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Secondary 3 Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Practice Paper (AI)

Subject: Physics
Level: Secondary 3
Paper: Practice Paper (Version 2 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60

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, $g = 10 \text{ m/s}^2$ .
Assume the density of water is $1000 \text{ kg/m}^3$ unless stated otherwise.

Section A: Structured Questions (40 Marks)

Answer all questions in this section.

1. A student measures the length of a pendulum string using a metre rule. The reading at the top of the string is $2.0 \text{ cm}$ and at the bottom is $82.0 \text{ cm}$ . (a) Calculate the length of the string in metres.
[1]
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(b) The student measures the time for 20 oscillations as $36.0 \text{ s}$ . Calculate the period of one oscillation.
[1]
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2. A car travels along a straight road. The velocity-time graph for the car’s motion is shown below.

(Imagine a graph: From t=0 to t=10s, velocity increases linearly from 0 to 20 m/s. From t=10s to t=30s, velocity is constant at 20 m/s. From t=30s to t=40s, velocity decreases linearly to 0.)

(a) Describe the motion of the car between $t = 10 \text{ s}$ and $t = 30 \text{ s}$ .
[1]
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(b) Calculate the acceleration of the car during the first 10 seconds.
[2]
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(c) Calculate the total distance travelled by the car in the first 40 seconds.
[3]
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3. A box of mass $15 \text{ kg}$ is pushed across a horizontal floor with a constant horizontal force of $60 \text{ N}$ . The box accelerates at $2.0 \text{ m/s}^2$ . (a) Calculate the resultant force acting on the box.
[2]
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(b) Calculate the frictional force acting on the box.
[2]
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(c) Explain why the box continues to move forward for a short distance after the pushing force is removed.
[1]
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4. A uniform metre rule is pivoted at the $50 \text{ cm}$ mark. A weight of $4.0 \text{ N}$ is hung at the $20 \text{ cm}$ mark. (a) Calculate the moment of the $4.0 \text{ N}$ weight about the pivot.
[2]
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(b) A second weight of $6.0 \text{ N}$ is hung on the right side of the pivot to balance the rule. Calculate the distance from the pivot where this weight must be placed.
[2]
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5. A diver is swimming at a depth of $12 \text{ m}$ in seawater. The density of seawater is $1030 \text{ kg/m}^3$ . Atmospheric pressure is $1.0 \times 10^5 \text{ Pa}$ . (a) Calculate the pressure due to the seawater at this depth.
[2]
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(b) Calculate the total pressure acting on the diver.
[1]
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6. A crane lifts a container of mass $2000 \text{ kg}$ vertically through a height of $15 \text{ m}$ in $25 \text{ s}$ . (a) Calculate the gain in gravitational potential energy of the container.
[2]
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(b) Calculate the useful power output of the crane.
[2]
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(c) The motor powering the crane has an efficiency of $40\%$ . Calculate the total energy input required from the power source.
[2]
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7. A skydiver falls from a stationary helicopter. (a) State the two main forces acting on the skydiver during the fall.
[1]
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(b) Explain, in terms of forces, why the skydiver reaches a constant terminal velocity.
[3]
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8. A block of ice at $0^\circ\text{C}$ is placed in a warm room. (a) Describe what happens to the arrangement and motion of the water molecules as the ice melts into liquid water.
[2]
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(b) The specific latent heat of fusion of ice is $3.34 \times 10^5 \text{ J/kg}$ . Calculate the thermal energy required to melt $0.5 \text{ kg}$ of ice at $0^\circ\text{C}$ .
[2]
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Section B: Free Response Questions (20 Marks)

Answer all questions in this section.

