A Level H2 Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI) - Version 2

Subject: Physics H2
Level: A-Level
Paper: Structured Questions (Practice Paper)
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 a scientific calculator.
4. Physical constants:
   • Acceleration of free fall, $g = 9.81 \text{ m s}^{-2}$
   • Gravitational constant, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
   • Mass of Earth, $M_E = 5.97 \times 10^{24} \text{ kg}$
   • Radius of Earth, $R_E = 6.37 \times 10^6 \text{ m}$

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Section A: Newtonian Mechanics

Question 1 A block of mass $2.0 \text{ kg}$ is released from rest at the top of a rough inclined plane making an angle of $30^\circ$ to the horizontal. The coefficient of kinetic friction between the block and the plane is $0.15$. (a) Draw a free-body diagram of the block as it slides down the plane. [2] (b) Calculate the acceleration of the block. [3] (c) Determine the distance the block travels before coming to rest if it were launched up the plane with an initial velocity of $5.0 \text{ m s}^{-1}$. [4] [9 marks]

Question 2 A small sphere of mass $0.10 \text{ kg}$ is attached to a string of length $0.50 \text{ m}$ and whirled in a vertical circle. (a) State the condition for the string to remain taut at the highest point of the circle. [1] (b) Calculate the minimum speed of the sphere at the top of the circle to maintain circular motion. [2] (c) If the speed at the bottom of the circle is $7.0 \text{ m s}^{-1}$, calculate the tension in the string at the lowest point. [3] (d) Explain, with reference to centripetal force, why the tension in the string is greater at the bottom than at the top. [2] [8 marks]

Question 3 Two trolleys, A and B, of masses $1.5 \text{ kg}$ and $2.5 \text{ kg}$ respectively, move towards each other on a smooth horizontal track. Trolley A has a velocity of $3.0 \text{ m s}^{-1}$ and Trolley B has a velocity of $2.0 \text{ m s}^{-1}$ in the opposite direction. (a) State the Principle of Conservation of Linear Momentum. [2] (b) The trolleys collide and stick together. Calculate the common velocity of the trolleys after the collision. [3] (c) Calculate the loss in kinetic energy during the collision. [3] (d) Explain why the kinetic energy is not conserved in this collision. [2] [10 marks]

Question 4 A satellite of mass $m$ is in a circular orbit around a planet of mass $M$ and radius $R$. The satellite orbits at a height $h$ above the surface. (a) Show that the orbital period $T$ is given by $T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$. [4] (b) If the height $h$ is increased, state and explain the effect on the orbital speed of the satellite. [2] (c) Calculate the escape velocity from the surface of the planet in terms of $G, M,$ and $R$. [3] [9 marks]

Question 5 A mass $m$ is suspended by a spring of spring constant $k$. The mass is displaced by a distance $X_0$ from its equilibrium position and released. (a) Show that the angular frequency $\omega$ of the resulting simple harmonic motion is $\sqrt{k/m}$. [3] (b) If $m = 0.20 \text{ kg}$ and $k = 20 \text{ N m}^{-1}$, calculate the maximum acceleration of the mass if $X_0 = 0.10 \text{ m}$. [3] (c) Describe the relationship between the energy of the oscillator and the amplitude of the motion. [2] [8 marks]

Section B: Integrated Mechanics and Energy

Question 6 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 reached by the projectile. [3] (b) Find the condition for the horizontal range to be maximized. [2] (c) A projectile is launched at $25 \text{ m s}^{-1}$ at $35^\circ$. Calculate the time taken to reach the maximum height. [3] [8 marks]

Question 7 A $0.5 \text{ kg}$ block slides down a frictionless curved track from a height $H = 2.0 \text{ m}$ and enters a rough horizontal region. (a) Calculate the speed of the block at the bottom of the curve. [2] (b) The block then enters a rough patch with a coefficient of friction $\mu = 0.30$. Calculate the distance it slides before stopping. [4] (c) Compare the work done by friction to the initial potential energy of the block. [2] [8 marks]

Question 8 A particle of mass $m$ moves in a gravitational field. (a) Define gravitational potential at a point. [2] (b) Calculate the work done in moving a $1000 \text{ kg}$ satellite from a radius of $7.0 \times 10^6 \text{ m}$ to $8.0 \times 10^6 \text{ m}$ from the center of the Earth. [4] (c) Explain why the gravitational potential is defined as zero at infinity. [2] [8 marks]

Question 9 A system consists of two masses $m_1 = 0.5 \text{ kg}$ and $m_2 = 1.5 \text{ kg}$ connected by a light inextensible string passing over a smooth pulley. (a) Calculate the acceleration of the system when released from rest. [4] (b) Calculate the tension in the string during the motion. [3] (c) Determine the velocity of the masses after $1.0 \text{ s}$. [2] [9 marks]

Question 10 A ball of mass $0.2 \text{ kg}$ is dropped from a height of $5.0 \text{ m}$ onto a concrete floor. It rebounds to a height of $3.0 \text{ m}$. (a) Calculate the velocity of the ball immediately before it hits the floor. [2] (b) Calculate the velocity of the ball immediately after it leaves the floor. [2] (c) Calculate the average force exerted by the floor on the ball if the contact time is $0.05 \text{ s}$. [4] [8 marks]

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TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key - Version 2

