A Level H2 Physics Mechanics Quiz

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A-Level Physics H2 Quiz - Mechanics

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 60

Duration: 60 minutes
Total Marks: 60

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all working clearly. Marks are awarded for correct reasoning and steps, not just the final answer.
4. Use $g = 9.81 \text{ m s}^{-2}$ where appropriate.
5. The use of an approved scientific calculator is expected.

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Section A: Kinematics and Dynamics (Questions 1–5)

1. A car accelerates uniformly from rest along a straight horizontal road. It travels a distance of $120 \text{ m}$ in $8.0 \text{ s}$. (a) Calculate the acceleration of the car. [2]

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(b) Determine the velocity of the car at the end of the $8.0 \text{ s}$ interval. [1]

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2. A stone is thrown vertically upwards from the edge of a cliff with an initial velocity of $15 \text{ m s}^{-1}$. The cliff is $40 \text{ m}$ above sea level. Air resistance is negligible. (a) Calculate the maximum height reached by the stone above the point of projection. [2]

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(b) Calculate the time taken for the stone to hit the sea surface. [3]

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3. State Newton’s Second Law of Motion in terms of momentum. [1]

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4. A box of mass $25 \text{ kg}$ is pulled along a rough horizontal floor by a force of $100 \text{ N}$ acting at an angle of $30^\circ$ above the horizontal. The box moves at a constant velocity. (a) Calculate the magnitude of the frictional force acting on the box. [2]

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(b) Calculate the normal contact force exerted by the floor on the box. [2]

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5. Two objects, A and B, are dropped from the same height in a vacuum. Object A has a mass of $1 \text{ kg}$ and Object B has a mass of $5 \text{ kg}$. Explain, using Newton’s laws, why both objects accelerate at the same rate despite their different masses. [2]

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Section B: Work, Energy, and Power (Questions 6–10)

6. Define the work done by a force. [1]

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7. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards at a constant speed of $2.0 \text{ m s}^{-1}$. (a) Calculate the power output of the crane motor. [2]

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(b) Explain why the kinetic energy of the load remains constant during this process. [1]

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8. A block of mass $2.0 \text{ kg}$ slides down a smooth inclined plane from rest. The vertical height of the incline is $3.0 \text{ m}$. (a) Using the principle of conservation of energy, calculate the speed of the block at the bottom of the incline. [2]

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(b) If the plane were rough and the block reached the bottom with a speed of $6.0 \text{ m s}^{-1}$, calculate the work done against friction. [2]

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9. The graph below shows the variation of force $F$ with displacement $s$ for a spring being extended. (Imagine a linear graph starting from origin (0,0) to point (0.10 m, 20 N))

Calculate the elastic potential energy stored in the spring when the extension is $0.10 \text{ m}$. [2]

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10. A car of mass $1200 \text{ kg}$ travels up a hill inclined at $5^\circ$ to the horizontal at a constant speed of $20 \text{ m s}^{-1}$. The total resistive force (air resistance and friction) is $500 \text{ N}$. Calculate the power developed by the car’s engine. [3]

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Section C: Momentum and Collisions (Questions 11–15)

11. State the Principle of Conservation of Linear Momentum. [2]

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12. A trolley of mass $0.5 \text{ kg}$ moving at $1.2 \text{ m s}^{-1}$ collides with a stationary trolley of mass $1.5 \text{ kg}$. The two trolleys stick together after the collision. (a) Calculate the common velocity of the trolleys after the collision. [2]

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(b) Determine whether this collision is elastic or inelastic. Show your working. [2]

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13. A gas molecule of mass $m$ moves with speed $v$ perpendicular to a wall and rebounds elastically with the same speed. (a) State the change in momentum of the molecule. [1]

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(b) If the time of contact with the wall is $\Delta t$, derive an expression for the average force exerted by the molecule on the wall. [2]

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14. In a game of billiards, a white ball strikes a stationary red ball. After the collision, the white ball moves off at an angle of $30^\circ$ to its original direction, and the red ball moves at an angle of $60^\circ$ to the original direction of the white ball. Explain why the sum of the kinetic energies of the two balls after the collision is less than the initial kinetic energy of the white ball, assuming the collision is not perfectly elastic. [1]

