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A Level H2 Physics Mechanics Quiz

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Questions

A-Level Physics H2 Quiz - Mechanics

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Numerical answers should be given to an appropriate number of significant figures.
The use of an approved scientific calculator is expected.
Where necessary, take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ .

Section A: Structured Questions (25 Marks)

1. State the Principle of Conservation of Linear Momentum.
[2]

2. A ball of mass $0.15 \text{ kg}$ undergoes simple harmonic motion with an amplitude of $4.0 \text{ cm}$ and a frequency of $2.5 \text{ Hz}$ . Calculate the maximum acceleration of the ball.
[3]

3. In an experiment to determine the acceleration of free fall $g$ , a student drops a steel ball from rest through a vertical distance $h$ and measures the time $t$ taken. (a) State one precaution the student should take to improve the accuracy of the measurement of $h$ .
[1]

(b) State one precaution the student should take to improve the accuracy of the measurement of $t$ .
[1]

4. A car of mass $1200 \text{ kg}$ travels at a constant speed of $20 \text{ m s}^{-1}$ along a horizontal circular track of radius $50 \text{ m}$ . (a) Calculate the magnitude of the centripetal force acting on the car.
[2]

(b) State the physical origin of this centripetal force.
[1]

5. Define the gravitational field strength at a point.
[1]

6. A satellite orbits the Earth in a circular orbit. Explain why the satellite is considered to be in a state of "free fall" despite maintaining a constant altitude.
[2]

7. A block of mass $2.0 \text{ kg}$ slides down a rough inclined plane at a constant velocity. The angle of inclination is $30^\circ$ . (a) Draw a free-body diagram showing all forces acting on the block.
[2]

(b) Calculate the magnitude of the frictional force acting on the block.
[2]

8. State Newton’s Second Law of Motion in terms of momentum.
[1]

9. A projectile is fired horizontally from a cliff with a speed of $15 \text{ m s}^{-1}$ . Air resistance is negligible. (a) Describe the horizontal component of the projectile's velocity during its flight.
[1]

(b) Describe the vertical component of the projectile's acceleration during its flight.
[1]

10. Two objects, A and B, collide inelastically. Object A has mass $m$ and velocity $2v$ . Object B has mass $2m$ and is initially at rest. After the collision, they stick together. Calculate the final velocity of the combined mass in terms of $v$ .
[3]

11. A spring obeys Hooke’s Law. When a load of $10 \text{ N}$ is applied, the extension is $4.0 \text{ cm}$ . Calculate the elastic potential energy stored in the spring.
[2]

12. Distinguish between a scalar quantity and a vector quantity, giving one example of each relevant to mechanics.
[2]

13. A body is in equilibrium under the action of three coplanar forces. State the condition required for these forces to be in equilibrium in terms of their vector sum.
[1]

14. Explain why the work done by the centripetal force on an object moving in uniform circular motion is zero.
[2]

15. A rocket accelerates vertically upwards. Explain, using Newton’s Third Law, how the rocket achieves upward acceleration.
[2]

Section B: Data Analysis and Application (20 Marks)

16. A student investigates the relationship between the period $T$ of a simple pendulum and its length $L$ . The student plots a graph of $T^2$ against $L$ . (a) The theoretical relationship is $T = 2\pi \sqrt{\frac{L}{g}}$ . Show that the gradient of the graph of $T^2$ against $L$ is equal to $\frac{4\pi^2}{g}$ .
[2]

(b) The student obtains a gradient of $4.05 \text{ s}^2 \text{ m}^{-1}$ . Calculate the value of $g$ determined from this experiment.
[2]

(c) The student repeats the experiment but uses a heavier bob. State and explain the effect, if any, on the value of the gradient.
[2]

17. A car of mass $1500 \text{ kg}$ accelerates from rest to a speed of $25 \text{ m s}^{-1}$ in $10 \text{ s}$ . The average resistive force during this time is $400 \text{ N}$ . (a) Calculate the average acceleration of the car.
[1]

(b) Calculate the average driving force exerted by the engine.
[3]

18. A ball is dropped from a height of $2.0 \text{ m}$ onto a hard surface. It rebounds to a height of $1.5 \text{ m}$ . (a) Calculate the speed of the ball just before it hits the ground.
[2]

(b) Calculate the speed of the ball just after it leaves the ground.
[2]

