A Level H2 Physics Practice Paper 4

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI) - Version 4

Subject: Physics H2
Level: A-Level
Paper: Structured Questions (Practice Set)
Duration: 2 hours
Total Marks: 80
Name: ____________________ Class: __________ Date: __________

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Instructions to Candidates

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Use $g = 9.81 \text{ m s}^{-2}$ and other constants as provided in the data booklet.
4. Show all working clearly.

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Section A: Foundational Mechanics and Fields

Question 1 (a) State the principle of conservation of linear momentum. [2]

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(b) A smooth horizontal surface is used for a collision experiment between two trolleys of masses $0.50\text{ kg}$ and $0.80\text{ kg}$. The first trolley moves at $2.0\text{ m s}^{-1}$ and strikes the second, which is initially stationary. After the collision, the first trolley moves at $0.50\text{ m s}^{-1}$ at an angle of $30^\circ$ to its original direction. (i) Calculate the velocity of the second trolley after the collision. [4] (ii) Determine whether the collision is elastic. Justify your answer with calculations. [3]

Question 2 A satellite of mass $m$ is in a stable circular orbit around a planet of mass $M$ at a height $h$ above the planet's surface. The radius of the planet is $R$. (a) Show that the orbital period $T$ is given by $T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$. [4] (b) If the satellite's altitude is increased, explain how the orbital speed and the period are affected. [2] (c) Calculate the escape velocity from the surface of the planet in terms of $G, M,$ and $R$. [3]

Question 3 A mass $m = 0.20\text{ kg}$ is attached to a vertical spring with spring constant $k = 80\text{ N m}^{-1}$. The mass is pulled down $0.05\text{ m}$ from its equilibrium position and released. (a) Calculate the angular frequency $\omega$ of the resulting oscillation. [2] (b) Determine the maximum acceleration of the mass. [2] (c) State the total energy of the system in terms of $k$ and the amplitude $X_0$. [2]

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Section B: Dynamics and Energy

Question 4 A block of mass $2.0\text{ kg}$ is pushed against a horizontal spring (constant $k = 500\text{ N m}^{-1}$) and compressed by $0.10\text{ m}$. The block is released and slides across a rough surface with a coefficient of kinetic friction $\mu = 0.20$. (a) Calculate the initial kinetic energy of the block the moment it leaves the spring. [2] (b) Calculate the distance the block slides before coming to rest. [4] (c) Discuss how the distance would change if the surface were lubricated to reduce friction. [2]

Question 5 Two particles, $A$ and $B$, are connected by a light inextensible string passing over a smooth frictionless pulley. Particle $A$ ($m_A = 3.0\text{ kg}$) sits on a rough horizontal table ($\mu = 0.30$), and particle $B$ ($m_B = 2.0\text{ kg}$) hangs vertically. (a) Draw a free-body diagram for particle $A$. [2] (b) Calculate the acceleration of the system when released from rest. [4] (c) Calculate the tension in the string. [2]

Question 6 A particle moves in a vertical circle of radius $1.5\text{ m}$. At the lowest point of the circle, the velocity of the particle is $5.0\text{ m s}^{-1}$. (a) Calculate the normal force acting on the particle at the lowest point. [3] (b) Determine the minimum velocity required at the lowest point so that the particle just maintains contact with the circle at the highest point. [4]

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Section C: Integrated Analysis and Practical Application

Question 7 An experiment is conducted to determine the acceleration of free fall $g$ using a falling object and an electronic timer. (a) Describe the precautions that should be taken to improve the accuracy of the measurement of the fall distance. [3] (b) Explain how the effect of air resistance can be minimized in this experiment. [2] (c) If the timer has a systematic error of $+0.01\text{ s}$, explain the effect on the calculated value of $g$. [3]

