre rule is pivoted at the 50 cm mark. A weight of 2 N is hung at the 20 cm mark. Where must a weight of 3 N be hung to balance the rule horizontally?
[2]

9. Define the moment of a force.
[1]

10. A hydraulic press consists of two pistons, A and B. Piston A has an area of $0.01 \text{ m}^2$ and Piston B has an area of $0.1 \text{ m}^2$ . A force of 100 N is applied to Piston A.
(a) Calculate the pressure transmitted through the liquid.
[1]

(b) Calculate the output force exerted by Piston B.
[1]

Section B: Structured Questions [30 marks]

11. A student investigates the motion of a trolley rolling down an inclined plane. The trolley starts from rest. The distance $s$ traveled by the trolley is recorded at different times $t$ .

Time $t$ (s)	0.0	0.5	1.0	1.5	2.0
Distance $s$ (m)	0.0	0.2	0.8	1.8	3.2

(a) Plot a graph of distance $s$ (y-axis) against time squared $t^2$ (x-axis) on the grid provided below.
[3]

(Grid space for plotting)

(b) Determine the gradient of the graph.
[2]

12. A box of mass 12 kg is pulled along a horizontal floor by a rope inclined at $30^\circ$ to the horizontal. The tension in the rope is 50 N. The box moves at a constant velocity.

(a) Draw a free-body diagram showing all the forces acting on the box. Label the forces clearly (Weight, Normal Reaction, Tension, Friction).
[3]

(b) Calculate the horizontal component of the tension.
[1]

(d) Calculate the vertical component of the tension.
[1]

(e) Hence, calculate the normal reaction force exerted by the floor on the box.
[2]

13. A crane lifts a load of mass 500 kg vertically upwards through a height of 20 m in 10 seconds.

(a) Calculate the work done by the crane in lifting the load.
[2]

(b) Calculate the power developed by the crane.
[2]

(c) The crane motor consumes 150,000 J of electrical energy to perform this lift. Calculate the efficiency of the crane.
[2]

14. A diver jumps from a platform 10 m above the water surface. Assume air resistance is negligible.

(a) State the principle of conservation of energy.
[1]

(b) Calculate the speed of the diver just before he hits the water.
[3]

15. Two blocks, A (mass 2 kg) and B (mass 3 kg), are connected by a light inextensible string on a smooth horizontal surface. A force of 20 N is applied to block B, pulling both blocks to the right.

(a) Calculate the acceleration of the system.
[2]

(b) Calculate the tension in the string connecting the two blocks.
[2]

(c) If the surface was rough and a frictional force of 5 N acted on each block, would the acceleration increase, decrease, or remain the same? Explain your answer.
[2]

End of Paper

Answers

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.

(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.