9. A student investigates the relationship between the extension of a spring and the load applied. The results are plotted on a graph of Extension (cm) against Load (N). The graph is a straight line passing through the origin up to a load of $8 \text{ N}$ . Beyond $8 \text{ N}$ , the graph curves. (a) State the law that describes the linear region of the graph.
[1]
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(b) Calculate the spring constant $k$ if a load of $4 \text{ N}$ causes an extension of $2 \text{ cm}$ . Give your answer in $\text{N/m}$ .
[3]
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(c) Explain what is meant by the "limit of proportionality" in this context.
[1]
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10. Two trolleys, A and B, are on a smooth horizontal track. Trolley A has a mass of $2 \text{ kg}$ and moves with a velocity of $3 \text{ m/s}$ towards Trolley B, which has a mass of $1 \text{ kg}$ and is stationary. They collide and stick together. (a) Calculate the total momentum of the system before the collision.
[2]
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(b) Calculate the velocity of the combined trolleys after the collision.
[3]
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(c) State whether kinetic energy is conserved in this collision and explain your answer.
[2]
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11. A hydraulic press consists of two pistons connected by a tube filled with oil. The small piston has an area of $0.01 \text{ m}^2$ and the large piston has an area of $0.5 \text{ m}^2$ . A force of $50 \text{ N}$ is applied to the small piston. (a) Calculate the pressure transmitted through the oil.
[2]
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(b) Calculate the force exerted by the large piston.
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(c) Explain why liquids are used in hydraulic systems instead of gases.
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12. A ball is thrown vertically upwards with an initial speed of $20 \text{ m/s}$ . Air resistance is negligible. (a) Calculate the maximum height reached by the ball.
[3]
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(b) Determine the time taken for the ball to return to the thrower’s hand.
[2]
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End of Paper

Answers

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Version 2 of 5

Section A: Structured Questions

1. (a) Length = $82.0 \text{ cm} - 2.0 \text{ cm} = 80.0 \text{ cm}$ .
Convert to metres: $80.0 / 100 = 0.80 \text{ m}$ .
[1]

(b) Period $T = \text{Total Time} / \text{Number of Oscillations}$ .
$T = 36.0 \text{ s} / 20 = 1.8 \text{ s}$ .
[1]

2. (a) The car moves at a constant velocity (or constant speed) of $20 \text{ m/s}$ .
[1]

(b) Acceleration $a = \Delta v / \Delta t$ .
$a = (20 - 0) / (10 - 0) = 20 / 10 = 2.0 \text{ m/s}^2$ .
[2] (1 for formula/substitution, 1 for answer with unit)

(c) Distance = Area under the graph.
Area 1 (Triangle, 0-10s): $0.5 \times 10 \times 20 = 100 \text{ m}$ .
Area 2 (Rectangle, 10-30s): $20 \times 20 = 400 \text{ m}$ .
Area 3 (Triangle, 30-40s): $0.5 \times 10 \times 20 = 100 \text{ m}$ .
Total Distance = $100 + 400 + 100 = 600 \text{ m}$ .
[3] (1 for each correct area calculation or method)

3. (a) Resultant Force $F_{net} = ma$ .
$F_{net} = 15 \text{ kg} \times 2.0 \text{ m/s}^2 = 30 \text{ N}$ .
[2]

(b) $F_{net} = F_{applied} - F_{friction}$ .
$30 \text{ N} = 60 \text{ N} - F_{friction}$ .
$F_{friction} = 60 - 30 = 30 \text{ N}$ .
[2]

(c) The box has inertia (or mass). It resists the change in its state of motion, so it continues moving until friction brings it to rest.
[1]

4. (a) Moment = Force $\times$ Perpendicular Distance from Pivot.
Distance = $50 \text{ cm} - 20 \text{ cm} = 30 \text{ cm} = 0.30 \text{ m}$ .
Moment = $4.0 \text{ N} \times 0.30 \text{ m} = 1.2 \text{ Nm}$ .
[2] (1 for distance, 1 for calculation)

(b) Principle of Moments: Clockwise Moment = Anticlockwise Moment.
$1.2 \text{ Nm} = 6.0 \text{ N} \times d$ .
$d = 1.2 / 6.0 = 0.20 \text{ m}$ (or $20 \text{ cm}$ ).
[2]

5. (a) Pressure due to liquid $P = h\rho g$ .
$P = 12 \text{ m} \times 1030 \text{ kg/m}^3 \times 10 \text{ N/kg}$ .
$P = 123,600 \text{ Pa}$ (or $1.236 \times 10^5 \text{ Pa}$ ).
[2]

(b) Total Pressure = Atmospheric Pressure + Liquid Pressure.
$P_{total} = 100,000 \text{ Pa} + 123,600 \text{ Pa} = 223,600 \text{ Pa}$ (or $2.236 \times 10^5 \text{ Pa}$ ).
[1]

6. (a) GPE = $mgh$ .
GPE = $2000 \text{ kg} \times 10 \text{ N/kg} \times 15 \text{ m} = 300,000 \text{ J}$ (or $300 \text{ kJ}$ ).
[2]

