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

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TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Secondary School (AI)

Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper (Version 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.
Take the acceleration of free fall, $g$ , to be $10 \text{ m/s}^2$ .

Section A: Multiple Choice & Short Structured Questions (20 Marks)

Answer all questions in this section.

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 aligns with the 32nd division (where 1 division = 0.01 mm). What is the diameter of the wire? A. 2.532 mm B. 2.82 mm C. 5.70 mm D. 25.32 mm [1]

2. Which of the following pairs consists of one scalar quantity and one vector quantity? A. Speed and Velocity B. Mass and Weight C. Distance and Displacement D. All of the above [1]

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 round trip? A. 48 km/h B. 50 km/h C. 55 km/h D. 100 km/h [1]

4. The velocity-time graph below shows the motion of a lift. (Imagine a graph: Velocity increases linearly from 0 to 4 m/s in 2s, stays constant at 4 m/s for 4s, then decreases linearly to 0 in 2s.) What is the total distance traveled by the lift? A. 16 m B. 20 m C. 24 m D. 32 m [1]

5. A block of mass 5 kg rests on a rough horizontal surface. A horizontal force of 20 N is applied, but the block 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]

6. Newton’s Third Law states that for every action, there is an equal and opposite reaction. Which of the following is a correct action-reaction pair? A. The weight of a book and the normal force from the table. B. The force of gravity on the Earth by the Moon and the force of gravity on the Moon by the Earth. C. The thrust of a rocket and the air resistance on the rocket. D. The pull of a spring and the weight of the mass attached to it. [1]

7. A child of mass 40 kg slides down a vertical rope with a constant acceleration of $2 \text{ m/s}^2$ . Taking $g = 10 \text{ m/s}^2$ , what is the frictional force exerted by the rope on the child? A. 80 N B. 320 N C. 400 N D. 480 N [1]

8. A uniform meter rule is pivoted at the 50 cm mark. A 2 N weight is hung at the 20 cm mark. Where must a 3 N weight be hung to balance the rule horizontally? A. 30 cm mark B. 60 cm mark C. 70 cm mark D. 80 cm mark [1]

9. Why does a wider base and a lower center of gravity increase the stability of an object? A. It increases the weight of the object. B. It requires a larger tilt angle for the line of action of the weight to fall outside the base. C. It reduces the friction between the object and the ground. D. It increases the moment of inertia. [1]

10. A hydraulic press has a small piston of area $0.01 \text{ m}^2$ and a large piston of area $0.5 \text{ m}^2$ . If a force of 50 N is applied to the small piston, what is the maximum load that can be lifted by the large piston (ignoring friction)? A. 1 N B. 100 N C. 2500 N D. 5000 N [1]

11. A diver is at a depth of 20 m in seawater (density $1030 \text{ kg/m}^3$ ). Calculate the pressure due to the seawater alone at this depth. ( $g = 10 \text{ m/s}^2$ ) [2]

12. Define the term moment of a force. [1]

13. State the Principle of Moments. [1]

14. A box of mass 10 kg is pushed up a rough inclined plane. The length of the plane is 5 m and the vertical height gained is 3 m. Calculate the gain in gravitational potential energy of the box. ( $g = 10 \text{ m/s}^2$ ) [2]

15. In the scenario in Q14, if the force applied parallel to the plane is 80 N, calculate the work done by this applied force. [2]

Section B: Structured Questions (30 Marks)

Answer all questions in this section.

16. A student investigates the motion of a trolley using a ticker-tape timer. The trolley is pulled by a string attached to a hanging mass.

(a) The ticker-tape shows dots that are getting further apart at a constant rate. Describe the motion of the trolley. [1]

(b) The student plots a velocity-time graph for the trolley. The graph is a straight line starting from the origin with a gradient of $1.5 \text{ m/s}^2$ . (i) What physical quantity does the gradient of the velocity-time graph represent? [1]

(ii) Calculate the displacement of the trolley after 4 seconds. [2]

(d) If the hanging mass provides a pulling force of 2.0 N, calculate the frictional force acting on the trolley. [2]

17. Figure 17.1 shows a uniform beam AB of length 2.0 m and weight 50 N. The beam is hinged at A and supported by a vertical cable at B. A load of 100 N is placed on the beam at a distance of 0.5 m from A.

(Diagram Description: Horizontal beam AB. Hinge at left end A. Cable at right end B pulling upwards. Load W=100N acting downwards at 0.5m from A. Weight of beam 50N acting downwards at center, 1.0m from A.)

(a) On Figure 17.1, draw and label the forces acting on the beam. Include the weight of the beam, the load, the tension in the cable, and the reaction force at the hinge. [2]

(b) By taking moments about point A, calculate the tension $T$ in the cable. [3]

18. A car of mass 1200 kg is traveling at a constant speed of 20 m/s on a level road. The engine provides a driving force of 800 N.

(a) State the magnitude of the resistive forces acting on the car. Explain your answer. [2]

(b) The driver accelerates the car by increasing the driving force to 2000 N. Assume the resistive forces remain constant at the value calculated in (a). (i) Calculate the resultant force on the car. [1]

(ii) Calculate the acceleration of the car. [2]

(d) Calculate the distance traveled by the car during these 10 seconds. [2]

19. A crane lifts a container of mass 5000 kg vertically upwards from the ground to a height of 15 m in 30 seconds.

(a) Calculate the work done by the crane in lifting the container. ( $g = 10 \text{ m/s}^2$ ) [2]

(b) Calculate the useful power output of the crane. [2]

(d) Explain why the efficiency of the crane is less than 100%. [1]

20. A block of ice of mass 2 kg slides down a smooth curved track from rest at a height of 5 m. At the bottom of the track, it moves onto a rough horizontal surface where it comes to a stop after traveling 10 m. ( $g = 10 \text{ m/s}^2$ )

(a) Calculate the speed of the block at the bottom of the curved track (just before it hits the rough surface). Assume no energy loss on the curved track. [3]

(b) Calculate the average frictional force acting on the block on the horizontal surface. [3]

(c) If the track was not smooth and there was friction on the curved part as well, how would the speed at the bottom compare to your answer in (a)? Explain. [2]

[End of Paper]

Answers

TuitionGoWhere Practice Paper - Physics Secondary 3

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

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]

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).

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]

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]

$v = u + at$ [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]

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]

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]