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A Level H2 Physics Practice Paper 1

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Questions

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Subject: Physics H2
Level: A-Level
Paper: Practice Paper 1 (Version 1 of 5)
Topic: Mechanics
Duration: 1 hour 15 minutes
Total Marks: 40

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates:

Answer all questions.
Write your answers in the spaces provided.
All working must be clearly shown.
The use of an approved scientific calculator is expected.
Take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ .

Section A: Structured Questions (20 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. A student sets up an experiment to determine the acceleration of free fall $g$ using a free-fall apparatus. The ball is dropped from rest, and the time $t$ taken to fall a distance $h$ is recorded.

(a) State one precaution that should be taken to ensure the accuracy of the measurement of $h$ .
[1]

(b) State one precaution that should be taken to ensure the accuracy of the measurement of $t$ .
[1]

4. Explain what is meant by the binding energy of a nucleus.
[2]

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

Section B: Calculation and Application (20 Marks)

6. Two trolleys, A and B, move along a straight horizontal frictionless track. Trolley A has a mass of $2.0 \text{ kg}$ and moves with a velocity of $3.0 \text{ m s}^{-1}$ to the right. Trolley B has a mass of $1.0 \text{ kg}$ and is initially at rest. The trolleys collide and stick together.

(a) Calculate the velocity of the combined trolleys immediately after the collision.
[3]

(b) Calculate the loss in kinetic energy during the collision.
[3]

7. A projectile is launched from ground level with an initial velocity of $30 \text{ m s}^{-1}$ at an angle of $40^\circ$ to the horizontal. Air resistance is negligible.

(a) Calculate the maximum height reached by the projectile.
[3]

(b) Calculate the horizontal range of the projectile.
[3]

8. A block of mass $5.0 \text{ kg}$ is pulled up a rough inclined plane by a constant force of $40 \text{ N}$ parallel to the slope. The plane is inclined at $30^\circ$ to the horizontal. The block moves at a constant speed.

(a) Calculate the component of the weight of the block acting down the slope.
[2]

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

9. A satellite orbits the Earth in a circular orbit of radius $r$ . The gravitational force provides the centripetal force.

(a) Show that the orbital speed $v$ of the satellite is given by $v = \sqrt{\frac{GM}{r}}$ , where $M$ is the mass of the Earth and $G$ is the gravitational constant.
[2]

(b) State and explain how the orbital speed changes if the radius of the orbit increases.
[2]

Section C: Data Analysis and Reasoning (20 Marks)

10. In an experiment to verify the relationship between the period $T$ of a simple pendulum and its length $L$ , a student obtains the following data:

$L$ (m)	$T$ (s)
0.20	0.90
0.40	1.27
0.60	1.55
0.80	1.79
1.00	2.01

The relationship is given by $T = 2\pi \sqrt{\frac{L}{g}}$ .

(a) Plot a graph of $T^2$ against $L$ on the grid provided below.
[4]

(Note: In a real exam, a grid would be provided. Here, describe the expected plot.)

Calculate $T^2$ for each value.
Plot points.
Draw line of best fit.

(b) Determine the gradient of the graph.
[2]

11. A spring obeys Hooke's Law. When a load of $2.0 \text{ N}$ is applied, the extension is $4.0 \text{ cm}$ .

(a) Calculate the spring constant $k$ .
[2]

(b) Calculate the elastic potential energy stored in the spring when the extension is $4.0 \text{ cm}$ .
[2]

(c) The load is increased to $4.0 \text{ N}$ . State whether the elastic potential energy stored doubles, quadruples, or increases by a different factor. Explain your answer.
[2]

12. A car accelerates uniformly from rest to a speed of $25 \text{ m s}^{-1}$ in $10 \text{ s}$ .

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

(b) Calculate the distance travelled by the car during this time.
[2]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Version 1 of 5)
Topic: Mechanics

Section A: Structured Questions

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].
Marking Notes:
- 1 mark for "closed/isolated system" or "no external forces".
- 1 mark for "total momentum before = total momentum after".

