A Level H1 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI) - Version 1

Subject: Physics H1
Level: A-Level
Paper: Practice Paper 1 (Comprehensive)
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 $g = 9.81\text{ m s}^{-2}$ unless otherwise stated.
4. Show all working clearly for calculation questions.

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Section A: Short Answer and Structured Questions (40 Marks)

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

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(b) A $0.5\text{ kg}$ block moving at $4.0\text{ m s}^{-1}$ collides with a stationary $0.3\text{ kg}$ block. After the collision, the blocks stick together. Calculate the final velocity of the combined mass. [3]




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Question 2 A uniform beam of length $4.0\text{ m}$ and mass $20\text{ kg}$ is supported by two vertical pillars at its ends. A $60\text{ kg}$ person stands $1.0\text{ m}$ from the left pillar. (a) Draw a free-body diagram of the beam, labeling all forces. [3]


(b) Calculate the reaction force exerted by the right pillar. [4]




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Question 3 (a) Define the term terminal velocity in the context of an object falling through a viscous fluid. [2]

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(b) Sketch a graph of acceleration $a$ against time $t$ for a spherical ball dropped from a height in air. [2]


(c) Explain the shape of the graph sketched in (b). [3]

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Question 4 A battery of EMF $12.0\text{ V}$ and internal resistance $1.5\text{ }\Omega$ is connected to a variable resistor $R$. (a) Explain, using the concept of a potential divider, why the terminal voltage of the battery decreases as $R$ is decreased. [3]

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(b) Calculate the value of $R$ such that the terminal voltage is $9.0\text{ V}$. [3]




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Question 5 A monochromatic light source of wavelength $550\text{ nm}$ is used in a Young's double-slit experiment. The slits are separated by $0.2\text{ mm}$ and the screen is $1.5\text{ m}$ away. (a) Calculate the distance between the central maximum and the first-order bright fringe. [3]


(b) If the entire apparatus is immersed in water (refractive index $1.33$), state and explain the change in the fringe spacing. [3]

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Question 6 (a) A metal surface has a work function of $2.2\text{ eV}$. Calculate the threshold frequency of incident light. [3]


(b) When light of frequency $8.0 \times 10^{14}\text{ Hz}$ is incident on the surface, calculate the maximum kinetic energy of the emitted photoelectrons. [4]




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Question 7 A sample of a radioactive isotope has an initial activity of $1.2 \times 10^4\text{ Bq}$. After $48\text{ hours}$, the activity is measured to be $1.5 \times 10^3\text{ Bq}$. (a) Determine the half-life of the isotope. [3]


(b) Calculate the decay constant $\lambda$ for this isotope. [2]


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Section B: Extended Response and Application (40 Marks)

Question 8 A projectile is launched from the ground with an initial velocity of $30\text{ m s}^{-1}$ at an angle of $40^\circ$ to the horizontal. (a) Calculate the maximum height reached by the projectile. [4]


(b) Determine the horizontal range of the projectile. [4]


(c) A wall of height $10\text{ m}$ is located $40\text{ m}$ from the launch point. Determine whether the projectile clears the wall. [6]




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Question 9 Two parallel current-carrying wires, A and B, are separated by $5.0\text{ cm}$. Wire A carries a current of $3.0\text{ A}$ and Wire B carries a current of $5.0\text{ A}$ in the opposite direction. (a) State the direction of the force exerted by Wire A on Wire B. [2]

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(b) Calculate the magnitude of the force per unit length acting between the wires. [4]


(c) If the current in Wire B is increased, describe the effect on the force and explain your answer. [4]

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Question 10 A motor is used to lift a $150\text{ kg}$ crate vertically at a constant speed of $0.8\text{ m s}^{-1}$. The motor has an input power of $1.5\text{ kW}$. (a) Calculate the useful power output of the motor. [4]


(b) Calculate the efficiency of the motor. [3]


(c) Suggest two ways the efficiency of this system could be improved. [3]

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Question 11 A $0.2\text{ kg}$ mass $M_1$ moving at $5.0\text{ m s}^{-1}$ collides with a stationary $0.3\text{ kg}$ mass $M_2$. After the collision, $M_1$ moves at $2.0\text{ m s}^{-1}$ at an angle of $30^\circ$ to the original line of motion. (a) Using the conservation of momentum in two dimensions, calculate the final velocity (magnitude and direction) of $M_2$. [8]







(b) Determine whether the collision is elastic or inelastic. Justify your answer with calculations. [8]







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TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key - Version 1

Section A

Question 1 (a) In a closed/isolated system, the total linear momentum remains constant provided no external forces act. [2] (b) $m_1v_1 + m_2v_2 = (m_1+m_2)V$ $(0.5)(4.0) + (0.3)(0) = (0.5+0.3)V$ $2.0 = 0.8V \rightarrow V = 2.5\text{ m s}^{-1}$ [3]

Question 2 (a) Diagram should show: Weight of beam (20kg $\times$ 9.81) at center (2m); Weight of person (60kg $\times$ 9.81) at 1m; Reaction $R_L$ at left end; Reaction $R_R$ at right end. [3] (b) Take moments about left pillar: $\sum \tau = 0$ $(20 \times 9.81)(2.0) + (60 \times 9.81)(1.0) - R_R(4.0) = 0$ $392.4 + 588.6 = 4R_R$ $981 = 4R_R \rightarrow R_R = 245.25\text{ N}$ [4]

