A Level H1 Physics Mechanics Quiz

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A-Level Physics H1 Quiz - Mechanics

Name: ____________________ Class: __________ Date: __________ Score: ________ / 60

Duration: 90 Minutes
Total Marks: 60
Instructions: Answer all questions. Show all working clearly. Use $g = 9.81 \text{ m s}^{-1}$ unless otherwise stated.

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Section A: Kinematics and Dynamics (Questions 1–7)

1. State the principle of conservation of linear momentum. [2]
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2. A particle has a horizontal momentum of $12.0 \text{ N}\cdot\text{s}$ and a kinetic energy of $36.0 \text{ J}$. Calculate the mass and velocity of the particle. [3]
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3. A ball is launched from the ground with an initial velocity of $25.0 \text{ m s}^{-1}$ at an angle of $35^\circ$ to the horizontal. Calculate the maximum height reached by the ball. [3]
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4. A projectile is launched from a cliff of height $20.0 \text{ m}$ with a horizontal velocity of $15.0 \text{ m s}^{-1}$. Determine the time taken for the projectile to hit the ground. [2]
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5. Describe the motion of an object falling vertically through a viscous fluid, specifically explaining why it eventually reaches a terminal velocity. [3]
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6. Sketch a graph of acceleration $a$ against time $t$ for an object falling from rest with air resistance. Label the terminal velocity point on the $t$-axis. [3]
   
   
   
   
   
   
   
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7. A $0.50 \text{ kg}$ block is pushed across a rough horizontal surface with a constant force of $10.0 \text{ N}$. If the block accelerates at $4.0 \text{ m s}^{-1}$, calculate the magnitude of the frictional force. [2]
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Section B: Momentum and Collisions (Questions 8–13)

8. Distinguish between an elastic collision and an inelastic collision in terms of kinetic energy. [2]
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9. A $0.20 \text{ kg}$ trolley moving at $3.0 \text{ m s}^{-1}$ collides with a stationary $0.30 \text{ kg}$ trolley. The two trolleys stick together after the collision. Calculate their common velocity. [3]
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10. A $0.10 \text{ kg}$ mass moving at $5.0 \text{ m s}^{-1}$ hits a stationary $0.10 \text{ kg}$ mass. After the collision, the first mass moves at $2.0 \text{ m s}^{-1}$ at an angle of $45^\circ$ to the original path. Calculate the final velocity of the second mass. [4]
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11. Explain why the total momentum of a system is conserved during a collision even though the kinetic energy may not be. [2]
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12. A $0.05 \text{ kg}$ ball hits a wall at $12.0 \text{ m s}^{-1}$ and rebounds perpendicularly at $8.0 \text{ m s}^{-1}$. Calculate the impulse delivered to the ball. [3]
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13. A $2.0 \text{ kg}$ object is moving at $4.0 \text{ m s}^{-1}$ when it collides elastically with a $1.0 \text{ kg}$ object moving in the opposite direction at $2.0 \text{ m s}^{-1}$. Determine the total kinetic energy of the system before the collision. [3]
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Section C: Forces, Work, and Energy (Questions 14–20)

14. A uniform plank AB of length $4.0 \text{ m}$ and weight $100 \text{ N}$ is supported by two vertical pillars at its ends. A $60 \text{ kg}$ person stands $1.0 \text{ m}$ from end A. Draw a free-body diagram of the plank, labeling all forces. [3]
    
    
    
    
    
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15. Using the scenario in Question 14, calculate the reaction force at support A. [3]
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16. Define the term "Power" and state its SI unit. [2]
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17. A motor lifts a $200 \text{ kg}$ crate at a constant speed of $0.80 \text{ m s}^{-1}$. If the motor's input power is $2000 \text{ W}$, calculate the efficiency of the motor. [3]
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18. A $0.50 \text{ kg}$ block is released from rest at the top of a frictionless inclined plane at an angle of $30^\circ$ to the horizontal. Calculate the velocity of the block after it has slid $5.0 \text{ m}$ down the plane. [3]
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19. A spring with a force constant $k = 500 \text{ N m}^{-1}$ is compressed by $0.10 \text{ m}$. Calculate the elastic potential energy stored in the spring. [2]
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20. A $1.0 \text{ kg}$ object is launched vertically upwards with a speed of $15.0 \text{ m s}^{-1}$. Calculate the height at which its kinetic energy is equal to half of its initial kinetic energy. [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.
• [B1] Provided no external forces act on the system (or it is a closed/isolated system).

