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A Level H1 Physics Practice Paper 1

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Questions

TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

Practice Paper 1 (Version 1 of 5)

Subject: Physics
Level: H1 A-Level
Paper: Practice Paper 1 (Mechanics Focus)
Duration: 1 Hour
Total Marks: 40

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your Name, Class, and Date in the spaces provided.
Answer all questions.
You are advised to spend about 60 minutes on this paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
Show your working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
Use $g = 9.81 \, \text{m s}^{-2}$ unless otherwise stated.

Section A: Structured Questions (20 Marks)

Answer all questions in this section.

1. State the principle of conservation of linear momentum.
[2]

2. A car of mass $1200 \, \text{kg}$ travels at a constant speed of $25 \, \text{m s}^{-1}$ along a horizontal road. The total resistive force acting on the car is $800 \, \text{N}$ .
(a) Calculate the power developed by the car’s engine.
[2]

(b) Explain, in terms of energy transformations, why the power calculated in (a) is not equal to the rate of increase of the car’s kinetic energy.
[2]

3. A ball is thrown vertically upwards with an initial velocity of $15 \, \text{m s}^{-1}$ . Air resistance is negligible.
(a) Calculate the maximum height reached by the ball.
[2]

(b) On the axes below, sketch the variation with time $t$ of the velocity $v$ of the ball from the moment it is thrown until it returns to the starting height. Take upward velocity as positive.
[2]

      v / m s⁻¹
      ^
      |
      |
      |
      |
      |
      +--------------------------> t / s
      |
      |
      |
      |
      v

4. Two trolleys, A and B, move along a smooth horizontal 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 common velocity of the trolleys after the collision.
[3]

(b) Determine whether the collision is elastic or inelastic. Show your working.
[3]

5. A uniform plank AB of length $4.0 \, \text{m}$ and weight $200 \, \text{N}$ rests on two supports. Support X is at end A, and support Y is $1.0 \, \text{m}$ from end B. A boy of weight $400 \, \text{N}$ stands on the plank at a distance of $1.5 \, \text{m}$ from A.
(a) Draw a free-body diagram showing all the vertical forces acting on the plank. Label the forces clearly.
[2]

(b) Calculate the reaction force at support Y.
[4]

Section B: Data and Context Questions (20 Marks)

Answer all questions in this section.

6. A skydiver of mass $80 \, \text{kg}$ jumps from a stationary helicopter.
(a) Explain why the skydiver’s acceleration decreases as his speed increases, before he reaches terminal velocity.
[3]

(b) The skydiver reaches a terminal velocity of $50 \, \text{m s}^{-1}$ . Calculate the magnitude of the air resistance force acting on him at this speed.
[2]

(c) The skydiver opens his parachute. Describe and explain the change in his motion immediately after the parachute opens.
[3]

7. In a laboratory experiment, a student investigates the relationship between the force applied to a spring and its extension. The results are plotted on a graph of Force ( $F$ ) against Extension ( $x$ ). The graph is a straight line passing through the origin with a gradient of $25 \, \text{N m}^{-1}$ .
(a) State what physical quantity is represented by the gradient of this graph.
[1]

(b) Calculate the work done in extending the spring by $0.20 \, \text{m}$ .
[2]

(c) The student repeats the experiment with a second spring of the same material but twice the length. State and explain how the gradient of the new graph compares to the original gradient.
[2]

8. A projectile is launched from ground level with an initial velocity of $20 \, \text{m s}^{-1}$ at an angle of $30^\circ$ to the horizontal. Air resistance is negligible.
(a) Calculate the horizontal component of the initial velocity.
[1]

(b) Calculate the time taken for the projectile to reach its maximum height.
[2]

9. 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 $20^\circ$ to the horizontal. The block moves at a constant speed.
(a) Draw a free-body diagram for the block, showing the weight, normal contact force, applied force, and frictional force.
[2]

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

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

Practice Paper 1 (Version 1 of 5) - Answer Key & Marking Scheme

Subject: Physics
Level: H1 A-Level
Total Marks: 40

Section A: Structured Questions

1. State the principle of conservation of linear momentum. [2]

B1: In a closed system (or isolated system / system with no external forces),
B1: the total linear momentum remains constant (or total momentum before collision = total momentum after collision).

2. Car Power and Energy [4]

(a) Calculate power: [2]
- Since speed is constant, driving force $F = \text{resistive force} = 800 \, \text{N}$ . [M1]
- $P = Fv = 800 \times 25 = 20,000 \, \text{W}$ (or $20 \, \text{kW}$ ). [A1]
(b) Explain energy transformation: [2]
- B1: The work done by the engine is used to overcome resistive forces (friction/air resistance).
- B1: Energy is dissipated as heat/thermal energy (and sound), so there is no net increase in kinetic energy (since speed is constant).

3. Vertical Motion [4]

(a) Maximum height: [2]
- Using $v^2 = u^2 + 2as$ with $v=0, u=15, a=-9.81$ . [M1]
- $0 = 15^2 + 2(-9.81)s \Rightarrow s = \frac{225}{19.62} = 11.47 \, \text{m}$ .
- Answer: $11.5 \, \text{m}$ (3 s.f.). [A1]
(b) Velocity-time graph: [2]
- B1: Straight line with negative gradient (constant acceleration due to gravity).
- B1: Line starts at positive $v$ ( $+15$ ), crosses time axis ( $v=0$ ), and ends at negative $v$ ( $-15$ ) at twice the time to peak.

