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A Level H1 Physics Practice Paper 1

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Questions

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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 working clearly. Use g=9.81 m s−2g = 9.81 \text{ m s}^{-2} unless otherwise stated.

Section A: Fundamental Concepts (Questions 1–5)

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



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  2. Write down the expressions for: (a) Momentum pp in terms of mass mm and velocity vv. [1] (b) Kinetic energy KK in terms of mass mm and velocity vv. [1]

    (a) ________________________________________________________________________ (b) ________________________________________________________________________

  3. A small metal sphere has a horizontal momentum of 0.45 N s0.45 \text{ N s} and a kinetic energy of 1.2 J1.2 \text{ J}. Calculate the mass of the sphere. [3]



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  4. Define the term displacement and state whether it is a scalar or vector quantity. [2]



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  5. A ball is dropped from a height. Sketch the graph of vertical speed vs. time, taking into account the effect of air resistance. [2]




    (Space for sketch)

Section B: Kinematics and Dynamics (Questions 6–12)

  1. Explain the shape of the speed-time graph sketched in Question 5, specifically referring to the net force acting on the ball. [2]



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  2. A projectile is launched at an angle of 35∘35^\circ to the horizontal with an initial velocity of 25 m s−125 \text{ m s}^{-1}. Calculate the maximum height reached. [3]



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  3. A 0.5 kg0.5 \text{ kg} block slides down a rough inclined plane at 30∘30^\circ to the horizontal with a constant acceleration of 2.0 m s−22.0 \text{ m s}^{-2}. Calculate the magnitude of the frictional force acting on the block. [3]



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  4. Two trolleys, A (mass 2.0 kg2.0 \text{ kg}) and B (mass 3.0 kg3.0 \text{ kg}), move toward each other on a smooth track. A moves at 4.0 m s−14.0 \text{ m s}^{-1} and B moves at 2.0 m s−12.0 \text{ m s}^{-1}. They collide and stick together. Calculate the final velocity of the combined mass. [3]



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  5. Distinguish between an elastic collision and an inelastic collision. [2]



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  6. A car of mass 1200 kg1200 \text{ kg} decelerates from 30 m s−130 \text{ m s}^{-1} to 10 m s−110 \text{ m s}^{-1} in 5.0 s5.0 \text{ s}. Calculate the average net force acting on the car. [3]



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  7. A 2.0 kg2.0 \text{ kg} object is projected vertically upwards with a speed of 15 m s−115 \text{ m s}^{-1}. Calculate the time taken to reach the maximum height. [2]



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Section C: Equilibrium, Work, and Energy (Questions 13–20)

  1. A uniform plank AB of length 4.0 m4.0 \text{ m} and weight 100 N100 \text{ N} is placed across two supports. A person of weight 600 N600 \text{ N} stands 1.0 m1.0 \text{ m} from end A. Draw a free-body diagram of the plank, labeling all forces. [3]




    (Space for diagram)

  2. Using the scenario in Question 13, calculate the reaction force at support A if the other support is at end B. [3]



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  3. A 50 kg50 \text{ kg} crate is pulled 10 m10 \text{ m} across a floor by a force of 200 N200 \text{ N} acting at an angle of 20∘20^\circ to the horizontal. Calculate the work done by the pulling force. [3]



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  4. A pump lifts 20 kg20 \text{ kg} of water per minute from a well 15 m15 \text{ m} deep. Calculate the minimum power output of the pump. [3]



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  5. A ball of mass 0.2 kg0.2 \text{ kg} is dropped from a height of 5.0 m5.0 \text{ m}. If it bounces back to a height of 3.0 m3.0 \text{ m}, calculate the energy lost during the impact. [3]



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  6. A constant force of 15 N15 \text{ N} acts on a 3.0 kg3.0 \text{ kg} mass initially at rest. Calculate the velocity of the mass after it has moved 6.0 m6.0 \text{ m}. [3]



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  7. Explain why the efficiency of a real mechanical system is always less than 100%100\%. [2]



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  8. A 0.1 kg0.1 \text{ kg} block is pushed against a spring with spring constant k=500 N m−1k = 500 \text{ N m}^{-1}, compressing it by 0.05 m0.05 \text{ m}. When released, it slides on a frictionless surface. Calculate the speed of the block as it leaves the spring. [3]



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Answers

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Answer Key - A-Level Physics H1 Quiz: Mechanics

1. Conservation of Linear Momentum

  • [B1] In a closed/isolated system, the total momentum remains constant.
  • [B1] Provided no external forces act on the system.