(b) Power = Energy / Time.
$P = 300,000 \text{ J} / 25 \text{ s} = 12,000 \text{ W}$ (or $12 \text{ kW}$ ).
[2]

(c) Efficiency = Useful Output / Total Input.
$0.40 = 300,000 \text{ J} / E_{input}$ .
$E_{input} = 300,000 / 0.40 = 750,000 \text{ J}$ (or $750 \text{ kJ}$ ).
[2]

7. (a) Weight (gravity) and Air Resistance (drag).
[1]

(b)

Initially, weight is greater than air resistance, so the skydiver accelerates downwards.
As speed increases, air resistance increases.
Eventually, air resistance equals weight. The resultant force is zero, so acceleration becomes zero and velocity becomes constant (terminal velocity).
[3] (1 mark for each distinct point)

8. (a)

Arrangement: Molecules move from a fixed, regular lattice structure to a random, less ordered arrangement.
Motion: Molecules gain enough energy to overcome strong bonds and can slide past one another (increase in kinetic energy/potential energy).
[2]

(b) Energy $Q = mL$ .
$Q = 0.5 \text{ kg} \times 3.34 \times 10^5 \text{ J/kg}$ .
$Q = 167,000 \text{ J}$ (or $167 \text{ kJ}$ ).
[2]

Section B: Free Response Questions

9. (a) Hooke’s Law.
[1]

(b) Extension $x = 2 \text{ cm} = 0.02 \text{ m}$ . Force $F = 4 \text{ N}$ .
$F = kx \Rightarrow k = F/x$ .
$k = 4 \text{ N} / 0.02 \text{ m} = 200 \text{ N/m}$ .
[3] (1 for conversion, 1 for formula, 1 for answer)

(c) The limit of proportionality is the point beyond which the extension is no longer directly proportional to the load (the graph is no longer linear).
[1]

10. (a) Momentum $p = mv$ .
Momentum of A = $2 \text{ kg} \times 3 \text{ m/s} = 6 \text{ kg m/s}$ .
Momentum of B = $1 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s}$ .
Total Momentum = $6 + 0 = 6 \text{ kg m/s}$ .
[2]

(b) Conservation of Momentum: Total Momentum Before = Total Momentum After.
$6 \text{ kg m/s} = (m_A + m_B) \times v_{final}$ .
$6 = (2 + 1) \times v_{final}$ .
$6 = 3 v_{final}$ .
$v_{final} = 2 \text{ m/s}$ .
[3] (1 for principle, 1 for substitution, 1 for answer)

(c) Kinetic energy is not conserved.
This is an inelastic collision (objects stick together). Some kinetic energy is converted into other forms such as heat, sound, or deformation energy.
[2] (1 for "No", 1 for explanation)

11. (a) Pressure $P = F/A$ .
$P = 50 \text{ N} / 0.01 \text{ m}^2 = 5000 \text{ Pa}$ .
[2]

(b) Pascal’s Principle: Pressure is transmitted equally.
$F_{large} = P \times A_{large}$ .
$F_{large} = 5000 \text{ Pa} \times 0.5 \text{ m}^2 = 2500 \text{ N}$ .
[2]

(c) Liquids are virtually incompressible, whereas gases are compressible. This ensures that the force applied is transmitted effectively without loss of energy to compression.
[1]

12. (a) At maximum height, final velocity $v = 0 \text{ m/s}$ .
Using $v^2 = u^2 + 2as$ (taking up as positive, $a = -10 \text{ m/s}^2$ ):
$0 = 20^2 + 2(-10)s$ .
$0 = 400 - 20s$ .
$20s = 400$ .
$s = 20 \text{ m}$ .
(Alternative: Conservation of Energy: $1/2 mv^2 = mgh \Rightarrow h = v^2/2g = 400/20 = 20 \text{ m}$ )
[3] (1 for formula/method, 1 for substitution, 1 for answer)

(b) Time to reach max height: $v = u + at$ .
$0 = 20 + (-10)t$ .
$10t = 20 \Rightarrow t = 2 \text{ s}$ .
Total time to return = Time up + Time down. By symmetry, Time down = Time up.
Total time = $2 \text{ s} + 2 \text{ s} = 4 \text{ s}$ .
[2] (1 for time up, 1 for total time)