A Level H1 Physics Practice Paper 5

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A-Level Physics H1 Quiz - Mechanics

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 55

Duration: 90 Minutes
Total Marks: 55

Instructions:

• Answer all questions.
• Show all necessary working for calculation questions.
• Use $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.
• Write your answers in the spaces provided.

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Section A: Fundamentals & Kinematics

1. State the principle of conservation of linear momentum. [2]
   
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2. Write down the expressions for: (a) Momentum $p$ in terms of mass $m$ and velocity $v$. [1] (b) Kinetic energy $K$ in terms of mass $m$ and velocity $v$. [1]
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3. A small metal sphere has a horizontal momentum of $0.45 \text{ kg m s}^{-1}$ and a kinetic energy of $0.12 \text{ J}$. Calculate the mass of the sphere. [3]
   
   
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4. A ball is dropped from a height of $20 \text{ m}$. Calculate the velocity of the ball immediately before it hits the ground, assuming no air resistance. [2]
   
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5. A projectile is launched at an angle of $30^\circ$ to the horizontal with an initial velocity of $40 \text{ m s}^{-1}$. Determine the time taken to reach its maximum height. [2]
   
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6. Sketch a graph of vertical speed versus time for an object falling from a great height through a fluid, accounting for air resistance. [2]
   
   
   (Space for graph)

7. Explain the shape of the graph sketched in Question 6, specifically referring to the net force acting on the object. [2]
   
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Section B: Dynamics & Forces

8. State Newton's Second Law of Motion. [2]
   
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9. A $2.0 \text{ kg}$ block is pushed across a smooth horizontal surface by a constant force of $10 \text{ N}$ at an angle of $20^\circ$ to the horizontal. Calculate the acceleration of the block. [3]
   
   
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10. Define the term "impulse" and state its SI unit. [2]
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11. A $0.15 \text{ kg}$ tennis ball traveling at $25 \text{ m s}^{-1}$ hits a wall perpendicularly and rebounds at $20 \text{ m s}^{-1}$. Calculate the change in momentum of the ball. [3]
    
    
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12. Two trolleys, A ($1.0 \text{ kg}$) and B ($2.0 \text{ kg}$), are moving toward each other on a smooth track. A moves at $3.0 \text{ m s}^{-1}$ and B moves at $2.0 \text{ m s}^{-1}$. They collide and stick together. Calculate the final velocity of the combined mass. [3]
    
    
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13. Explain why the total kinetic energy is not conserved in an inelastic collision, even though momentum is conserved. [2]
    
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14. A uniform plank AB of length $4.0 \text{ m}$ and weight $100 \text{ N}$ is supported by two pivots at its ends. A $600 \text{ N}$ person stands $1.0 \text{ m}$ from end A. Draw a free-body diagram of the plank, labeling all forces. [3]
    
    
    (Space for diagram)

15. Using the scenario in Question 14, calculate the reaction force at pivot A. [3]
    
    
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Section C: Work, Energy & Power

16. Define "work done" by a force and state the condition under which no work is done despite a force being applied. [2]
    
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17. A $50 \text{ kg}$ crate is pulled up a rough ramp inclined at $30^\circ$ to the horizontal at a constant speed. If the coefficient of friction is $0.2$, calculate the work done by the pulling force over a distance of $5.0 \text{ m}$ along the ramp. [4]
    
    
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18. An electric motor lifts a $20 \text{ kg}$ mass vertically through a height of $3.0 \text{ m}$ in $4.0 \text{ s}$. Calculate the average power output of the motor. [3]
    
    
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19. A car of mass $1200 \text{ kg}$ accelerates from rest to $20 \text{ m s}^{-1}$ in $8.0 \text{ s}$. Calculate the average power delivered by the engine, assuming no friction. [3]
    
    
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20. A ball of mass $0.5 \text{ kg}$ is dropped from $10 \text{ m}$. It bounces back to a height of $7 \text{ m}$. Calculate the energy lost during the impact with the floor. [3]
    
    
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Answer Key - A-Level Physics H1 Quiz (Mechanics)

1. Principle of Conservation of Linear Momentum

   • [B1] Total momentum of a system remains constant / is conserved.
   • [B1] Provided no external forces act on the system / in a closed system.

