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Secondary 3 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Practice Paper (AI)

Subject: Physics
Level: Secondary 3
Paper: Practice Paper (Version 1 of 5)
Duration: 1 hour 30 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 of free fall, $g = 10 \text{ m/s}^2$ .

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

Answer all questions in this section.

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

2. A car travels 120 km due North in 2 hours, then turns and travels 60 km due East in 1 hour.
Calculate the average speed of the car for the entire journey.
[2]

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

(Imagine a graph: Velocity increases linearly from 0 to 10 m/s in 5s, stays constant at 10 m/s for 10s, then decreases linearly to 0 in 5s.)

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

4. Define acceleration.
[1]

5. A box of mass 50 kg is pushed across a horizontal floor. The frictional force opposing the motion is 20 N. If the box accelerates at $0.5 \text{ m/s}^2$ , calculate the applied force.
[2]

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

7. Explain why a passenger in a moving bus leans forward when the bus brakes suddenly.
[2]

8. A uniform meter 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?
[2]

9. Calculate the pressure exerted by a column of water 10 m deep.
(Density of water = $1000 \text{ kg/m}^3$ , $g = 10 \text{ N/kg}$ )
[2]

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, calculate the force exerted by the large piston.
[2]

Section B: Structured Questions (20 Marks)

Answer all questions in this section.

11. A stone is dropped from rest from the top of a cliff. It takes 4 seconds to reach the ground. (Ignore air resistance).
(a) Calculate the velocity of the stone just before it hits the ground.
[2]
(b) Calculate the height of the cliff.
[2]
(c) Sketch the displacement-time graph for the stone’s motion. Label the axes clearly.
[2]

12. A block of mass 10 kg is pulled up a rough inclined plane by a force of 80 N parallel to the plane. The plane is inclined at $30^\circ$ to the horizontal. The block moves at a constant speed.
(a) Draw a free-body diagram showing all forces acting on the block.
[2]
(b) Calculate the component of the weight acting down the slope.
[2]
(c) Calculate the frictional force acting on the block.
[2]

13. A crane lifts a load of mass 500 kg vertically to a height of 20 m in 10 seconds.
(a) Calculate the work done by the crane against gravity.
[2]
(b) Calculate the power developed by the crane.
[2]
(c) If the motor of the crane consumes 120,000 J of electrical energy, calculate the efficiency of the crane.
[2]

Section C: Free Response & Application (10 Marks)

Answer all questions in this section.

14. A skydiver jumps from a stationary helicopter.
(a) Describe and explain the motion of the skydiver from the moment she jumps until she reaches terminal velocity. In your answer, refer to the forces acting on her.
[4]
(b) The skydiver opens her parachute. Explain, in terms of forces, why her speed decreases rapidly after opening the parachute.
[3]
(c) State one factor that affects the magnitude of air resistance on the skydiver.
[1]

15. A student investigates the principle of moments using a uniform ruler pivoted at its center.
(a) State the Principle of Moments.
[2]
(b) The student hangs a 100 g mass at the 10 cm mark and a 200 g mass at the 80 cm mark on a 100 cm ruler pivoted at 50 cm.
(i) Show by calculation whether the ruler is in equilibrium.
[2]
(ii) If the ruler is not in equilibrium, state which way it will rotate.
[1]

End of Paper

Answers

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Version 1

Section A: Multiple Choice & Short Structured Questions

1. D
Explanation: Displacement has both magnitude and direction. Mass, speed, and distance are scalars. [1]

2. Total Distance = $120 + 60 = 180 \text{ km}$
Total Time = $2 + 1 = 3 \text{ hours}$
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{180}{3} = 60 \text{ km/h}$
[2] (1 for distance/time, 1 for answer)

3. Distance = Area under graph
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s)
$= (\frac{1}{2} \times 5 \times 10) + (10 \times 10) + (\frac{1}{2} \times 5 \times 10)$
$= 25 + 100 + 25 = 150 \text{ m}$
[2] (1 for method, 1 for answer)

4. Acceleration is the rate of change of velocity.
[1]

5. Resultant Force $F_{net} = ma = 50 \times 0.5 = 25 \text{ N}$
$F_{net} = F_{applied} - F_{friction}$
$25 = F_{applied} - 20$
$F_{applied} = 45 \text{ N}$
[2] (1 for net force, 1 for applied force)

6. An object remains at rest or moves with constant velocity in a straight line unless acted upon by a resultant external force.
[2] (1 for rest/constant velocity, 1 for resultant force condition)

7. The passenger has inertia [1]. When the bus brakes, the bus slows down, but the passenger’s body tends to continue moving forward at the original speed [1].