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

Given: $m = 0.15 \text{ kg}$ , $A = 4.0 \text{ cm} = 0.04 \text{ m}$ , $f = 2.5 \text{ Hz}$ .
Formula: $a_{\max} = \omega^2 A$ and $\omega = 2\pi f$ .
Working:
- $\omega = 2\pi(2.5) = 5\pi \approx 15.71 \text{ rad s}^{-1}$ [1]
- $a_{\max} = (15.71)^2 \times 0.04$ [1]
- $a_{\max} = 9.87 \text{ m s}^{-2}$ [1]
Answer: $9.9 \text{ m s}^{-2}$ (2 s.f.)

3. Precautions for free-fall experiment. (a) Accuracy of $h$ . [1]

Answer: Use a meter rule with mm graduations and ensure eye is level with the scale to avoid parallax error. OR Measure from the bottom of the ball to the trapdoor.
Marking Notes: Accept specific practical details. "Be careful" is not accepted.

(b) Accuracy of $t$ . [1]

Answer: Use an electronic timer triggered by the release mechanism and impact sensor to eliminate human reaction time error. OR Repeat the experiment and take the average.
Marking Notes: Must link to reducing error.

4. Explain what is meant by the binding energy of a nucleus. [2]

Answer: The energy required to completely separate a nucleus into its constituent protons and neutrons [1]. OR The energy released when protons and neutrons combine to form a nucleus [1]. It is equivalent to the mass defect via $E=mc^2$ [1].
Marking Notes: 1 mark for "separate constituents", 1 mark for "energy required/released".

5. Calculate the magnitude of the centripetal force. [2]

Given: $m = 1200 \text{ kg}$ , $v = 20 \text{ m s}^{-1}$ , $r = 50 \text{ m}$ .
Formula: $F = \frac{mv^2}{r}$
Working:
- $F = \frac{1200 \times 20^2}{50}$ [1]
- $F = \frac{1200 \times 400}{50} = 9600 \text{ N}$ [1]
Answer: $9600 \text{ N}$

Section B: Calculation and Application

6. Collision of trolleys. (a) Velocity after collision. [3]

Principle: Conservation of Momentum.
Working:
- $m_A u_A + m_B u_B = (m_A + m_B) v$ [1]
- $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0) v$
- $6.0 = 3.0 v$
- $v = 2.0 \text{ m s}^{-1}$ [1]
- Direction: To the right [1]
Answer: $2.0 \text{ m s}^{-1}$ to the right.

(b) Loss in kinetic energy. [3]

Working:
- $KE_{\text{initial}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ [1]
- $KE_{\text{final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$ [1]
- Loss $= 9.0 - 6.0 = 3.0 \text{ J}$ [1]
Answer: $3.0 \text{ J}$

7. Projectile Motion. (a) Maximum height. [3]

Given: $u = 30 \text{ m s}^{-1}$ , $\theta = 40^\circ$ .
Vertical component: $u_y = 30 \sin 40^\circ \approx 19.28 \text{ m s}^{-1}$ .
At max height: $v_y = 0$ .
Formula: $v_y^2 = u_y^2 - 2gh$
Working:
- $0 = (19.28)^2 - 2(9.81)h$ [1]
- $19.62 h = 371.7$
- $h = 18.9 \text{ m}$ [1]
- Answer to 2 or 3 s.f. [1]
Answer: $19 \text{ m}$ (2 s.f.)

(b) Horizontal range. [3]

Time of flight: $v_y = u_y - gt \Rightarrow 0 = 19.28 - 9.81 t_{\text{up}} \Rightarrow t_{\text{up}} = 1.965 \text{ s}$ $v_{y} = u_{y} - g t \Rightarrow 0 = 19.28 - 9.81 t_{up} \Rightarrow t_{up} = 1.965 s$ .
- Total time $T = 2 \times 1.965 = 3.93 \text{ s}$ . [1]
Horizontal component: $u_x = 30 \cos 40^\circ \approx 22.98 \text{ m s}^{-1}$ .
Range: $R = u_x T$ $R = u_{x} T$
- $R = 22.98 \times 3.93 = 90.3 \text{ m}$ [1]
- Correct unit and s.f. [1]
Answer: $90 \text{ m}$ (2 s.f.)