2. Expressions

   • (a) $p = mv$ [B1]
   • (b) $K = \frac{1}{2}mv^2$ [B1]

3. Mass Calculation

   • $p = 0.45$, $K = 0.12$
   • $K = \frac{p^2}{2m} \implies m = \frac{p^2}{2K}$ [M1]
   • $m = \frac{0.45^2}{2 \times 0.12} = \frac{0.2025}{0.24}$ [M1]
   • $m = 0.844 \text{ kg}$ [A1]

4. Velocity Calculation

   • $v^2 = u^2 + 2as \implies v^2 = 0 + 2(9.81)(20)$ [M1]
   • $v = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$ [A1]

5. Projectile Motion

   • $u_y = 40 \sin(30^\circ) = 20 \text{ m s}^{-1}$ [M1]
   • At max height $v_y = 0 \implies 0 = 20 - 9.81t \implies t = \frac{20}{9.81}$ [M1]
   • $t = 2.04 \text{ s}$ [A1] (Note: 2 marks allocated, accept $2.0$ if $g=10$ used)

6. Graph

   • [B1] Curve starting from origin, increasing gradient initially, then flattening to a horizontal asymptote.
   • [B1] Correct labeling of axes (Speed on y, Time on x).

7. Explanation

   • [B1] As speed increases, air resistance (drag) increases.
   • [B1] Net force ($W - F_{\text{drag}}$) decreases, so acceleration decreases until $F_{\text{drag}} = W$, resulting in terminal velocity.

8. Newton's Second Law

   • [B1] The rate of change of momentum of an object is directly proportional to the resultant force acting on it.
   • [B1] And takes place in the direction of the force. (Accept $F = ma$ if defined as resultant force).

9. Acceleration

   • $F_x = 10 \cos(20^\circ) = 9.40 \text{ N}$ [M1]
   • $a = \frac{F_x}{m} = \frac{9.40}{2.0}$ [M1]
   • $a = 4.70 \text{ m s}^{-2}$ [A1]

10. Impulse

    • [B1] The product of the force and the time interval over which it acts / change in momentum.
    • [B1] Unit: $\text{N s}$ or $\text{kg m s}^{-1}$.

11. Change in Momentum

    • $\Delta p = mv - mu = 0.15(20) - 0.15(-25)$ [M1] (Directional change)
    • $\Delta p = 3.0 + 3.75$ [M1]
    • $\Delta p = 6.75 \text{ kg m s}^{-1}$ [A1]

12. Collision

    • $m_A u_A + m_B u_B = (m_A + m_B)v$
    • $(1.0)(3.0) + (2.0)(-2.0) = (3.0)v$ [M1]
    • $3.0 - 4.0 = 3.0v \implies -1.0 = 3.0v$ [M1]
    • $v = -0.333 \text{ m s}^{-1}$ (opposite to A's initial direction) [A1]

13. Energy Conservation

    • [B1] In an inelastic collision, some kinetic energy is converted into other forms (e.g., heat, sound, internal deformation energy).
    • [B1] Momentum is conserved because no external forces act, but internal forces do work to deform the objects.

14. Free-Body Diagram

    • [B1] Weight of plank ($100 \text{ N}$) at center.
    • [B1] Weight of person ($600 \text{ N}$) at $1.0 \text{ m}$ from A.
    • [B1] Upward reaction forces at A and B.

15. Reaction Force

    • Moments about B: $(R_A \times 4.0) - (600 \times 3.0) - (100 \times 2.0) = 0$ [M1]
    • $4 R_A = 1800 + 200$ [M1]
    • $R_A = \frac{2000}{4} = 500 \text{ N}$ [A1]

16. Work Done

    • [B1] Product of the force and displacement in the direction of the force ($W = Fs \cos \theta$).
    • [B1] No work is done if the force is perpendicular to the displacement ($\theta = 90^\circ$).

17. Ramp Calculation

    • $F_{\text{parallel}} = mg \sin(30^\circ) + \mu mg \cos(30^\circ)$ [M1]
    • $F_{\text{parallel}} = (50)(9.81)(0.5) + (0.2)(50)(9.81)(0.866)$ [M1]
    • $F_{\text{parallel}} = 245.25 + 84.95 = 330.2 \text{ N}$ [M1]
    • $W = 330.2 \times 5.0 = 1651 \text{ J}$ [A1]

18. Power Output

    • $W = mgh = 20 \times 9.81 \times 3.0 = 588.6 \text{ J}$ [M1]
    • $P = \frac{W}{t} = \frac{588.6}{4.0}$ [M1]
    • $P = 147 \text{ W}$ [A1]

19. Average Power

    • $\Delta KE = \frac{1}{2}mv^2 = 0.5 \times 1200 \times 20^2 = 240,000 \text{ J}$ [M1]
    • $P = \frac{\Delta KE}{t} = \frac{240,000}{8.0}$ [M1]
    • $P = 30,000 \text{ W}$ or $30 \text{ kW}$ [A1]

20. Energy Loss

    • Initial $PE = 0.5 \times 9.81 \times 10 = 49.05 \text{ J}$ [M1]
    • Final $PE = 0.5 \times 9.81 \times 7 = 34.34 \text{ J}$ [M1]
    • Loss $= 49.05 - 34.34 = 14.71 \text{ J}$ [A1]