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A Level H1 Physics Practice Paper 5

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1 (8867)
Level: A-Level
Paper: Practice Paper 2 (Structured Questions)
Version: 5 of 5
Duration: 2 hours
Total Marks: 80

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this question paper.
You may lose marks if you do not show your working or if you do not use appropriate units.
The number of marks is given in brackets [ ] at the end of each question or part question.
Assume acceleration due to gravity $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A

Answer all questions in this section. This section focuses on Kinematics, Dynamics, and Forces.

1. A drone is used to deliver a package. It starts from rest and accelerates uniformly upwards. (a) Define acceleration. [1] _________________________________________________________________________ _________________________________________________________________________

(b) The drone reaches a vertical velocity of $12.0 \text{ m s}^{-1}$ in $4.0 \text{ s}$. Calculate the vertical displacement of the drone during this time. [2]

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(c) Sketch a velocity-time graph for this motion from $t=0$ to $t=4.0 \text{ s}$. Label the axes with appropriate values. [2]

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2. A car of mass $1200 \text{ kg}$ travels along a straight horizontal road. The engine provides a driving force of $3000 \text{ N}$ . The total resistive force acting on the car is constant at $600 \text{ N}$ . (a) Calculate the acceleration of the car. [2]

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(b) The car travels for $10 \text{ s}$ from rest. Calculate the distance travelled in this time. [2]

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(c) Explain, in terms of forces, why the car eventually reaches a constant maximum speed (terminal velocity) if the driving force remains constant but air resistance increases with speed. [2]
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3. A uniform beam $AB$ of length $4.0 \text{ m}$ and weight $200 \text{ N}$ is hinged at end $A$ to a vertical wall. The beam is held horizontal by a cable attached to end $B$ and to the wall at a point $3.0 \text{ m}$ vertically above $A$ . (a) Draw a free-body diagram for the beam, showing all forces acting on it. Label the forces clearly. [3]

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(b) Calculate the tension in the cable. [3]

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4. State the principle of conservation of linear momentum. [2] _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________

5. Two ice skaters, Skater X (mass $50 \text{ kg}$ ) and Skater Y (mass $70 \text{ kg}$ ), are initially at rest on frictionless ice. They push against each other and move apart. Skater X moves with a velocity of $2.1 \text{ m s}^{-1}$ to the left. (a) Calculate the velocity of Skater Y. [3]

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(b) Determine whether the collision (push) is elastic or inelastic. Show your working by comparing the total kinetic energy before and after the push. [3]

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Section B

Answer all questions in this section. This section focuses on Work, Energy, and Power.

6. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards at a constant speed of $0.5 \text{ m s}^{-1}$ . (a) Calculate the power output of the crane motor. [3]

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(b) The crane motor has an efficiency of $80\%$. Calculate the electrical power input required. [2]

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7. A ball of mass $0.2 \text{ kg}$ is dropped from a height of $5.0 \text{ m}$ onto a hard surface. It rebounds to a height of $3.2 \text{ m}$ . (a) Calculate the speed of the ball just before it hits the ground. [2]

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(b) Calculate the speed of the ball just after it leaves the ground. [2]

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(c) Calculate the loss in kinetic energy during the impact with the ground. [3]

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8. A block of mass $2.0 \text{ kg}$ slides down a rough inclined plane. The plane is inclined at $30^\circ$ to the horizontal. The block starts from rest and travels $5.0 \text{ m}$ down the slope, reaching a speed of $4.0 \text{ m s}^{-1}$ . (a) Calculate the loss in gravitational potential energy. [2]

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(b) Calculate the gain in kinetic energy. [2]

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(c) Determine the average frictional force acting on the block. [3]

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9. Define work done by a force. [1] _________________________________________________________________________ _________________________________________________________________________

10. A car engine exerts a constant driving force of $800 \text{ N}$ to move the car at a constant speed of $25 \text{ m s}^{-1}$ along a horizontal road. (a) Calculate the work done by the driving force in $10 \text{ s}$ . [2]

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(b) State the magnitude of the resistive forces acting on the car. [1]
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Section C

Answer all questions in this section. This section focuses on Momentum, Impulse, and Complex Mechanics Applications.