8. Block on Inclined Plane. (a) Component of weight down slope. [2]

Formula: $W_{\parallel} = mg \sin \theta$
Working:
- $W_{\parallel} = 5.0 \times 9.81 \times \sin 30^\circ$ [1]
- $W_{\parallel} = 49.05 \times 0.5 = 24.5 \text{ N}$ [1]
Answer: $24.5 \text{ N}$

(b) Frictional force. [2]

Reasoning: Constant speed means zero acceleration, so net force is zero.
Equation: $F_{\text{pull}} = W_{\parallel} + F_{\text{friction}}$
Working:
- $40 = 24.5 + F_{\text{friction}}$ [1]
- $F_{\text{friction}} = 40 - 24.5 = 15.5 \text{ N}$ [1]
Answer: $15.5 \text{ N}$

9. Satellite Orbit. (a) Show $v = \sqrt{\frac{GM}{r}}$ . [2]

Working:
- Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1]
- Cancel $m$ and one $r$ : $\frac{GM}{r} = v^2 \Rightarrow v = \sqrt{\frac{GM}{r}}$ [1]

(b) Change in orbital speed if radius increases. [2]

Answer: The orbital speed decreases [1].
Explanation: Since $v \propto \frac{1}{\sqrt{r}}$ , as $r$ increases, $v$ decreases [1].

Section C: Data Analysis and Reasoning

10. Simple Pendulum Experiment. (a) Plot $T^2$ against $L$ . [4]

Data Processing:
- $L=0.20, T^2=0.81$
- $L=0.40, T^2=1.61$
- $L=0.60, T^2=2.40$
- $L=0.80, T^2=3.20$
- $L=1.00, T^2=4.04$
Marking:
- 1 mark for correct labels and units ( $T^2 / \text{s}^2$ , $L / \text{m}$ ).
- 1 mark for suitable scales.
- 1 mark for all 5 points plotted correctly.
- 1 mark for straight line of best fit through origin.

(b) Determine the gradient. [2]

Working:
- Gradient $= \frac{\Delta T^2}{\Delta L}$
- Using points $(0,0)$ and $(1.00, 4.04)$ : Gradient $= \frac{4.04 - 0}{1.00 - 0} = 4.04 \text{ s}^2 \text{ m}^{-1}$ [1]
- Accept range $3.9 - 4.1$ based on line drawn. [1]

(c) Calculate $g$ . [3]

Formula: $T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g} L$ $T = 2 π \frac{L}{g} \Rightarrow T^{2} = \frac{4 π ^{2}}{g} L$ .
- Gradient $= \frac{4\pi^2}{g}$ [1]
- $g = \frac{4\pi^2}{\text{Gradient}}$ [1]
- $g = \frac{4\pi^2}{4.04} = 9.77 \text{ m s}^{-2}$ [1]
Answer: $9.8 \text{ m s}^{-2}$ (2 s.f.)

11. Hooke's Law. (a) Spring constant $k$ . [2]

Formula: $F = kx$
Working:
- $2.0 = k(0.04)$
- $k = \frac{2.0}{0.04} = 50 \text{ N m}^{-1}$ [1]
- Unit correct [1]
Answer: $50 \text{ N m}^{-1}$

(b) Elastic potential energy. [2]

Formula: $E = \frac{1}{2} k x^2$ OR $E = \frac{1}{2} F x$
Working:
- $E = \frac{1}{2}(50)(0.04)^2$ [1]
- $E = 0.04 \text{ J}$ [1]
Answer: $0.04 \text{ J}$

(c) Change in energy. [2]

Answer: Quadruples [1].
Explanation: $E \propto x^2$ (or $E \propto F^2$ ). Since load doubles, extension doubles. $2^2 = 4$ times the energy [1].

12. Uniform Acceleration. (a) Acceleration. [2]

Formula: $a = \frac{v - u}{t}$
Working:
- $a = \frac{25 - 0}{10} = 2.5 \text{ m s}^{-2}$ [1]
- Unit correct [1]
Answer: $2.5 \text{ m s}^{-2}$

(b) Distance travelled. [2]

Formula: $s = ut + \frac{1}{2}at^2$ OR Area under graph.
Working:
- $s = 0 + \frac{1}{2}(2.5)(10)^2$ [1]
- $s = 125 \text{ m}$ [1]
Answer: $125 \text{ m}$

(c) Velocity-time graph. [2]

Sketch:
- Axes labeled $v / \text{m s}^{-1}$ and $t / \text{s}$ [1].
- Straight line from $(0,0)$ to $(10, 25)$ [1].

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