2. Expressions

  • (a) p=mvp = mv [1]
  • (b) K=12mv2K = \frac{1}{2}mv^2 [1]

3. Mass Calculation

  • p=mv  ⟹  v=p/mp = mv \implies v = p/m [M1]
  • K=12mv2  ⟹  K=p22mK = \frac{1}{2}mv^2 \implies K = \frac{p^2}{2m} [M1]
  • m=p22K=0.4522×1.2=0.20252.4≈0.0844 kgm = \frac{p^2}{2K} = \frac{0.45^2}{2 \times 1.2} = \frac{0.2025}{2.4} \approx 0.0844 \text{ kg} [A1]

4. Displacement

  • Definition: The straight-line distance between the initial and final positions of a particle. [1]
  • Type: Vector quantity. [1]

5. Graph Sketch

  • [B1] Curve starts at origin, increases with a decreasing gradient (concave down).
  • [B1] Curve levels off to a horizontal line (terminal velocity).

6. Graph Explanation

  • [B1] As speed increases, the upward air resistance force increases.
  • [B1] The net downward force (W−RW - R) decreases, leading to a decrease in acceleration until R=WR = W (net force = 0).

7. Projectile Height

  • uy=25sin⁡(35∘)≈14.34 m s−1u_y = 25 \sin(35^\circ) \approx 14.34 \text{ m s}^{-1} [M1]
  • v2=u2+2as  ⟹  0=(14.34)2+2(−9.81)hv^2 = u^2 + 2as \implies 0 = (14.34)^2 + 2(-9.81)h [M1]
  • h=205.619.62≈10.5 mh = \frac{205.6}{19.62} \approx 10.5 \text{ m} [A1]

8. Frictional Force

  • Net force Fnet=mgsin⁡(30∘)−f=maF_{net} = mg \sin(30^\circ) - f = ma [M1]
  • 0.5(9.81)(0.5)−f=0.5(2.0)0.5(9.81)(0.5) - f = 0.5(2.0) [M1]
  • 2.45−f=1.0  ⟹  f=1.45 N2.45 - f = 1.0 \implies f = 1.45 \text{ N} [A1]

9. Collision Velocity

  • m1u1+m2u2=(m1+m2)vm_1u_1 + m_2u_2 = (m_1 + m_2)v [M1]
  • (2.0×4.0)+(3.0×−2.0)=(2.0+3.0)v(2.0 \times 4.0) + (3.0 \times -2.0) = (2.0 + 3.0)v [M1]
  • 8.0−6.0=5v  ⟹  v=0.4 m s−18.0 - 6.0 = 5v \implies v = 0.4 \text{ m s}^{-1} (in direction of A) [A1]

10. Collision Types

  • Elastic: Both momentum and kinetic energy are conserved. [1]
  • Inelastic: Only momentum is conserved; kinetic energy is lost (converted to heat/sound). [1]

11. Average Net Force

  • F=mΔvΔt=120010−305.0F = m \frac{\Delta v}{\Delta t} = 1200 \frac{10 - 30}{5.0} [M1]
  • F=1200×(−4.0)F = 1200 \times (-4.0) [M1]
  • F=−4800 NF = -4800 \text{ N} (or 4800 N4800 \text{ N} opposing motion) [A1]

12. Time to Max Height

  • v=u+at  ⟹  0=15+(−9.81)tv = u + at \implies 0 = 15 + (-9.81)t [M1]
  • t=159.81≈1.53 st = \frac{15}{9.81} \approx 1.53 \text{ s} [A1]

13. 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 (600 N600 \text{ N}) acting 1.0 m1.0 \text{ m} from A.
  • [B1] Upward reaction forces RAR_A and RBR_B at the supports.

14. Reaction Force RAR_A

  • Take moments about B: ∑MB=0\sum M_B = 0 [M1]
  • RA(4.0)−600(3.0)−100(2.0)=0R_A(4.0) - 600(3.0) - 100(2.0) = 0 [M1]
  • 4RA=1800+200  ⟹  RA=500 N4R_A = 1800 + 200 \implies R_A = 500 \text{ N} [A1]

15. Work Done

  • W=Fdcos⁡θ=200×10×cos⁡(20∘)W = Fd \cos \theta = 200 \times 10 \times \cos(20^\circ) [M1]
  • W=2000×0.9397W = 2000 \times 0.9397 [M1]
  • W≈1879 JW \approx 1879 \text{ J} [A1]

16. Power Output

  • P=Wt=mghtP = \frac{W}{t} = \frac{mgh}{t} [M1]
  • P=20×9.81×1560P = \frac{20 \times 9.81 \times 15}{60} [M1]
  • P=294360≈49.1 WP = \frac{2943}{60} \approx 49.1 \text{ W} [A1]

17. Energy Loss

  • $\Delta E = mgh_{initial} - mgh
<stage3_exam_answers_md>
# Answer Key - A-Level Physics H1 Quiz: Mechanics

**1. Conservation of Linear Momentum**
- [B1] In a closed/isolated system, the total momentum remains constant.
- [B1] Provided no external forces act on the system.