Question 3 (a) The constant maximum velocity reached by a falling object when the drag force equals the weight of the object. [2] (b) Graph: $a$ starts at $g$ (9.81) and curves exponentially downwards toward $a=0$. [2] (c) As speed increases, air resistance (drag) increases. [1] The net force ($W - D$) decreases, so acceleration decreases according to $F=ma$. [1] Eventually, $D=W$, net force is zero, and acceleration becomes zero. [1]

Question 4 (a) The battery acts as a potential divider between internal resistance $r$ and external resistance $R$. [1] As $R$ decreases, the proportion of EMF across $R$ decreases. [1] More voltage is dropped across $r$ (lost volts $Ir$ increases), reducing terminal voltage $V = E - Ir$. [1] (b) $V = E - Ir \rightarrow 9.0 = 12.0 - I(1.5)$ $3.0 = 1.5I \rightarrow I = 2.0\text{ A}$ $R = V/I = 9.0 / 2.0 = 4.5\text{ }\Omega$ [3]

Question 5 (a) $\beta = \lambda D / a = (550 \times 10^{-9} \times 1.5) / (0.2 \times 10^{-3}) = 4.125 \times 10^{-3}\text{ m}$ (or $4.13\text{ mm}$). [3] (b) Fringe spacing decreases. [1] In water, the wavelength $\lambda$ decreases ($\lambda_{water} = \lambda_{air}/n$). [1] Since $\beta \propto \lambda$, the spacing decreases. [1]

Question 6 (a) $\Phi = hf_0 \rightarrow f_0 = (2.2 \times 1.6 \times 10^{-19}) / (6.63 \times 10^{-34}) = 5.31 \times 10^{14}\text{ Hz}$. [3] (b) $K.E._{\max} = hf - \Phi$ $K.E._{\max} = (6.63 \times 10^{-34} \times 8.0 \times 10^{14}) - (2.2 \times 1.6 \times 10^{-19})$ $K.E._{\max} = 5.30 \times 10^{-19} - 3.52 \times 10^{-19} = 1.78 \times 10^{-19}\text{ J}$ (or $1.11\text{ eV}$). [4]

Question 7 (a) $1.2 \times 10^4 \rightarrow 6000 \rightarrow 3000 \rightarrow 1500$. This is 3 half-lives. $3t_{1/2} = 48\text{ hours} \rightarrow t_{1/2} = 16\text{ hours}$. [3] (b) $\lambda = \ln 2 / t_{1/2} = 0.693 / (16 \times 3600) = 1.20 \times 10^{-5}\text{ s}^{-1}$. [2]

Section B

Question 8 (a) $u_y = 30 \sin 40^\circ = 19.28\text{ m s}^{-1}$ $H = u_y^2 / 2g = (19.28)^2 / (2 \times 9.81) = 18.9\text{ m}$. [4] (b) $t_{flight} = 2u_y / g = (2 \times 19.28) / 9.81 = 3.93\text{ s}$ $u_x = 30 \cos 40^\circ = 22.98\text{ m s}^{-1}$ $Range = u_x \times t_{flight} = 22.98 \times 3.93 = 90.3\text{ m}$. [4] (c) Time to reach wall: $t = 40 / 22.98 = 1.74\text{ s}$ Height at $t=1.74$: $y = (19.28)(1.74) - 0.5(9.81)(1.74)^2$ $y = 33.55 - 14.84 = 18.71\text{ m}$ Since $18.71\text{ m} > 10\text{ m}$, it clears the wall. [6]

Question 9 (a) Repulsive (opposite currents repel). [2] (b) $F/L = (\mu_0 I_1 I_2) / (2\pi d) = (4\pi \times 10^{-7} \times 3.0 \times 5.0) / (2\pi \times 0.05)$ $F/L = (2 \times 10^{-7} \times 15) / 0.05 = 6.0 \times 10^{-5}\text{ N m}^{-1}$. [4] (c) Force increases. [2] Since $F \propto I_2$, increasing the current in Wire B increases the magnetic field produced by B at the location of A, and vice versa, increasing the Lorentz force. [2]

Question 10 (a) $P_{out} = Fv = (150 \times 9.81) \times 0.8 = 1177.2\text{ W}$. [4] (b) $\text{Efficiency} = (1177.2 / 1500) \times 100\% = 78.5\%$. [3] (c) Use a more efficient motor/better lubrication to reduce friction; use a pulley system to reduce the required input force. [3]

Question 11 (a) $x$-axis: $0.2(5.0) = 0.2(2.0 \cos 30^\circ) + 0.3(v_{2x})$ $1.0 = 0.173 + 0.3v_{2x} \rightarrow v_{2x} = 2.76\text{ m s}^{-1}$ $y$-axis: $0 = 0.2(2.0 \sin 30^\circ) + 0.3(v_{2y})$ $0 = 0.2 + 0.3v_{2y} \rightarrow v_{2y} = -0.67\text{ m s}^{-1}$ $v_2 = \sqrt{2.76^2 + (-0.67)^2} = 2.84\text{ m s}^{-1}$ $\theta = \tan^{-1}(-0.67/2.76) = -13.7^\circ$ (below original line). [8] (b) $KE_{initial} = 0.5(0.2)(5^2) = 2.5\text{ J}$ $KE_{final} = 0.5(0.2)(2^2) + 0.5(0.3)(2.84^2) = 0.4 + 1.21 = 1.61\text{ J}$ Since $KE_{initial} \neq KE_{final}$, the collision is inelastic. [8]