19. A uniform beam of length $4.0 \text{ m}$ and weight $200 \text{ N}$ is hinged at one end to a vertical wall. The beam is supported in a horizontal position by a cable attached to the other end, making an angle of $30^\circ$ with the beam. (a) Calculate the tension in the cable by taking moments about the hinge.
[3]

(b) State the direction of the vertical component of the force exerted by the hinge on the beam.
[1]

20. In a collision experiment, trolley A (mass $0.5 \text{ kg}$ ) moves at $0.6 \text{ m s}^{-1}$ towards stationary trolley B (mass $0.5 \text{ kg}$ ). They collide and move off together. (a) Calculate the total momentum before the collision.
[1]

(b) Calculate the kinetic energy lost during the collision.
[2]

Answers

A-Level Physics H2 Quiz - Mechanics (Answer Key)

1. State the Principle of Conservation of Linear Momentum. [2]

Answer: In a closed system (or isolated system) [1], the total momentum before an event (collision/explosion) is equal to the total momentum after the event, provided no external forces act [1].
Note: Must mention "closed/isolated system" or "no external forces" and "total momentum constant/before=after".

2. Calculate the maximum acceleration of the ball. [3]

Formula: $a_{max} = \omega^2 x_0$ or $a_{max} = (2\pi f)^2 x_0$ [1]
Substitution: $\omega = 2\pi(2.5) = 5\pi \text{ rad s}^{-1}$ . $x_0 = 0.04 \text{ m}$ . $a_{max} = (5\pi)^2 \times 0.04$ [1]
Calculation: $a_{max} \approx 246.74 \times 0.04 = 9.87 \text{ m s}^{-2}$ [1]
Accept: $9.9 \text{ m s}^{-2}$ .

3. Precautions for accuracy.

(a) Measurement of $h$ : Use a set square to ensure the ruler is vertical/parallel to drop path, or use a marker to define the start/end points clearly to avoid parallax error. [1]
(b) Measurement of $t$ : Use an electronic timer/light gate instead of a stopwatch to eliminate human reaction time error. [1]

4. Circular motion.

(a) Centripetal force: $F = \frac{mv^2}{r}$ [1] $F = \frac{1200 \times 20^2}{50} = \frac{1200 \times 400}{50} = 9600 \text{ N}$ [1]
(b) Origin: Friction between the tyres and the road. [1]

5. Define gravitational field strength. [1]

Answer: The gravitational force per unit mass acting on a small test mass placed at that point. [1]
Alternative: $g = F/m$ .

6. Satellite in free fall. [2]

Answer: The only force acting on the satellite is gravity (weight) [1]. This force provides the centripetal acceleration required to keep it in orbit, so it is constantly falling towards the Earth but has sufficient tangential velocity to miss it [1].

7. Block on inclined plane.

(a) Free-body diagram: [2]
- Weight ( $mg$ ) acting vertically downwards. [1]
- Normal contact force ( $N$ ) acting perpendicular to the plane. [1]
- Friction ( $f$ ) acting up the plane (parallel to surface). [1]
- Note: Award 2 marks if all three are correct in direction and label. Deduct 1 if one is wrong/missing.
(b) Frictional force: Since velocity is constant, forces are balanced. $f = mg \sin \theta$ [1] $f = 2.0 \times 9.81 \times \sin 30^\circ = 2.0 \times 9.81 \times 0.5 = 9.81 \text{ N}$ [1]

8. Newton’s Second Law in terms of momentum. [1]

Answer: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of the force. [1]
Formula: $F = \frac{dp}{dt}$ .

9. Projectile motion.

(a) Horizontal velocity: Constant (zero acceleration horizontally as air resistance is negligible). [1]
(b) Vertical acceleration: Constant, equal to $g$ ( $9.81 \text{ m s}^{-2}$ ) downwards. [1]

10. Inelastic collision. [3]

Conservation of Momentum: $m_A u_A + m_B u_B = (m_A + m_B) v$ [1]
Substitution: $m(2v) + 2m(0) = (m + 2m) v_{final}$ [1] $2mv = 3m v_{final}$ $v_{final} = \frac{2}{3}v$ [1]

11. Elastic potential energy. [2]

Formula: $E_p = \frac{1}{2} F x$ or $\frac{1}{2} k x^2$ [1]
Calculation: $E_p = \frac{1}{2} \times 10 \times 0.04 = 0.20 \text{ J}$ [1]

12. Scalar vs Vector. [2]

Distinction: A scalar has magnitude only [1], while a vector has both magnitude and direction [1].
Examples: Scalar: Mass, Speed, Energy. Vector: Displacement, Velocity, Force. (Must match category).

