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

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TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Secondary School (AI)
Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper (Version 4 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 calculator.
Take the acceleration due to gravity, $g$ , to be $10 \text{ m/s}^2$ .

Section A: Multiple Choice & Short Structured Questions

Answer all questions in this section. Questions 1–10 carry 1–2 marks each.

1. Which of the following is a vector quantity?
A. Mass
B. Speed
C. Weight
D. Energy
[1]

2. A car travels 120 km North in 2 hours, then turns and travels 60 km South in 1 hour. What is the average velocity of the car for the entire journey?
A. 20 km/h North
B. 60 km/h North
C. 60 km/h South
D. 180 km/h North
[1]

3. The graph below shows the velocity-time graph of a falling object.

(Imagine a graph where velocity increases linearly from 0 to 20 m/s in 2 seconds, then curves to become horizontal at 20 m/s)

What does the horizontal section of the graph represent?
A. The object has stopped moving.
B. The object is accelerating at $10 \text{ m/s}^2$ .
C. The object has reached terminal velocity.
D. Air resistance is zero.
[1]

4. 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?
A. 0 N
B. 20 N
C. 50 N
D. Greater than 20 N
[1]

5. Two forces, 3 N and 4 N, act on an object at right angles to each other. What is the magnitude of the resultant force?
A. 1 N
B. 5 N
C. 7 N
D. 12 N
[1]

6. A student pushes a box with a force of 50 N along a corridor for a distance of 10 m. The force is applied in the direction of motion. Calculate the work done by the student.
[1]

7. Define the term inertia.
[1]

8. 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 $0.1 \text{ m}^2$ . If a force of 100 N is applied to Piston A, calculate the force exerted by Piston B.
[2]

9. Explain why a sharp knife cuts better than a blunt knife, referring to the concept of pressure.
[2]

10. A ball is thrown vertically upwards. Describe the energy changes that occur from the moment it leaves the hand until it reaches its maximum height. (Ignore air resistance).
[2]

Section B: Structured Questions

Answer all questions in this section. Questions 11–15 carry 3–5 marks each.

11. A cyclist accelerates uniformly from rest to a speed of $12 \text{ m/s}$ in 4 seconds. She then maintains this constant speed for 10 seconds before braking uniformly to rest in 2 seconds.

(a) Calculate the acceleration of the cyclist during the first 4 seconds.
[2]

(b) Sketch the velocity-time graph for the entire motion described above. Label the axes with appropriate values.
[2]

12. A box of mass 20 kg is pulled up a rough inclined plane by a rope parallel to the slope. The box moves at a constant speed. The angle of inclination is $30^\circ$ . The frictional force acting on the box is 40 N.

(a) Draw a free-body diagram showing all the forces acting on the box. Label the forces clearly (Weight, Normal Contact Force, Friction, Tension).
[2]

(b) Calculate the component of the weight acting down the slope.
[2]

13. A uniform metre rule is pivoted at the 50 cm mark. A weight of 2 N is hung at the 10 cm mark. Another weight $W$ is hung at the 80 cm mark to balance the rule horizontally.

(a) State the Principle of Moments.
[1]

(b) Calculate the value of weight $W$ .
[2]

(c) The pivot is now moved to the 30 cm mark. The 2 N weight remains at the 10 cm mark. Where must weight $W$ be placed to balance the rule again? (Assume the weight of the rule is negligible).
[3]

14. A diver of mass 60 kg jumps from a diving board 10 m above the water surface.

(a) Calculate the gravitational potential energy of the diver relative to the water surface before jumping.
[2]

(b) Assuming negligible air resistance, calculate the speed of the diver just before entering the water.
[3]

(c) In reality, the diver enters the water at a speed slightly lower than calculated in (b). Explain why, in terms of energy conservation.
[2]

15. A gas is trapped in a cylinder by a movable piston. The volume of the gas is $0.002 \text{ m}^3$ and the pressure is $100,000 \text{ Pa}$ . The piston is pushed in slowly until the volume is $0.001 \text{ m}^3$ . The temperature of the gas remains constant.

(a) Calculate the new pressure of the gas.
[2]

(b) Explain, in terms of the kinetic particle model, why the pressure increases when the volume decreases.
[3]

Section C: Free Response & Application

Answer all questions in this section. Questions 16–20 carry 4–6 marks each.

16. A car of mass 1000 kg is traveling at $20 \text{ m/s}$ . The driver sees an obstacle and applies the brakes. The car comes to a stop in 5 seconds.

(a) Calculate the deceleration of the car.
[2]

(b) Calculate the braking force acting on the car.
[2]

17. A crane lifts a load of 500 kg vertically 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 motor of the crane consumes 120,000 J of electrical energy to perform this lift. Calculate the efficiency of the crane.
[2]

18. Two students, Ali and Beng, are discussing the motion of a satellite orbiting the Earth at constant speed.

Ali says: "Since the speed is constant, the velocity is constant, so there is no acceleration."
Beng says: "The velocity is changing, so there is acceleration."

(a) Who is correct? Explain your answer.
[2]

(b) Identify the force responsible for keeping the satellite in orbit.
[1]

(c) If the satellite were to move to a higher orbit, would its orbital speed increase, decrease, or remain the same? Explain briefly.
[2]

19. A U-tube manometer is connected to a gas supply. One end is open to the atmosphere. The difference in height between the two liquid columns is 0.2 m. The liquid used has a density of $13,600 \text{ kg/m}^3$ . Atmospheric pressure is $100,000 \text{ Pa}$ .

(a) Calculate the pressure difference between the gas supply and the atmosphere.
[2]

(b) Determine the absolute pressure of the gas supply if the level in the open arm is higher than the level in the arm connected to the gas.
[2]

(c) If water (density $1000 \text{ kg/m}^3$ ) were used instead of the original liquid, would the height difference be larger, smaller, or the same? Explain.
[2]

20. A block of ice at $0^\circ\text{C}$ is placed in a warm room.

(a) Describe what happens to the temperature of the ice as it melts.
[1]

(b) Explain, in terms of particle energy, why the temperature remains constant during melting even though heat is being absorbed.
[3]

(c) Once all the ice has melted, the water continues to absorb heat. Describe how the motion of the water particles changes as the temperature rises from $0^\circ\text{C}$ to $20^\circ\text{C}$ .
[2]

End of Paper

Answers

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)

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