Secondary 4 Pure Physics Mechanics Quiz

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Secondary 4 Pure Physics Quiz - Mechanics

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all working clearly. Numerical answers should be given to 3 significant figures unless otherwise stated.
4. Take the acceleration of free fall, $g = 10 \text{ m/s}^2$.

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Section A: Multiple Choice & Short Concepts (10 Marks)

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

2. A car travels 60 km North in 1 hour, then 60 km South in 1 hour. What is the average velocity of the car for the entire journey?
A. 0 km/h
B. 60 km/h North
C. 60 km/h South
D. 120 km/h
[1]

3. The graph below shows the velocity-time graph of a falling object.
(Imagine a graph where velocity increases linearly then becomes constant)
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. State the SI unit for Moment of a Force.
[1]

5. A block of mass 5 kg rests on a horizontal table. What is the weight of the block?
[1]

6. Define inertia.
[1]

7. Two forces act on an object: 3 N to the East and 4 N to the North. Calculate the magnitude of the resultant force.
[1]

8. A hydraulic press uses a liquid to transmit pressure. State the principle that explains how pressure is transmitted equally in all directions in an enclosed fluid.
[1]

9. A student pushes a box with a force of 20 N. The box moves 5 m in the direction of the force. Calculate the work done.
[1]

10. Why does a sharp knife cut better than a blunt knife? Explain in terms of pressure.
[1]

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Section B: Structured Questions (20 Marks)

11. A skydiver jumps from a stationary helicopter.
(a) Describe the motion of the skydiver from the moment he jumps until he reaches terminal velocity. In your answer, refer to the forces acting on him.
[3]

(b) The skydiver opens his parachute. Explain why his velocity decreases immediately after opening the parachute.
[2]

12. A uniform metre rule is pivoted at the 50 cm mark. A weight of 4.0 N is hung at the 20 cm mark.
(a) Calculate the moment of the 4.0 N weight about the pivot.
[2]

(b) A second weight, $W$, is hung at the 80 cm mark to balance the rule horizontally. Calculate the value of $W$.
[2]

(c) The pivot is moved to the 30 cm mark. The 4.0 N weight remains at the 20 cm mark. Where must weight $W$ be placed to balance the rule? (Assume the rule is light and its weight is negligible).
[3]

13. Figure 13.1 shows a velocity-time graph for a toy car moving along a straight track.

(Graph Description: From t=0 to t=4s, velocity increases uniformly from 0 to 8 m/s. From t=4s to t=10s, velocity remains constant at 8 m/s.)

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

(b) Calculate the total distance travelled by the car in the first 10 seconds.
[3]

(c) Sketch the corresponding displacement-time graph for the first 4 seconds. (No values required on the y-axis, but show the shape).
[2]

14. A diver is swimming at a depth of 20 m in sea water. The density of sea water is $1030 \text{ kg/m}^3$. Atmospheric pressure is $1.0 \times 10^5 \text{ Pa}$.
(a) Calculate the pressure due to the sea water at this depth.
[2]

(b) Calculate the total pressure acting on the diver.
[2]

15. A car of mass 1200 kg is travelling at a constant speed of 20 m/s on a horizontal road. The driving force provided by the engine is 800 N.
(a) State the magnitude of the resistive forces acting on the car. Explain your answer.
[2]

(b) The driver increases the driving force to 1400 N. Assuming the resistive forces remain constant, calculate the initial acceleration of the car.
[3]

(c) Explain why the acceleration will not remain constant as the car speeds up, even if the driver keeps the driving force at 1400 N.
[2]

16. A student investigates the stability of a bottle.
(a) Define centre of gravity.
[1]

(b) The student fills the bottle with water and places it on a rough surface. He then tilts it slowly. Explain, in terms of centre of gravity and base area, why the bottle eventually topples over.
[2]

17. A crane lifts a load of mass 500 kg vertically upwards at a constant speed.
(a) Calculate the tension in the cable lifting the load.
[2]

(b) The load is lifted through a height of 12 m. Calculate the work done by the tension in the cable.
[2]

18. A cyclist travels around a circular track of radius 50 m.
(a) If the cyclist completes one full lap in 40 seconds, calculate his average speed.
[2]

(b) Explain why the cyclist’s velocity is changing even though his speed is constant.
[2]

19. A spring extends by 4 cm when a load of 20 N is attached to it.
(a) Calculate the spring constant, $k$, of the spring.
[2]

(b) Calculate the elastic potential energy stored in the spring when it is extended by 4 cm.
[2]

20. A box of mass 10 kg is pushed across a rough floor.
(a) If the coefficient of friction is 0.5, calculate the frictional force acting on the box.
[2]

(b) If the applied force is 60 N, calculate the acceleration of the box.
[2]

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Secondary 4 Pure Physics Quiz - Mechanics (Answer Key)

1. D
Reasoning: Acceleration has both magnitude and direction. Speed, distance, and mass are scalars. [1]

2. A
Reasoning: Average velocity = Total Displacement / Total Time. Displacement is 0 km (returns to start). $0/2 = 0$ km/h. [1]

3. C
Reasoning: Constant velocity implies zero acceleration, meaning forces are balanced (Weight = Air Resistance). This is terminal velocity. [1]

4. Newton metre (Nm)
[1]

5. $W = mg = 5 \times 10 = 50 \text{ N}$
[1]

6. Inertia is the resistance of an object to change its state of rest or uniform motion.
[1]

7. $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \text{ N}$
[1]

