A Level H1 Physics Practice Paper 3

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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 $g = 9.81 \text{ m s}^{-2}$ where applicable.

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Section A: Fundamentals & Kinematics (Questions 1–6)

1. Write down the expressions for: (a) Linear momentum $p$ in terms of mass $m$ and velocity $v$. [1] (b) Kinetic energy $K$ in terms of mass $m$ and velocity $v$. [1]

   (a) ______________________________________________________ (b) ______________________________________________________

2. A small sphere has a horizontal momentum of $12.0 \text{ N s}$ and a kinetic energy of $36.0 \text{ J}$. Calculate the mass of the sphere. [3]

   
   
   
   Answer: ____________________

3. A ball is dropped from a height of $20 \text{ m}$ in a vacuum. Calculate the time taken to reach the ground. [2]

   
   
   
   Answer: ____________________

4. A ball is dropped from a height of $20 \text{ m}$ in air. Sketch the graph of vertical speed $v$ against time $t$ until it reaches terminal velocity. [2]

   [Space for Graph]
   
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5. With reference to your graph in Question 4, explain why the gradient of the curve decreases over time. [2]

   
   
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6. A projectile is launched at an angle $\theta$ to the horizontal. State the acceleration of the projectile in the horizontal direction, assuming air resistance is negligible. [1]

   Answer: ____________________

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Section B: Dynamics & Momentum (Questions 7–13)

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

   
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8. Two trolleys, A and B, of masses $1.5 \text{ kg}$ and $2.5 \text{ kg}$ respectively, move towards each other on a smooth horizontal track. Trolley A has a velocity of $2.0 \text{ m s}^{-1}$ and Trolley B has a velocity of $1.0 \text{ m s}^{-1}$. They collide and stick together. Calculate the final velocity of the combined mass. [3]

   
   
   
   Answer: ____________________

9. In the collision described in Question 8, determine whether the collision is elastic or inelastic. Justify your answer by calculating the change in kinetic energy. [4]

   
   
   
   Conclusion: ____________________

10. A $0.2 \text{ kg}$ ball hits a wall at $15 \text{ m s}^{-1}$ and rebounds at $10 \text{ m s}^{-1}$ in the opposite direction. Calculate the impulse exerted by the wall on the ball. [3]

    
    
    
    Answer: ____________________

11. A force $F$ acts on a body of mass $m$ for a time $t$. Show that the change in momentum is equal to the impulse. [2]

    
    
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12. A $50 \text{ g}$ bullet is fired into a $2.0 \text{ kg}$ wooden block resting on a smooth surface. The bullet becomes embedded in the block, and the system moves forward at $1.5 \text{ m s}^{-1}$. Calculate the initial velocity of the bullet. [3]

    
    
    
    Answer: ____________________

13. Explain why the total kinetic energy is not conserved in the collision described in Question 12. [2]

    
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Section C: Forces, Energy & Equilibrium (Questions 14–20)

14. A uniform plank AB of length $4.0 \text{ m}$ and weight $200 \text{ N}$ is supported by two pivots at its ends. A person of weight $600 \text{ N}$ stands $1.0 \text{ m}$ from end A. Draw a free-body diagram of the plank, labeling all forces. [3]

    [Space for Diagram]
    
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15. Using the scenario in Question 14, calculate the reaction force at pivot A. [3]

    
    
    
    Answer: ____________________

16. A block of mass $5.0 \text{ kg}$ is pushed up a smooth incline of $30^\circ$ to the horizontal with a constant velocity. Calculate the magnitude of the force applied parallel to the incline. [3]

    
    
    
    Answer: ____________________

17. A $2.0 \text{ kg}$ object is lifted vertically through a height of $1.5 \text{ m}$ in $0.8 \text{ s}$. Calculate the average power output. [3]

    
    
    
    Answer: ____________________

18. A spring with force constant $k = 500 \text{ N m}^{-1}$ is compressed by $0.04 \text{ m}$. Calculate the elastic potential energy stored in the spring. [2]

    
    
    
    Answer: ____________________

19. A car of mass $1200 \text{ kg}$ accelerates from rest to $25 \text{ m s}^{-1}$. Calculate the work done by the engine, assuming no friction. [3]

    
    
    
    Answer: ____________________

20. A body is in equilibrium under the action of three concurrent forces. State the two conditions that must be satisfied for this equilibrium to exist. [2]

    
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A-Level Physics H1 Quiz - Mechanics (Answer Key)

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

2. $p = mv \implies v = p/m$ $K = \frac{1}{2}mv^2 = \frac{1}{2}m(p/m)^2 = p^2 / 2m$ $m = p^2 / 2K = (12.0)^2 / (2 \times 36.0) = 144 / 72 = 2.0 \text{ kg}$ [M1 for formula/substitution, M1 for rearrangement, A1 for 2.0 kg]