8. Moment clockwise = Moment anticlockwise
Pivot at 50 cm.
2 N weight is at 20 cm, so distance from pivot $d_1 = 50 - 20 = 30 \text{ cm}$ .
Moment $_1 = 2 \times 30 = 60 \text{ N cm}$ .
Let distance of 3 N weight from pivot be $d_2$ .
$3 \times d_2 = 60$
$d_2 = 20 \text{ cm}$ .
Position = $50 + 20 = 70 \text{ cm}$ mark (or $50 - 20 = 30$ cm mark, but usually opposite side). Assuming opposite side to balance: 70 cm mark.
[2] (1 for moment equation, 1 for position)

9. $P = h \rho g$
$P = 10 \times 1000 \times 10 = 100,000 \text{ Pa}$ (or $100 \text{ kPa}$ )
[2] (1 for formula/substitution, 1 for answer with unit)

10. Pressure is transmitted equally.
$P_1 = P_2 \Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2}$
$\frac{50}{0.01} = \frac{F_2}{0.5}$
$5000 = \frac{F_2}{0.5}$
$F_2 = 2500 \text{ N}$
[2] (1 for principle/equation, 1 for answer)

Section B: Structured Questions

11.
(a) $v = u + at$
$v = 0 + (10 \times 4) = 40 \text{ m/s}$
[2]
(b) $s = ut + \frac{1}{2}at^2$
$s = 0 + \frac{1}{2}(10)(4^2) = 5 \times 16 = 80 \text{ m}$
[2]
(c) Graph: Curve starting from origin, getting steeper (parabola opening upwards).
Y-axis: Displacement (m), X-axis: Time (s).
[2] (1 for shape, 1 for labels)

12.
(a) Diagram showing:

Weight ( $mg$ ) acting vertically downwards.
Normal contact force acting perpendicular to the plane.
Applied force (80 N) acting up the slope.
Friction acting down the slope.
[2] (1 for correct directions, 1 for all 4 forces)
(b) Component of weight down slope = $mg \sin \theta$
$= 10 \times 10 \times \sin(30^\circ)$
$= 100 \times 0.5 = 50 \text{ N}$
[2]
(c) Since speed is constant, acceleration is 0, so resultant force is 0.
Forces up slope = Forces down slope
$F_{applied} = F_{friction} + \text{Weight component}$
$80 = F_{friction} + 50$
$F_{friction} = 30 \text{ N}$
[2] (1 for equilibrium condition, 1 for answer)

13.
(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 $10 \text{ kW}$ )
[2]
(c) Efficiency = $\frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%$
$= \frac{100,000}{120,000} \times 100\% = 83.3\%$
[2] (1 for substitution, 1 for answer)

Section C: Free Response & Application

14.
(a)

Initially, weight is greater than air resistance [1].
There is a resultant downward force, so the skydiver accelerates downwards [1].
As speed increases, air resistance increases [1].
Eventually, air resistance equals weight. Resultant force is zero, and she falls at constant terminal velocity [1].
(b)
Opening the parachute greatly increases the surface area, causing a large increase in air resistance [1].
Air resistance becomes much larger than weight [1].
There is a resultant upward force, causing deceleration (upward acceleration) [1].
(c) Speed / Surface area / Shape / Density of air. (Any one) [1]

15.
(a) For an object in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot. [2]
(b)
(i) Pivot at 50 cm.
Mass 1 (100g = 0.1kg, Weight = 1N) at 10 cm. Distance $d_1 = 40 \text{ cm}$ .
Moment $_1 = 1 \times 40 = 40 \text{ N cm}$ (Anticlockwise).
Mass 2 (200g = 0.2kg, Weight = 2N) at 80 cm. Distance $d_2 = 30 \text{ cm}$ .
Moment $_2 = 2 \times 30 = 60 \text{ N cm}$ (Clockwise).
$40 \neq 60$ , so not in equilibrium. [2]
(ii) Clockwise moment is larger, so it will rotate clockwise. [1]