Question 8 A projectile is launched from ground level with an initial velocity $u$ at an angle $\theta$ to the horizontal. (a) Derive the expression for the maximum height $H$ reached by the projectile. [3] (b) Show that the horizontal range $R$ is maximized when $\theta = 45^\circ$. [3] (c) A ball is launched at $20\text{ m s}^{-1}$ at $30^\circ$. Calculate the time of flight. [3]

Question 9 A system consists of two masses $m_1 = 1.0\text{ kg}$ and $m_2 = 2.0\text{ kg}$ moving towards each other on a frictionless surface. $m_1$ moves at $4.0\text{ m s}^{-1}$ and $m_2$ moves at $2.0\text{ m s}^{-1}$. They collide and stick together. (a) Calculate the final velocity of the combined mass. [3] (b) Calculate the loss in kinetic energy during the collision. [3] (c) Explain why the total momentum is conserved but kinetic energy is not. [2]

Question 10 A conical pendulum consists of a bob of mass $m$ rotating in a horizontal circle of radius $r$ at a constant speed $v$, attached to a string of length $L$ making an angle $\theta$ with the vertical. (a) Resolve the forces acting on the bob to find an expression for the tension $T$ in terms of $m, g,$ and $\theta$. [3] (b) Derive an expression for the period $T_{period}$ of the rotation. [4] (c) What happens to the period if the mass of the bob is doubled? Explain. [2]

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Answer Key - TuitionGoWhere Practice Paper (AI) Version 4

Section A

Question 1 (a) In a closed system (or isolated system), the total momentum before an event equals the total momentum after the event, provided no external forces act. [2] (b) (i) x-axis: $(0.5 \times 2.0) = (0.5 \times 0.5 \cos 30^\circ) + (0.8 \times v_{2x}) \implies 1.0 = 0.2165 + 0.8v_{2x} \implies v_{2x} = 0.98\text{ m s}^{-1}$ y-axis: $0 = (0.5 \times 0.5 \sin 30^\circ) + (0.8 \times v_{2y}) \implies 0 = 0.125 + 0.8v_{2y} \implies v_{2y} = -0.156\text{ m s}^{-1}$ $v_2 = \sqrt{0.98^2 + (-0.156)^2} = 0.99\text{ m s}^{-1}$ [4] (ii) $KE_{initial} = 0.5 \times 0.5 \times 2^2 = 1.0\text{ J}$ $KE_{final} = (0.5 \times 0.5 \times 0.5^2) + (0.5 \times 0.8 \times 0.99^2) = 0.0625 + 0.392 = 0.45\text{ J}$ $KE_{initial} \neq KE_{final}$, therefore the collision is inelastic. [3]

Question 2 (a) Centripetal force = Gravitational force: $mv^2/(R+h) = GMm/(R+h)^2 \implies v^2 = GM/(R+h)$. $T = 2\pi(R+h)/v = 2\pi(R+h) / \sqrt{GM/(R+h)} = 2\pi \sqrt{(R+h)^3/GM}$. [4] (b) Orbital speed $v \propto 1/\sqrt{R+h}$, so speed decreases. Period $T \propto (R+h)^{3/2}$, so period increases. [2] (c) $KE + PE = 0 \implies 0.5mv^2 - GMm/R = 0 \implies v = \sqrt{2GM/R}$. [3]

Question 3 (a) $\omega = \sqrt{k/m} = \sqrt{80/0.2} = \sqrt{400} = 20\text{ rad s}^{-1}$. [2] (b) $a_{\max} = \omega^2 X_0 = 20^2 \times 0.05 = 400 \times 0.05 = 20\text{ m s}^{-2}$. [2] (c) $E = 0.5 k X_0^2$. [2]

Section B

Question 4 (a) $KE = 0.5 k X_0^2 = 0.5 \times 500 \times 0.1^2 = 2.5\text{ J}$. [2] (b) Work done by friction = Initial Energy $\implies \mu mg d = 2.5 \implies 0.2 \times 2 \times 9.81 \times d = 2.5 \implies 3.924d = 2.5 \implies d = 0.64\text{ m}$. [4] (c) Lubrication reduces $\mu$, therefore the distance $d$ increases as less energy is dissipated per unit distance. [2]

