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A Level H1 Physics Mechanics Quiz

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Questions

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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 m s−1g = 9.81 \text{ m s}^{-1} unless otherwise stated.

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 N⋅s12.0 \text{ N}\cdot\text{s} and a kinetic energy of 36.0 J36.0 \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 m s−125.0 \text{ m s}^{-1} at an angle of 35∘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 m20.0 \text{ m} with a horizontal velocity of 15.0 m s−115.0 \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 aa against time tt for an object falling from rest with air resistance. Label the terminal velocity point on the tt-axis. [3]







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  7. A 0.50 kg0.50 \text{ kg} block is pushed across a rough horizontal surface with a constant force of 10.0 N10.0 \text{ N}. If the block accelerates at 4.0 m s−14.0 \text{ m s}^{-1}, calculate the magnitude of the frictional force. [2]
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Section B: Momentum and Collisions (Questions 8–13)

  1. Distinguish between an elastic collision and an inelastic collision in terms of kinetic energy. [2]
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  2. A 0.20 kg0.20 \text{ kg} trolley moving at 3.0 m s−13.0 \text{ m s}^{-1} collides with a stationary 0.30 kg0.30 \text{ kg} trolley. The two trolleys stick together after the collision. Calculate their common velocity. [3]
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  3. A 0.10 kg0.10 \text{ kg} mass moving at 5.0 m s−15.0 \text{ m s}^{-1} hits a stationary 0.10 kg0.10 \text{ kg} mass. After the collision, the first mass moves at 2.0 m s−12.0 \text{ m s}^{-1} at an angle of 45∘45^\circ to the original path. Calculate the final velocity of the second mass. [4]
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  4. 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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  5. A 0.05 kg0.05 \text{ kg} ball hits a wall at 12.0 m s−112.0 \text{ m s}^{-1} and rebounds perpendicularly at 8.0 m s−18.0 \text{ m s}^{-1}. Calculate the impulse delivered to the ball. [3]
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  6. A 2.0 kg2.0 \text{ kg} object is moving at 4.0 m s−14.0 \text{ m s}^{-1} when it collides elastically with a 1.0 kg1.0 \text{ kg} object moving in the opposite direction at 2.0 m s−12.0 \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)

  1. A uniform plank AB of length 4.0 m4.0 \text{ m} and weight 100 N100 \text{ N} is supported by two vertical pillars at its ends. A 60 kg60 \text{ kg} person stands 1.0 m1.0 \text{ m} from end A. Draw a free-body diagram of the plank, labeling all forces. [3]





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  2. Using the scenario in Question 14, calculate the reaction force at support A. [3]
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  3. Define the term "Power" and state its SI unit. [2]
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  4. A motor lifts a 200 kg200 \text{ kg} crate at a constant speed of 0.80 m s−10.80 \text{ m s}^{-1}. If the motor's input power is 2000 W2000 \text{ W}, calculate the efficiency of the motor. [3]
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  5. A 0.50 kg0.50 \text{ kg} block is released from rest at the top of a frictionless inclined plane at an angle of 30∘30^\circ to the horizontal. Calculate the velocity of the block after it has slid 5.0 m5.0 \text{ m} down the plane. [3]
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  6. A spring with a force constant k=500 N m−1k = 500 \text{ N m}^{-1} is compressed by 0.10 m0.10 \text{ m}. Calculate the elastic potential energy stored in the spring. [2]
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  7. A 1.0 kg1.0 \text{ kg} object is launched vertically upwards with a speed of 15.0 m s−115.0 \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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Answers

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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→v=p/mp = mv \rightarrow v = p/m
  • K=12mv2→K=12m(p/m)2=p2/2mK = \frac{1}{2}mv^2 \rightarrow K = \frac{1}{2}m(p/m)^2 = p^2/2m
  • m=p2/2K=(12)2/(2×36)=144/72=2.0 kgm = p^2 / 2K = (12)^2 / (2 \times 36) = 144 / 72 = 2.0 \text{ kg} [M1, A1]
  • v=p/m=12/2=6.0 m s−1v = p/m = 12 / 2 = 6.0 \text{ m s}^{-1} [A1]