8. Pascal’s Principle
[1]

9. $W = F \times d = 20 \times 5 = 100 \text{ J}$
[1]

10. A sharp knife has a smaller surface area. For the same force, Pressure = Force/Area is higher, allowing it to penetrate easier.
[1]

11.
(a) Initially, weight is greater than air resistance, so there is a resultant downward force. The skydiver accelerates downwards. As speed increases, air resistance increases. The resultant force decreases, so acceleration decreases. Eventually, air resistance equals weight, resultant force is zero, and he moves at constant terminal velocity. [3]
(b) Opening the parachute greatly increases the surface area, causing a large increase in air resistance. The upward air resistance becomes greater than the downward weight, creating a resultant upward force that decelerates the skydiver. [2]

12.
(a) Distance from pivot = $50 - 20 = 30 \text{ cm} = 0.3 \text{ m}$.
Moment = $4.0 \text{ N} \times 0.3 \text{ m} = 1.2 \text{ Nm}$. [2]
(b) Clockwise Moment = Anticlockwise Moment.
Distance of W from pivot = $80 - 50 = 30 \text{ cm} = 0.3 \text{ m}$.
$W \times 0.3 = 1.2$
$W = 1.2 / 0.3 = 4.0 \text{ N}$. [2]
(c) New pivot at 30 cm.
4.0 N weight is at 20 cm. Distance = $30 - 20 = 10 \text{ cm} = 0.1 \text{ m}$.
Anticlockwise Moment = $4.0 \times 0.1 = 0.4 \text{ Nm}$.
W must provide Clockwise Moment of 0.4 Nm.
$W = 4.0 \text{ N}$ (from previous part, assuming W is the same object).
$4.0 \times d = 0.4$
$d = 0.1 \text{ m} = 10 \text{ cm}$ from pivot.
Position = $30 \text{ cm} + 10 \text{ cm} = 40 \text{ cm}$ mark. [3]

13.
(a) Acceleration = Gradient = $\frac{\Delta v}{\Delta t} = \frac{8 - 0}{4 - 0} = 2 \text{ m/s}^2$. [2]
(b) Distance = Area under graph.
Area 1 (Triangle) = $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$.
Area 2 (Rectangle) = $(10 - 4) \times 8 = 6 \times 8 = 48 \text{ m}$.
Total Distance = $16 + 48 = 64 \text{ m}$. [3]
(c) Curve starting from origin with increasing gradient (parabolic shape) up to t=4s. [2]

14.
(a) $P = h \rho g = 20 \times 1030 \times 10 = 206,000 \text{ Pa}$ (or $2.06 \times 10^5 \text{ Pa}$). [2]
(b) Total Pressure = Atmospheric + Liquid Pressure
$P_{total} = 1.0 \times 10^5 + 2.06 \times 10^5 = 3.06 \times 10^5 \text{ Pa}$. [2]

15.
(a) 800 N. Since the car is moving at constant speed, acceleration is zero. By Newton's First Law, forces are balanced, so Resistive Force = Driving Force. [2]
(b) Resultant Force = Driving Force - Resistive Force = $1400 - 800 = 600 \text{ N}$.
$F = ma \Rightarrow 600 = 1200 \times a$
$a = 600 / 1200 = 0.5 \text{ m/s}^2$. [3]
(c) As speed increases, air resistance (a resistive force) increases. This reduces the resultant force ($F_{driving} - F_{resistive}$), and since $a = F/m$, the acceleration decreases. [2]

16.
(a) The point through which the entire weight of the object appears to act. [1]
(b) As the bottle tilts, the line of action of its weight (acting from the centre of gravity) moves towards the edge of the base. When the line of action falls outside the base area, there is no normal contact force to counteract the moment of the weight, causing the bottle to topple. [2]

17.
(a) Since speed is constant, acceleration is 0. Tension = Weight.
$T = mg = 500 \times 10 = 5000 \text{ N}$. [2]
(b) Work Done = Force $\times$ Distance.
$W = 5000 \times 12 = 60,000 \text{ J}$ (or $60 \text{ kJ}$). [2]

18.
(a) Circumference $C = 2 \pi r = 2 \times \pi \times 50 = 100\pi \text{ m}$.
Speed = Distance / Time = $100\pi / 40 = 2.5\pi \approx 7.85 \text{ m/s}$. [2]
(b) Velocity is a vector quantity consisting of speed and direction. Although the speed is constant, the direction of motion is continuously changing as the cyclist moves around the circle. Therefore, the velocity is changing. [2]

19.
(a) Hooke's Law: $F = kx$.
$x = 4 \text{ cm} = 0.04 \text{ m}$.
$k = F / x = 20 / 0.04 = 500 \text{ N/m}$. [2]
(b) Elastic Potential Energy $E = \frac{1}{2} k x^2$.
$E = 0.5 \times 500 \times (0.04)^2 = 250 \times 0.0016 = 0.4 \text{ J}$. [2]

20.
(a) Normal Reaction Force $R = mg = 10 \times 10 = 100 \text{ N}$.
Frictional Force $f = \mu R = 0.5 \times 100 = 50 \text{ N}$. [2]
(b) Resultant Force $F_{net} = F_{applied} - f = 60 - 50 = 10 \text{ N}$.
$F_{net} = ma \Rightarrow 10 = 10 \times a$.
$a = 1 \text{ m/s}^2$. [2]