Question 5 (a) Forces: Tension $T$ (right), Friction $f$ (left), Weight $m_Ag$ (down), Normal force $N$ (up). [2] (b) For A: $T - \mu m_Ag = m_Aa \implies T - (0.3 \times 3 \times 9.81) = 3a \implies T - 8.829 = 3a$ For B: $m_Bg - T = m_Ba \implies (2 \times 9.81) - T = 2a \implies 19.62 - T = 2a$ Adding: $19.62 - 8.829 = 5a \implies 10.791 = 5a \implies a = 2.16\text{ m s}^{-2}$. [4] (c) $T = 3(2.16) + 8.829 = 15.31\text{ N}$. [2]

Question 6 (a) $F_{net} = T - mg = mv^2/r \implies T = m(g + v^2/r) = m(9.81 + 25/1.5) = m(9.81 + 16.67) = 26.48m$. (Since $m$ not given, answer as $26.5m$ or assume $m=1$ for $26.5\text{ N}$). [3] (b) At top: $T + mg = mv^2/r$. For min velocity, $T=0 \implies mg = mv_{top}^2/r \implies v_{top} = \sqrt{gr} = \sqrt{9.81 \times 1.5} = 3.84\text{ m s}^{-1}$. Using energy: $0.5mv_{bot}^2 - mg(2r) = 0.5mv_{top}^2 \implies v_{bot}^2 = v_{top}^2 + 4gr = gr + 4gr = 5gr$. $v_{bot} = \sqrt{5 \times 9.81 \times 1.5} = \sqrt{73.575} = 8.58\text{ m s}^{-1}$. [4]

Section C

Question 7 (a) Use a set square to ensure the ruler is vertical; use a fiducial marker for the start/end points. [3] (b) Use a denser, smaller object (e.g., steel ball bearing) to increase the ratio of weight to surface area. [2] (c) $t_{measured} = t_{actual} + 0.01$. Since $g = 2s/t^2$, a larger $t$ results in a smaller calculated $g$. [3]

Question 8 (a) $v_y = u \sin\theta - gt$. At max height $v_y = 0 \implies t = (u \sin\theta)/g$. $H = (u \sin\theta)t - 0.5gt^2 = (u^2 \sin^2\theta)/g - 0.5(u^2 \sin^2\theta)/g = (u^2 \sin^2\theta)/(2g)$. [3] (b) $R = (u^2 \sin 2\theta)/g$. $R$ is max when $\sin 2\theta = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ$. [3] (c) $t = (2u \sin\theta)/g = (2 \times 20 \times \sin 30^\circ)/9.81 = 20/9.81 = 2.04\text{ s}$. [3]

Question 9 (a) $m_1u_1 + m_2u_2 = (m_1+m_2)v \implies (1 \times 4) + (2 \times -2) = 3v \implies 4 - 4 = 3v \implies v = 0\text{ m s}^{-1}$. [3] (b) $KE_{initial} = 0.5(1)(4^2) + 0.5(2)(2^2) = 8 + 4 = 12\text{ J}$. $KE_{final} = 0$. Loss = $12\text{ J}$. [3] (c) Momentum is conserved because there are no external forces. KE is not conserved because the collision is perfectly inelastic; energy is converted to heat/sound/deformation. [2]

Question 10 (a) Vertical: $T \cos\theta = mg \implies T = mg / \cos\theta$. [3] (b) Horizontal: $T \sin\theta = mv^2/r \implies (mg/\cos\theta)\sin\theta = mv^2/r \implies g \tan\theta = v^2/r$. $v = \sqrt{gr \tan\theta}$. $T_{period} = 2\pi r / v = 2\pi r / \sqrt{gr \tan\theta} = 2\pi \sqrt{r / (g \tan\theta)}$. [4] (c) No change. The expression for $T_{period}$ is independent of mass $m$. [2]