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15. A rocket in deep space (no gravity) ejects fuel at a constant rate. Explain, in terms of momentum, why the rocket accelerates. [2]

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Section D: Circular Motion and Gravitation (Questions 16–20)

16. An object moves in a horizontal circle at constant speed. (a) Explain why the object is accelerating even though its speed is constant. [1]

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(b) State the direction of this acceleration. [1]

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17. A conical pendulum consists of a bob of mass $0.2 \text{ kg}$ attached to a string of length $0.5 \text{ m}$. The bob moves in a horizontal circle such that the string makes an angle of $30^\circ$ with the vertical. (a) Draw a free-body diagram showing the forces acting on the bob. [1]

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(b) Calculate the tension in the string. [2]

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(c) Calculate the centripetal force acting on the bob. [2]

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18. Define gravitational field strength at a point. [1]

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19. The Earth has a mass of $5.97 \times 10^{24} \text{ kg}$ and a radius of $6.37 \times 10^6 \text{ m}$. Calculate the gravitational field strength at the surface of the Earth. [2] (Gravitational constant $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$)

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20. A satellite orbits the Earth in a circular orbit of radius $r$. Show that the orbital speed $v$ of the satellite is given by $v = \sqrt{\frac{GM}{r}}$, where $M$ is the mass of the Earth. [2]

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A-Level Physics H2 Quiz - Mechanics (Answer Key)

1. (a) Using $s = ut + \frac{1}{2}at^2$: $120 = 0(8) + \frac{1}{2}a(8^2)$ $120 = 32a$ $a = 3.75 \text{ m s}^{-2}$ [2] (b) Using $v = u + at$: $v = 0 + 3.75(8) = 30 \text{ m s}^{-1}$ [1]

2. (a) At max height, $v=0$. Using $v^2 = u^2 + 2as$: $0 = 15^2 + 2(-9.81)s$ $19.61s = 225$ $s = 11.47 \text{ m} \approx 11.5 \text{ m}$ [2] (b) Displacement $s = -40 \text{ m}$ (taking up as positive). Using $s = ut + \frac{1}{2}at^2$: $-40 = 15t + \frac{1}{2}(-9.81)t^2$ $4.905t^2 - 15t - 40 = 0$ Using quadratic formula: $t = \frac{15 \pm \sqrt{225 - 4(4.905)(-40)}}{9.81}$ $t = \frac{15 \pm \sqrt{225 + 784.8}}{9.81} = \frac{15 \pm 31.78}{9.81}$ Taking positive root: $t = \frac{46.78}{9.81} = 4.77 \text{ s}$ [3]

3. The resultant force acting on an object is equal to the rate of change of its momentum. [1] (Or $F = \frac{dp}{dt}$)

4. (a) Horizontal component of pulling force $F_x = 100 \cos 30^\circ = 86.6 \text{ N}$. Since velocity is constant, acceleration is zero, so net force is zero. Friction $f = F_x = 86.6 \text{ N}$ [2] (b) Vertical forces balance: $N + F_y = mg$. $F_y = 100 \sin 30^\circ = 50 \text{ N}$. $N + 50 = 25(9.81)$ $N = 245.25 - 50 = 195.25 \text{ N} \approx 195 \text{ N}$ [2]

5. Force on object is weight $W = mg$. By Newton’s 2nd Law, $F = ma$. So, $mg = ma$. Mass $m$ cancels out, leaving $a = g$. Thus, acceleration is independent of mass. [2]

6. Work done is the product of the force and the displacement moved in the direction of the force. [1]

7. (a) Power $P = Fv$. Since speed is constant, $F = mg = 500(9.81) = 4905 \text{ N}$. $P = 4905 \times 2.0 = 9810 \text{ W} \approx 9.8 \text{ kW}$ [2] (b) Kinetic energy $KE = \frac{1}{2}mv^2$. Since mass and speed are constant, KE is constant. [1]