3. $s = ut + \frac{1}{2}at^2 \implies 20 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = 40 / 9.81 \approx 4.077$ $t = 2.02 \text{ s}$ [M1 for equation, A1 for 2.02 s]

4. Graph should show $v$ starting at 0, increasing with a decreasing gradient (concave down), and leveling off to a horizontal line (terminal velocity). [2]

5. As speed increases, the upward air resistance (drag) increases. [1] The net downward force ($W - \text{Drag}$) decreases, resulting in a decrease in acceleration. [1]

6. $0 \text{ m s}^{-2}$ (No horizontal forces act on the projectile). [1]

7. In a closed/isolated system [1], the total linear momentum remains constant provided no external forces act. [1]

8. Total momentum before = Total momentum after $(1.5 \times 2.0) + (2.5 \times -1.0) = (1.5 + 2.5)v$ $3.0 - 2.5 = 4.0v$ $0.5 = 4.0v \implies v = 0.125 \text{ m s}^{-1}$ (in direction of A) [M1 for momentum conservation, M1 for substitution, A1 for 0.125 m/s]

9. $K_{\text{initial}} = \frac{1}{2}(1.5)(2)^2 + \frac{1}{2}(2.5)(1)^2 = 3.0 + 1.25 = 4.25 \text{ J}$ $K_{\text{final}} = \frac{1}{2}(4.0)(0.125)^2 = 0.03125 \text{ J}$ $\Delta K = 4.25 - 0.03125 = 4.21875 \text{ J}$ (Loss of energy) Conclusion: Inelastic collision. [M1 for initial KE, M1 for final KE, M1 for comparison, A1 for "Inelastic"]

10. $\text{Impulse} = \Delta p = m(v - u)$ Taking direction of impact as positive: $u = 15, v = -10$ $\text{Impulse} = 0.2(-10 - 15) = 0.2(-25) = -5.0 \text{ N s}$ Magnitude = $5.0 \text{ N s}$ [M1 for $\Delta p$ formula, M1 for correct sign/substitution, A1 for 5.0 N s]

11. $F = ma$ $F = m(dv/dt)$ $Ft = m(v - u) = \Delta p$ [M1 for $F=ma$, A1 for linking to $\Delta p$]

12. $m_1u_1 + m_2u_2 = (m_1 + m_2)v$ $(0.050 \times u) + (2.0 \times 0) = (2.05) \times 1.5$ $0.050u = 3.075$ $u = 61.5 \text{ m s}^{-1}$ [M1 for conservation of momentum, M1 for substitution, A1 for 61.5 m/s]

13. The collision is inelastic. [1] Energy is dissipated as heat/sound or used in the deformation of the wooden block. [1]

14. Diagram must show:

• Weight of plank ($200 \text{ N}$) acting at $2.0 \text{ m}$ (center). [1]
• Weight of person ($600 \text{ N}$) acting at $1.0 \text{ m}$ from A. [1]
• Upward reaction forces $R_A$ and $R_B$ at the ends. [1]

15. Take moments about pivot B: $\sum \text{Clockwise} = \sum \text{Anti-clockwise}$ $R_A(4.0) = (600 \times 3.0) + (200 \times 2.0)$ $4R_A = 1800 + 400 = 2200$ $R_A = 550 \text{ N}$ [M1 for moment equation, M1 for substitution, A1 for 550 N]

16. Constant velocity $\implies$ Net force = 0. $F = mg \sin(30^\circ)$ $F = 5.0 \times 9.81 \times 0.5 = 24.525 \text{ N}$ [M1 for $mg \sin\theta$, M1 for substitution, A1 for 24.5 N]

17. $W = mgh = 2.0 \times 9.81 \times 1.5 = 29.43 \text{ J}$ $P = W/t = 29.43 / 0.8 = 36.7875 \text{ W}$ [M1 for work done, M1 for power formula, A1 for 36.8 W]

18. $E_p = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.04)^2$ $E_p = 250 \times 0.0016 = 0.4 \text{ J}$ [M1 for formula, A1 for 0.4 J]

19. $W = \Delta KE = \frac{1}{2}mv^2 - 0$ $W = \frac{1}{2}(1200)(25)^2 = 600 \times 625 = 375,000 \text{ J}$ [M1 for $\frac{1}{2}mv^2$, M1 for substitution, A1 for $3.75 \times 10^5 \text{ J}$]

20.

1. The vector sum of all forces must be zero ($\sum F = 0$). [1]
2. The sum of moments about any point must be zero ($\sum \tau = 0$). [1]