Section A: Newtonian Mechanics

Question 1 (a) Diagram should show: Weight ($mg$) acting downwards, Normal reaction ($R$) perpendicular to plane, Friction ($f$) acting up the plane. [2] (b) $F_{net} = mg \sin 30^\circ - \mu mg \cos 30^\circ = ma$ $a = g(\sin 30^\circ - 0.15 \cos 30^\circ) = 9.81(0.5 - 0.13) = 3.63 \text{ m s}^{-2}$ [3] (c) $v^2 = u^2 + 2as$. Here $u=5, v=0$. $a = -g(\sin 30^\circ + \mu \cos 30^\circ) = -9.81(0.5 + 0.13) = -6.18 \text{ m s}^{-2}$ $0 = 5^2 + 2(-6.18)s \implies s = 25 / 12.36 = 2.02 \text{ m}$ [4]

Question 2 (a) The tension $T$ must be $\ge 0$ (or the centripetal force must be provided by gravity and tension). [1] (b) At top: $T + mg = mv^2/r$. For min speed, $T=0 \implies v = \sqrt{gr} = \sqrt{9.81 \times 0.5} = 2.21 \text{ m s}^{-1}$ [2] (c) At bottom: $T - mg = mv^2/r \implies T = m(g + v^2/r) = 0.10(9.81 + 7^2/0.5) = 0.10(9.81 + 98) = 10.78 \text{ N}$ [3] (d) At the bottom, both weight and centripetal force act in the same vertical line, and tension must overcome weight AND provide the centripetal acceleration. [2]

Question 3 (a) In a closed system, the total momentum before an event equals the total momentum after the event, provided no external forces act. [2] (b) $m_1u_1 + m_2u_2 = (m_1+m_2)v$ $(1.5 \times 3.0) + (2.5 \times -2.0) = (1.5+2.5)v$ $4.5 - 5.0 = 4v \implies v = -0.125 \text{ m s}^{-1}$ (opposite to A's initial direction) [3] (c) $KE_{initial} = 0.5(1.5)(3^2) + 0.5(2.5)(2^2) = 6.75 + 5.0 = 11.75 \text{ J}$ $KE_{final} = 0.5(4.0)(-0.125)^2 = 0.03125 \text{ J}$ Loss $= 11.75 - 0.03 = 11.72 \text{ J}$ [3] (d) The collision is inelastic; energy is dissipated as heat/sound due to internal deformation of the trolleys. [2]

Question 4 (a) $F_c = F_g \implies mv^2/r = GMm/r^2$ where $r = R+h$. $v = \sqrt{GM/r}$. Since $v = 2\pi r / T$, then $T = 2\pi r / v = 2\pi r / \sqrt{GM/r} = 2\pi \sqrt{r^3/GM}$. [4] (b) Orbital speed decreases. As $r$ increases, the gravitational pull $GMm/r^2$ decreases, requiring a lower speed to maintain a stable circular orbit. [2] (c) $KE + PE = 0 \implies 0.5mv_{esc}^2 - GMm/R = 0 \implies v_{esc} = \sqrt{2GM/R}$ [3]

Question 5 (a) $F = -kx$. From $F=ma$, $ma = -kx \implies a = -(k/m)x$. Since $a = -\omega^2 x$ for SHM, $\omega^2 = k/m \implies \omega = \sqrt{k/m}$. [3] (b) $a_{max} = \omega^2 X_0 = (k/m)X_0 = (20/0.20) \times 0.10 = 10 \text{ m s}^{-2}$ [3] (c) Energy is proportional to the square of the amplitude ($E \propto X_0^2$). [2]

Section B: Integrated Mechanics and Energy

Question 6 (a) $v_y^2 = u_y^2 + 2ay \implies 0 = (u \sin \theta)^2 - 2gH \implies H = (u^2 \sin^2 \theta) / 2g$ [3] (b) Range $R = (u^2 \sin 2\theta) / g$. $R$ is max when $\sin 2\theta = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ$. [2] (c) $v_y = u \sin \theta - gt \implies 0 = 25 \sin 35^\circ - 9.81t \implies t = 14.34 / 9.81 = 1.46 \text{ s}$ [3]

Question 7 (a) $mgh = 0.5mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 2.0} = 6.26 \text{ m s}^{-1}$ [2] (b) Work done by friction $= \Delta KE \implies \mu mg s = mgh$ $0.30 \times 9.81 \times s = 9.81 \times 2.0 \implies s = 2.0 / 0.30 = 6.67 \text{ m}$ [4] (c) They are equal. All the initial gravitational potential energy is converted to kinetic energy and then dissipated as work done against friction. [2]

Question 8 (a) The work done per unit mass in bringing a small test mass from infinity to that point. [2] (b) $W = \Delta PE = GMm(1/r_1 - 1/r_2) = (6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)(1/7 \times 10^6 - 1/8 \times 10^6)$ $W = 3.98 \times 10^{17} \times (1.78 \times 10^{-8}) = 7.1 \times 10^9 \text{ J}$ [4] (c) To provide a consistent reference point where the gravitational force is zero. [2]

Question 9 (a) $a = (m_2 - m_1)g / (m_1 + m_2) = (1.5 - 0.5)9.81 / (1.5 + 0.5) = 0.5 \times 9.81 / 2 = 2.45 \text{ m s}^{-2}$ [4] (b) $T = m_1(g + a) = 0.5(9.81 + 2.45) = 6.13 \text{ N}$ [3] (c) $v = at = 2.45 \times 1.0 = 2.45 \text{ m s}^{-1}$ [2]

Question 10 (a) $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 5.0} = 9.90 \text{ m s}^{-1}$ [2] (b) $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 3.0} = 7.67 \text{ m s}^{-1}$ [2] (c) $\Delta p = m(v_f - v_i) = 0.2(7.67 - (-9.90)) = 0.2(17.57) = 3.51 \text{ kg m s}^{-1}$ $F_{avg} = \Delta p / \Delta t = 3.51 / 0.05 = 70.2 \text{ N}$ [4]