8. (a) Loss in GPE = Gain in KE. $mgh = \frac{1}{2}mv^2$ $gh = \frac{1}{2}v^2 \Rightarrow v = \sqrt{2gh}$ $v = \sqrt{2(9.81)(3.0)} = \sqrt{58.86} = 7.67 \text{ m s}^{-1}$ [2] (b) Initial Energy (GPE) $= mgh = 2(9.81)(3) = 58.86 \text{ J}$. Final Energy (KE) $= \frac{1}{2}mv^2 = 0.5(2)(6^2) = 36 \text{ J}$. Work done against friction = Energy Loss $= 58.86 - 36 = 22.86 \text{ J} \approx 22.9 \text{ J}$ [2]

9. Energy stored = Area under graph. Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$ $E = \frac{1}{2} \times 0.10 \times 20 = 1.0 \text{ J}$ [2]

10. Force required to overcome gravity component: $F_g = mg \sin \theta = 1200(9.81)\sin 5^\circ = 1025.5 \text{ N}$. Total driving force $F_D = F_g + F_{resistive} = 1025.5 + 500 = 1525.5 \text{ N}$. Power $P = F_D v = 1525.5 \times 20 = 30510 \text{ W} \approx 30.5 \text{ kW}$ [3]

11. In a closed system (no external forces), the total momentum before an interaction is equal to the total momentum after the interaction. [2]

12. (a) Conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $0.5(1.2) + 1.5(0) = (0.5 + 1.5)v$ $0.6 = 2.0v$ $v = 0.3 \text{ m s}^{-1}$ [2] (b) Initial KE $= \frac{1}{2}(0.5)(1.2)^2 = 0.36 \text{ J}$. Final KE $= \frac{1}{2}(2.0)(0.3)^2 = 0.09 \text{ J}$. Since $KE_{initial} \neq KE_{final}$ (KE is lost), the collision is inelastic. [2]

13. (a) Initial momentum $= mv$. Final momentum $= -mv$ (rebound). Change $= p_f - p_i = -mv - mv = -2mv$. Magnitude is $2mv$. [1] (b) Force $F = \frac{\Delta p}{\Delta t}$. $F = \frac{2mv}{\Delta t}$ [2]

14. Some kinetic energy is converted into other forms such as sound, heat, or deformation energy during the collision. [1]

15. The rocket ejects fuel backwards, giving the fuel backward momentum. To conserve total momentum of the system (rocket + fuel), the rocket must gain an equal and opposite forward momentum. This change in momentum over time results in a forward force (thrust) and thus acceleration. [2]

16. (a) Velocity is a vector quantity (speed + direction). Since the direction changes continuously, the velocity changes. A change in velocity implies acceleration. [1] (b) Towards the center of the circle. [1]

17. (a) Diagram should show: Weight ($mg$) acting vertically downwards, Tension ($T$) acting along the string towards the pivot. [1] (b) Vertical equilibrium: $T \cos 30^\circ = mg$. $T = \frac{0.2 \times 9.81}{\cos 30^\circ} = \frac{1.962}{0.866} = 2.265 \text{ N} \approx 2.27 \text{ N}$ [2] (c) Centripetal force is the horizontal component of Tension. $F_c = T \sin 30^\circ = 2.265 \times 0.5 = 1.13 \text{ N}$ (Alternatively $F_c = mg \tan 30^\circ = 1.962 \times 0.577 = 1.13 \text{ N}$) [2]

18. Gravitational field strength at a point is the gravitational force per unit mass acting on a small test mass placed at that point. [1]

19. $g = \frac{GM}{r^2}$ $g = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2}$ $g = \frac{3.982 \times 10^{14}}{4.058 \times 10^{13}} = 9.81 \text{ N kg}^{-1}$ [2]

20. Gravitational force provides centripetal force. $\frac{GMm}{r^2} = \frac{mv^2}{r}$ Cancel $m$ and one $r$: $\frac{GM}{r} = v^2$ $v = \sqrt{\frac{GM}{r}}$ [2]