2. Mass and Velocity

• $p = mv \rightarrow v = p/m$
• $K = \frac{1}{2}mv^2 \rightarrow K = \frac{1}{2}m(p/m)^2 = p^2/2m$
• $m = p^2 / 2K = (12)^2 / (2 \times 36) = 144 / 72 = 2.0 \text{ kg}$ [M1, A1]
• $v = p/m = 12 / 2 = 6.0 \text{ m s}^{-1}$ [A1]

3. Maximum Height

• $u_y = 25 \sin 35^\circ = 14.34 \text{ m s}^{-1}$ [M1]
• At max height, $v_y = 0$. Use $v^2 = u^2 + 2as \rightarrow 0 = (14.34)^2 + 2(-9.81)h$
• $h = (14.34)^2 / (2 \times 9.81) = 205.6 / 19.62 = 10.5 \text{ m}$ [A1]

4. Time to hit ground

• Vertical displacement $s = -20.0 \text{ m}$, $u_y = 0$, $a = -9.81 \text{ m s}^{-2}$
• $s = ut + \frac{1}{2}at^2 \rightarrow -20 = 0 + 0.5(-9.81)t^2$
• $t = \sqrt{20 / 4.905} = 2.02 \text{ s}$ [A1]

5. Terminal Velocity

• [B1] As object falls, speed increases, causing upward drag force (air resistance) to increase.
• [B2] Net force ($W - D$) decreases, so acceleration decreases.
• [B3] Eventually, drag equals weight ($D = W$), net force is zero, and object moves at constant speed (terminal velocity).

6. Graph $a$ vs $t$

• [B1] Y-intercept at $g$ ($9.81 \text{ m s}^{-2}$).
• [B2] Curve decaying exponentially/smoothly towards the x-axis.
• [B3] Asymptotically approaches $a = 0$ as $t \rightarrow \infty$.

7. Frictional Force

• $F_{\text{net}} = ma \rightarrow 10.0 - f = (0.50)(4.0)$
• $10.0 - f = 2.0 \rightarrow f = 8.0 \text{ N}$ [A1]

8. Elastic vs Inelastic

• [B1] In elastic collisions, total kinetic energy is conserved.
• [B1] In inelastic collisions, total kinetic energy is not conserved (some converted to heat/sound).

9. Common Velocity

• $m_1u_1 + m_2u_2 = (m_1 + m_2)v$
• $(0.20)(3.0) + (0.30)(0) = (0.20 + 0.30)v$
• $0.6 = 0.5v \rightarrow v = 1.2 \text{ m s}^{-1}$ [A1]

10. 2D Collision

• $x$-axis: $0.1(5) = 0.1(2\cos 45^\circ) + 0.1v_{2x} \rightarrow 0.5 = 0.141 + 0.1v_{2x} \rightarrow v_{2x} = 3.59 \text{ m s}^{-1}$ [M1]
• $y$-axis: $0 = 0.1(2\sin 45^\circ) + 0.1v_{2y} \rightarrow 0 = 0.141 + 0.1v_{2y} \rightarrow v_{2y} = -1.41 \text{ m s}^{-1}$ [M1]
• $v_2 = \sqrt{3.59^2 + (-1.41)^2} = 3.86 \text{ m s}^{-1}$ [A1]

11. Momentum vs Energy

• [B1] Momentum is conserved because there are no external forces acting on the system (Newton's 3rd Law).
• [B1] Kinetic energy is not conserved because internal work is done (deformation/heat) during the collision.

12. Impulse

• $\Delta p = m(v - u) = 0.05(8 - (-12)) = 0.05(20) = 1.0 \text{ N s}$ [A1]

13. Total Kinetic Energy

• $K_{\text{total}} = \frac{1}{2}(2.0)(4.0)^2 + \frac{1}{2}(1.0)(2.0)^2 = 16.0 + 2.0 = 18.0 \text{ J}$ [A1]

14. Free Body Diagram

• [B1] Weight of plank ($100 \text{ N}$) acting at center ($2.0 \text{ m}$ from A).
• [B1] Weight of person ($60 \times 9.81 = 588.6 \text{ N}$) acting $1.0 \text{ m}$ from A.
• [B1] Upward reaction forces $R_A$ and $R_B$ at ends.

15. Reaction Force A

• Sum of moments about B $= 0$:
• $R_A(4.0) - 588.6(3.0) - 100(2.0) = 0$
• $4R_A = 1765.8 + 200 = 1965.8 \rightarrow R_A = 491.5 \text{ N}$ [A1]

16. Power

• [B1] Rate of doing work or rate of energy transfer.
• [B1] Unit: Watt (W).

17. Efficiency

• $P_{\text{out}} = Fv = (200 \times 9.81) \times 0.80 = 1569.6 \text{ W}$ [M1]
• $\text{Eff} = (1569.6 / 2000) \times 100\% = 78.5\%$ [A1]

18. Velocity on Incline

• $a = g \sin 30^\circ = 9.81 \times 0.5 = 4.905 \text{ m s}^{-2}$ [M1]
• $v^2 = 0 + 2(4.905)(5.0) = 49.05 \rightarrow v = 7.0 \text{ m s}^{-1}$ [A1]

19. Elastic Potential Energy

• $U = \frac{1}{2}kx^2 = 0.5(500)(0.10)^2 = 0.5(500)(0.01) = 2.5 \text{ J}$ [A1]

20. Height for Half KE

• $K_0 = \frac{1}{2}(1.0)(15)^2 = 112.5 \text{ J}$
• $K_h = 56.25 \text{ J}$. Loss in $K = 56.25 \text{ J}$.
• Gain in $PE = mgh \rightarrow 56.25 = (1.0)(9.81)h$
• $h = 56.25 / 9.81 = 5.73 \text{ m}$ [A1]