11. A tennis ball of mass $0.06 \text{ kg}$ is moving horizontally towards a racket at $20 \text{ m s}^{-1}$ . It is struck by the racket and leaves horizontally in the opposite direction at $30 \text{ m s}^{-1}$ . The contact time is $0.01 \text{ s}$ . (a) Calculate the change in momentum of the ball. [3]

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(b) Calculate the average force exerted by the racket on the ball. [2]

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12. Explain why airbags in cars reduce the risk of injury to passengers during a collision, referring to the concepts of impulse and force. [3] _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________

13. A rocket of mass $1000 \text{ kg}$ (including fuel) is stationary on a launch pad. It ejects gas downwards at a speed of $2000 \text{ m s}^{-1}$ relative to the rocket. The rate of mass ejection is $5.0 \text{ kg s}^{-1}$ . (a) Calculate the thrust force produced by the rocket engine. [2]

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(b) Determine if the rocket will lift off immediately. Justify your answer with a calculation of the weight. [3]

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14. A student investigates the relationship between the extension of a spring and the load applied. The spring obeys Hooke's Law. (a) State Hooke's Law. [1] _________________________________________________________________________ _________________________________________________________________________

(b) The spring constant is $50 \text{ N m}^{-1}$. Calculate the elastic potential energy stored in the spring when it is extended by $0.2 \text{ m}$. [2]

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15. A projectile is launched from ground level with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $45^\circ$ to the horizontal. Air resistance is negligible. (a) Calculate the horizontal component of the initial velocity. [1]

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(b) Calculate the time taken to reach the maximum height. [2]

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(c) Calculate the horizontal range of the projectile. [2]

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16. Two trolleys, A and B, move on a frictionless track. Trolley A (mass $1.0 \text{ kg}$ ) moves at $2.0 \text{ m s}^{-1}$ towards stationary Trolley B (mass $2.0 \text{ kg}$ ). They collide and stick together. (a) Calculate the common velocity after the collision. [3]

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(b) Calculate the fraction of the initial kinetic energy that is lost in the collision. [3]

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17. A lift (elevator) of mass $800 \text{ kg}$ carries passengers of total mass $200 \text{ kg}$ . The lift accelerates upwards at $1.5 \text{ m s}^{-2}$ . (a) Calculate the tension in the cable supporting the lift. [3]

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(b) The lift then moves upwards at a constant speed. State how the tension in the cable compares to the weight of the lift and passengers. [1]
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18. A ball is thrown vertically upwards. (a) Describe the energy changes that occur from the moment the ball leaves the hand until it reaches its maximum height. [2] _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________

(b) At the maximum height, state the value of the ball's kinetic energy. [1]
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19. A car of mass $1000 \text{ kg}$ travels around a circular bend of radius $50 \text{ m}$ at a constant speed of $15 \text{ m s}^{-1}$ . (a) Calculate the centripetal acceleration of the car. [2]

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(b) Calculate the centripetal force required to keep the car on the circular path. [2]

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(c) Identify the force that provides this centripetal force. [1]
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20. A force $F$ varies with distance $d$ as shown in the graph below (description: linear increase from $0 \text{ N}$ at $0 \text{ m}$ to $10 \text{ N}$ at $5 \text{ m}$ ). (a) Calculate the work done by the force over the distance of $5 \text{ m}$ . [2]

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(b) If this work is done on a $2.0 \text{ kg}$ object initially at rest, calculate its final speed. [3]

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End of Paper

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Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme Version: 5 of 5

Section A

1. (a) Rate of change of velocity. [B1] (b) Using $s = \frac{1}{2}(u+v)t$ or $s = ut + \frac{1}{2}at^2$ . $u=0, v=12.0, t=4.0$ . $s = \frac{1}{2}(0 + 12.0)(4.0) = 24.0 \text{ m}$ . [M1, A1] (Alternatively: $a = 12/4 = 3 \text{ m s}^{-2}$ . $s = 0 + 0.5(3)(16) = 24 \text{ m}$ .) (c) Graph: Straight line starting from origin $(0,0)$ and ending at $(4.0, 12.0)$ . [B1 for shape, B1 for labels/values]

2. (a) Resultant Force $F_{net} = 3000 - 600 = 2400 \text{ N}$ . [M1] $a = F/m = 2400 / 1200 = 2.0 \text{ m s}^{-2}$ . [A1] (b) $s = ut + \frac{1}{2}at^2$ . $u=0, a=2.0, t=10$ . $s = 0 + 0.5(2.0)(100) = 100 \text{ m}$ . [M1, A1] (c) As speed increases, air resistance increases. [B1] Resultant force ( $Driving - Resistance$ ) decreases, so acceleration decreases. When resistance equals driving force, resultant force is zero and speed is constant. [B1]