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

**3. Mass Calculation**
- $p = mv \implies v = p/m$ [M1]
- $K = \frac{1}{2}mv^2 \implies K = \frac{p^2}{2m}$ [M1]
- $m = \frac{p^2}{2K} = \frac{0.45^2}{2 \times 1.2} = \frac{0.2025}{2.4} \approx 0.0844 \text{ kg}$ [A1]

**4. Displacement**
- Definition: The straight-line distance between the initial and final positions of a particle. [1]
- Type: Vector quantity. [1]

**5. Graph Sketch**
- [B1] Curve starts at origin, increases with a decreasing gradient (concave down).
- [B1] Curve levels off to a horizontal line (terminal velocity).

**6. Graph Explanation**
- [B1] As speed increases, the upward air resistance force increases.
- [B1] The net downward force ($W - R$) decreases, leading to a decrease in acceleration until $R = W$ (net force = 0).

**7. Projectile Height**
- $u_y = 25 \sin(35^\circ) \approx 14.34 \text{ m s}^{-1}$ [M1]
- $v^2 = u^2 + 2as \implies 0 = (14.34)^2 + 2(-9.81)h$ [M1]
- $h = \frac{205.6}{19.62} \approx 10.5 \text{ m}$ [A1]

**8. Frictional Force**
- Net force $F_{net} = mg \sin(30^\circ) - f = ma$ [M1]
- $0.5(9.81)(0.5) - f = 0.5(2.0)$ [M1]
- $2.45 - f = 1.0 \implies f = 1.45 \text{ N}$ [A1]

**9. Collision Velocity**
- $m_1u_1 + m_2u_2 = (m_1 + m_2)v$ [M1]
- $(2.0 \times 4.0) + (3.0 \times -2.0) = (2.0 + 3.0)v$ [M1]
- $8.0 - 6.0 = 5v \implies v = 0.4 \text{ m s}^{-1}$ (in direction of A) [A1]

**10. Collision Types**
- Elastic: Both momentum and kinetic energy are conserved. [1]
- Inelastic: Only momentum is conserved; kinetic energy is lost (converted to heat/sound). [1]

**11. Average Net Force**
- $F = m \frac{\Delta v}{\Delta t} = 1200 \frac{10 - 30}{5.0}$ [M1]
- $F = 1200 \times (-4.0)$ [M1]
- $F = -4800 \text{ N}$ (or $4800 \text{ N}$ opposing motion) [A1]

**12. Time to Max Height**
- $v = u + at \implies 0 = 15 + (-9.81)t$ [M1]
- $t = \frac{15}{9.81} \approx 1.53 \text{ s}$ [A1]

**13. Free Body Diagram**
- [B1] Weight of plank ($100 \text{ N}$) acting at center ($2.0 \text{ m}$ from A).
- [B1] Weight of person ($600 \text{ N}$) acting $1.0 \text{ m}$ from A.
- [B1] Upward reaction forces $R_A$ and $R_B$ at the supports.

**14. Reaction Force $R_A$**
- Take moments about B: $\sum M_B = 0$ [M1]
- $R_A(4.0) - 600(3.0) - 100(2.0) = 0$ [M1]
- $4R_A = 1800 + 200 \implies R_A = 500 \text{ N}$ [A1]

**15. Work Done**
- $W = Fd \cos \theta = 200 \times 10 \times \cos(20^\circ)$ [M1]
- $W = 2000 \times 0.9397$ [M1]
- $W \approx 1879 \text{ J}$ [A1]

**16. Power Output**
- $P = \frac{W}{t} = \frac{mgh}{t}$ [M1]
- $P = \frac{20 \times 9.81 \times 15}{60}$ [M1]
- $P = \frac{2943}{60} \approx 49.1 \text{ W}$ [A1]

**17. Energy Loss**
- $\Delta E = mgh_{initial} - mgh_{final} = 0.2(9.81)(5.0 - 3.0)$ [M1]
- $\Delta E = 0.2 \times 9.81 \times 2.0$ [M1]
- $\Delta E = 3.92 \text{ J}$ [A1]

**18. Velocity Calculation**
- Work done = $\Delta K \implies Fs = \frac{1}{2}mv^2$ [M1]
- $15 \times 6.0 = \frac{1}{2}(3.0)v^2$ [M1]
- $90 = 1.5v^2 \implies v^2 = 60 \implies v \approx 7.75 \text{ m s}^{-1}$ [A1]

**19. Efficiency**
- [B1] Energy is always lost to the surroundings (e.g., as heat due to friction or sound).
- [B1] Therefore, useful energy output is always less than total energy input.

**20. Spring Speed**
- $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ [M1]
- $500(0.05)^2 = 0.1v^2$ [M1]
- $1.25 = 0.1v^2 \implies v^2 = 12.5 \implies v \approx 3.54 \text{ m s}^{-1}$ [A1]