4. Collision of Trolleys [6]

(a) Common velocity: [3]
- Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ . [M1]
- $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v$ .
- $6.0 = 3.0v \Rightarrow v = 2.0 \, \text{m s}^{-1}$ . [A1] (Direction: to the right). [B1]
(b) Elastic or Inelastic: [3]
- Calculate KE before: $KE_i = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \, \text{J}$ . [M1]
- Calculate KE after: $KE_f = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \, \text{J}$ . [M1]
- Since $KE_i \neq KE_f$ (KE is lost), the collision is inelastic. [A1]

5. Equilibrium of Plank [6]

(a) Free-body diagram: [2]
- B1: Weight of plank ( $200 \, \text{N}$ ) acting downwards at center ( $2.0 \, \text{m}$ from A).
- B1: Weight of boy ( $400 \, \text{N}$ ) acting downwards at $1.5 \, \text{m}$ from A.
- (Note: Reaction forces $R_A$ at A and $R_Y$ at Y must also be shown upwards for full completeness, but question asks for "all vertical forces". Acceptable if $R_A$ and $R_Y$ are included. If only weights are drawn, max 1 mark. Ideally: 4 arrows total.)
- Correction for marking: To get 2 marks, student must show:
  1. Downward forces at correct positions.
  2. Upward reaction forces at supports A and Y.
(b) Reaction at Y: [4]
- Take moments about support A (to eliminate $R_A$ ). [M1]
- Clockwise moments = Anti-clockwise moments.
- Moment of Boy: $400 \times 1.5 = 600 \, \text{N m}$ .
- Moment of Plank Weight: $200 \times 2.0 = 400 \, \text{N m}$ (Center is at $2.0 \, \text{m}$ ).
- Total Clockwise Moment = $1000 \, \text{N m}$ . [M1]
- Anti-clockwise Moment from $R_Y$ : $R_Y \times 3.0$ (Since Y is $1.0 \, \text{m}$ from B, and length is $4.0 \, \text{m}$ , distance from A is $3.0 \, \text{m}$ ). [M1]
- $3.0 R_Y = 1000 \Rightarrow R_Y = 333.3 \, \text{N}$ .
- Answer: $333 \, \text{N}$ (3 s.f.). [A1]

Section B: Data and Context Questions

6. Skydiver [8]

(a) Explain decreasing acceleration: [3]
- B1: As speed increases, air resistance (drag) increases.
- B1: The resultant downward force ( $W - \text{Drag}$ ) decreases.
- B1: Since $F=ma$ , as resultant force decreases, acceleration decreases.
(b) Air resistance at terminal velocity: [2]
- At terminal velocity, acceleration is zero, so forces are balanced. [M1]
- Air resistance = Weight = $mg = 80 \times 9.81 = 784.8 \, \text{N}$ .
- Answer: $785 \, \text{N}$ (3 s.f.). [A1]
(c) Effect of opening parachute: [3]
- B1: Air resistance increases significantly (becomes much larger than weight).
- B1: Resultant force is now upwards (opposite to motion).
- B1: The skydiver decelerates (slows down) rapidly until a new, lower terminal velocity is reached.

7. Spring Experiment [5]

(a) Gradient represents: [1]
- B1: Spring constant ( $k$ ).
(b) Work done: [2]
- Work done = Area under graph = $\frac{1}{2}Fx$ or $\frac{1}{2}kx^2$ . [M1]
- $W = \frac{1}{2}(25)(0.20)^2 = 0.5 \times 25 \times 0.04 = 0.5 \, \text{J}$ . [A1]
(c) Second spring comparison: [2]
- B1: The gradient will be smaller (half the original).
- B1: For a spring of the same material and cross-section, $k$ is inversely proportional to length ( $k \propto 1/L$ ). Doubling length halves the stiffness.

8. Projectile Motion [6]

(a) Horizontal component: [1]
- $u_x = 20 \cos 30^\circ = 17.32 \, \text{m s}^{-1}$ .
- Answer: $17.3 \, \text{m s}^{-1}$ . [A1]
(b) Time to max height: [2]
- Vertical component $u_y = 20 \sin 30^\circ = 10 \, \text{m s}^{-1}$ .
- At max height, $v_y = 0$ . Using $v = u + at$ : $0 = 10 - 9.81t$ . [M1]
- $t = \frac{10}{9.81} = 1.019 \, \text{s}$ .
- Answer: $1.02 \, \text{s}$ . [A1]
(c) Horizontal distance (Range): [3]
- Total time of flight = $2 \times t_{\text{max height}} = 2 \times 1.019 = 2.038 \, \text{s}$ . [M1]
- Horizontal distance = $u_x \times t_{\text{total}}$ . [M1]
- $d = 17.32 \times 2.038 = 35.3 \, \text{m}$ . [A1]

9. Inclined Plane [4]

(a) Free-body diagram: [2]
- B1: Weight ( $mg$ ) acting vertically downwards.
- B1: Normal contact force ( $N$ ) perpendicular to the slope.
- (Also required for full correctness but marks often focused on orientation: Applied force $40 \, \text{N}$ up slope, Friction $f$ down slope. If student draws Weight and Normal correctly, award 1 mark. If all 4 forces are present and correctly oriented, award 2 marks.)
- Refined Marking:
  - 1 mark for Weight (vertical) and Normal (perpendicular to slope).
  - 1 mark for Applied Force (up slope) and Friction (down slope, opposing motion).
(b) Frictional force: [2]
- Since speed is constant, forces parallel to the slope are balanced. [M1]
- $F_{\text{applied}} = mg \sin \theta + f$ .
- $40 = (5.0)(9.81) \sin 20^\circ + f$ .
- $40 = 16.77 + f \Rightarrow f = 40 - 16.77 = 23.23 \, \text{N}$ .
- Answer: $23.2 \, \text{N}$ . [A1]

End of Marking Scheme

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