3. Maximum Height

  • uy=25sin⁡35∘=14.34 m s−1u_y = 25 \sin 35^\circ = 14.34 \text{ m s}^{-1} [M1]
  • At max height, vy=0v_y = 0. Use v2=u2+2as→0=(14.34)2+2(−9.81)hv^2 = u^2 + 2as \rightarrow 0 = (14.34)^2 + 2(-9.81)h
  • h=(14.34)2/(2×9.81)=205.6/19.62=10.5 mh = (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 ms = -20.0 \text{ m}, uy=0u_y = 0, a=−9.81 m s−2a = -9.81 \text{ m s}^{-2}
  • s=ut+12at2→−20=0+0.5(−9.81)t2s = ut + \frac{1}{2}at^2 \rightarrow -20 = 0 + 0.5(-9.81)t^2
  • t=20/4.905=2.02 st = \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−DW - D) decreases, so acceleration decreases.
  • [B3] Eventually, drag equals weight (D=WD = W), net force is zero, and object moves at constant speed (terminal velocity).

6. Graph aa vs tt

  • [B1] Y-intercept at gg (9.81 m s−29.81 \text{ m s}^{-2}).
  • [B2] Curve decaying exponentially/smoothly towards the x-axis.
  • [B3] Asymptotically approaches a=0a = 0 as t→∞t \rightarrow \infty.

7. Frictional Force

  • Fnet=ma→10.0−f=(0.50)(4.0)F_{\text{net}} = ma \rightarrow 10.0 - f = (0.50)(4.0)
  • 10.0−f=2.0→f=8.0 N10.0 - 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

  • m1u1+m2u2=(m1+m2)vm_1u_1 + m_2u_2 = (m_1 + m_2)v
  • (0.20)(3.0)+(0.30)(0)=(0.20+0.30)v(0.20)(3.0) + (0.30)(0) = (0.20 + 0.30)v
  • 0.6=0.5v→v=1.2 m s−10.6 = 0.5v \rightarrow v = 1.2 \text{ m s}^{-1} [A1]

10. 2D Collision

  • xx-axis: 0.1(5)=0.1(2cos⁡45∘)+0.1v2x→0.5=0.141+0.1v2x→v2x=3.59 m s−10.1(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]
  • yy-axis: 0=0.1(2sin⁡45∘)+0.1v2y→0=0.141+0.1v2y→v2y=−1.41 m s−10 = 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]
  • v2=3.592+(−1.41)2=3.86 m s−1v_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

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

13. Total Kinetic Energy

  • Ktotal=12(2.0)(4.0)2+12(1.0)(2.0)2=16.0+2.0=18.0 JK_{\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 N100 \text{ N}) acting at center (2.0 m2.0 \text{ m} from A).
  • [B1] Weight of person (60×9.81=588.6 N60 \times 9.81 = 588.6 \text{ N}) acting 1.0 m1.0 \text{ m} from A.
  • [B1] Upward reaction forces RAR_A and RBR_B at ends.

15. Reaction Force A

  • Sum of moments about B =0= 0:
  • RA(4.0)−588.6(3.0)−100(2.0)=0R_A(4.0) - 588.6(3.0) - 100(2.0) = 0
  • 4RA=1765.8+200=1965.8→RA=491.5 N4R_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

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

18. Velocity on Incline

  • a=gsin⁡30∘=9.81×0.5=4.905 m s−2a = g \sin 30^\circ = 9.81 \times 0.5 = 4.905 \text{ m s}^{-2} [M1]
  • v2=0+2(4.905)(5.0)=49.05→v=7.0 m s−1v^2 = 0 + 2(4.905)(5.0) = 49.05 \rightarrow v = 7.0 \text{ m s}^{-1} [A1]

19. Elastic Potential Energy

  • U=12kx2=0.5(500)(0.10)2=0.5(500)(0.01)=2.5 JU = \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

  • K0=12(1.0)(15)2=112.5 JK_0 = \frac{1}{2}(1.0)(15)^2 = 112.5 \text{ J}
  • Kh=56.25 JK_h = 56.25 \text{ J}. Loss in K=56.25 JK = 56.25 \text{ J}.
  • Gain in PE=mgh→56.25=(1.0)(9.81)hPE = mgh \rightarrow 56.25 = (1.0)(9.81)h
  • h=56.25/9.81=5.73 mh = 56.25 / 9.81 = 5.73 \text{ m} [A1]