13. Equilibrium condition. [1]

Answer: The vector sum of the three forces is zero. [1]
Alternative: They form a closed triangle of forces.

14. Work done by centripetal force. [2]

Answer: The centripetal force acts perpendicular to the direction of motion (velocity) [1]. Since work done $W = F s \cos \theta$ and $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , so work done is zero [1].

15. Rocket acceleration (Newton's 3rd Law). [2]

Answer: The rocket exerts a downward force on the expelled gases [1]. By Newton's Third Law, the gases exert an equal and opposite upward force on the rocket [1], which causes the upward acceleration.

16. Pendulum Graph Analysis.

(a) Show gradient: $T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = 4\pi^2 \frac{L}{g} = \left(\frac{4\pi^2}{g}\right) L$ [1] Comparing to $y = mx$ , where $y=T^2$ and $x=L$ , the gradient $m = \frac{4\pi^2}{g}$ [1].
(b) Calculate $g$ : Gradient $= 4.05$ . $4.05 = \frac{4\pi^2}{g} \Rightarrow g = \frac{4\pi^2}{4.05}$ [1] $g \approx 9.75 \text{ m s}^{-2}$ [1]
(c) Effect of heavier bob: No change [1]. The period of a simple pendulum is independent of mass (as seen in the formula) [1].

17. Car Acceleration.

(a) Average acceleration: $a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m s}^{-2}$ [1]
(b) Driving force: Resultant Force $F_{res} = ma = 1500 \times 2.5 = 3750 \text{ N}$ [1] $F_{drive} - F_{resistive} = F_{res}$ $F_{drive} - 400 = 3750$ [1] $F_{drive} = 4150 \text{ N}$ [1]
(c) Average power: Average velocity $v_{avg} = \frac{0 + 25}{2} = 12.5 \text{ m s}^{-1}$ [1] $P = F_{drive} \times v_{avg} = 4150 \times 12.5 = 51,875 \text{ W}$ (or $51.9 \text{ kW}$ ) [1] Note: Using $P = \frac{\text{Work}}{\text{time}}$ is also acceptable. Work = Change in KE + Work against friction.

18. Rebound Heights.

(a) Speed before impact ( $u$ ): $v^2 = u^2 + 2as \Rightarrow u = \sqrt{2gh_1}$ $u = \sqrt{2 \times 9.81 \times 2.0} = \sqrt{39.24} \approx 6.26 \text{ m s}^{-1}$ [2]
(b) Speed after impact ( $v$ ): $v = \sqrt{2gh_2} = \sqrt{2 \times 9.81 \times 1.5} = \sqrt{29.43} \approx 5.42 \text{ m s}^{-1}$ [2]
(c) Coefficient of restitution ( $e$ ): $e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{5.42}{6.26}$ [1] $e \approx 0.87$ [1]

19. Uniform Beam Moments.

(a) Tension in cable: Take moments about the hinge. Clockwise Moment (Weight) = Anti-clockwise Moment (Vertical component of Tension) Weight acts at center ( $2.0 \text{ m}$ from hinge). $200 \times 2.0 = (T \sin 30^\circ) \times 4.0$ [1] $400 = T \times 0.5 \times 4.0$ $400 = 2T$ $T = 200 \text{ N}$ [2]
(b) Vertical component of hinge force: Upwards [1]. (Since $T_y = 100 \text{ N}$ upwards and Weight $= 200 \text{ N}$ downwards, the hinge must provide $100 \text{ N}$ upwards to balance vertical forces).

20. Collision Energy Loss.

(a) Total momentum before: $p = m_A u_A = 0.5 \times 0.6 = 0.30 \text{ kg m s}^{-1}$ [1]
(b) Kinetic energy lost: $KE_{initial} = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(0.5)(0.6)^2 = 0.09 \text{ J}$ [1] Velocity after collision ( $v$ ): $0.30 = (0.5+0.5)v \Rightarrow v = 0.3 \text{ m s}^{-1}$ . $KE_{final} = \frac{1}{2} (1.0) (0.3)^2 = 0.045 \text{ J}$ [1] Loss $= 0.09 - 0.045 = 0.045 \text{ J}$ [1]