3. (a) Diagram should show: 1. Weight ( $200 \text{ N}$ ) acting downwards at the center of the beam ( $2.0 \text{ m}$ from A). [B1] 2. Tension ( $T$ ) acting at B, along the cable towards the wall. [B1] 3. Reaction force at hinge A (can be shown as vertical/horizontal components or a single resultant vector). [B1] (b) Take moments about A. Clockwise moment = Weight $\times$ distance = $200 \times 2.0 = 400 \text{ N m}$ . [M1] Anticlockwise moment = Vertical component of Tension $\times$ length. Geometry: Triangle with base 4, height 3. Hypotenuse = 5. $\sin(\theta) = 3/5 = 0.6$ (where $\theta$ is angle at B with horizontal? No, angle of cable with beam). Let $\alpha$ be angle between cable and beam. $\tan \alpha = 3/4$ . $\sin \alpha = 3/5 = 0.6$ . Vertical component of $T = T \sin \alpha = 0.6 T$ . Moment = $(0.6 T) \times 4.0 = 2.4 T$ . [M1] Equilibrium: $2.4 T = 400 \Rightarrow T = 400 / 2.4 = 166.7 \text{ N}$ . [A1] (Accept 167 N)

4. In a closed/isolated system [B1], the total linear momentum remains constant (or sum of momentum before = sum of momentum after) provided no external forces act. [B1]

5. (a) Total initial momentum = 0. $m_X v_X + m_Y v_Y = 0$ . $50(-2.1) + 70(v_Y) = 0$ (taking left as negative). $-105 + 70 v_Y = 0$ . $v_Y = 105 / 70 = 1.5 \text{ m s}^{-1}$ . [M1, A1] Direction: To the right. [B1] (b) $KE_{initial} = 0$ . [B1] $KE_{final} = \frac{1}{2}(50)(2.1)^2 + \frac{1}{2}(70)(1.5)^2$ . $KE_{final} = 110.25 + 78.75 = 189 \text{ J}$ . [M1] Since $KE_{final} > KE_{initial}$ , kinetic energy is not conserved (it increased due to chemical energy from muscles). Thus, it is not an elastic collision in the passive sense, but technically "inelastic" usually implies KE loss. However, strictly speaking, elastic requires KE conservation. Here KE is not conserved. [A1] (Note: In push-apart scenarios, KE is generated. It is not an elastic collision because KE is not conserved.)

Section B

6. (a) Force required to lift at constant speed = Weight = $mg = 500 \times 9.81 = 4905 \text{ N}$ . [M1] Power $P = Fv = 4905 \times 0.5 = 2452.5 \text{ W}$ . [M1] Answer: $2450 \text{ W}$ (or $2.45 \text{ kW}$ ). [A1] (b) Efficiency = Output / Input. $0.80 = 2452.5 / P_{in}$ . $P_{in} = 2452.5 / 0.80 = 3065.6 \text{ W}$ . [M1] Answer: $3070 \text{ W}$ (or $3.07 \text{ kW}$ ). [A1]

7. (a) $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ . $v = \sqrt{2 \times 9.81 \times 5.0} = \sqrt{98.1} = 9.90 \text{ m s}^{-1}$ . [M1, A1] (b) $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 3.2} = \sqrt{62.784} = 7.92 \text{ m s}^{-1}$ . [M1, A1] (c) $KE_{before} = \frac{1}{2}(0.2)(9.90)^2 = 9.81 \text{ J}$ . $KE_{after} = \frac{1}{2}(0.2)(7.92)^2 = 6.27 \text{ J}$ . Loss = $9.81 - 6.27 = 3.54 \text{ J}$ . [M1, A1] (Alternatively: Loss in GPE = $mg(h_1 - h_2) = 0.2 \times 9.81 \times (5.0 - 3.2) = 3.53 \text{ J}$ . Accept 3.5 J.) [M1, A1]

8. (a) Vertical height drop $h = 5.0 \sin 30^\circ = 2.5 \text{ m}$ . Loss in GPE = $mgh = 2.0 \times 9.81 \times 2.5 = 49.05 \text{ J}$ . [M1, A1] (b) Gain in KE = $\frac{1}{2}mv^2 = 0.5 \times 2.0 \times (4.0)^2 = 16.0 \text{ J}$ . [M1, A1] (c) Work done against friction = Loss in GPE - Gain in KE. $W_f = 49.05 - 16.0 = 33.05 \text{ J}$ . [M1] $W_f = F_f \times d \Rightarrow 33.05 = F_f \times 5.0$ . $F_f = 33.05 / 5.0 = 6.61 \text{ N}$ . [A1]

9. Product of the force and the distance moved in the direction of the force. [B1]

10. (a) Distance $d = vt = 25 \times 10 = 250 \text{ m}$ . [M1] Work = $Fd = 800 \times 250 = 200,000 \text{ J}$ ( $200 \text{ kJ}$ ). [A1] (b) Since speed is constant, forces are balanced. Resistive force = Driving force = $800 \text{ N}$ . [B1]

Section C

11. (a) Take direction towards racket as positive. $u = +20 \text{ m s}^{-1}$ , $v = -30 \text{ m s}^{-1}$ . $\Delta p = m(v - u) = 0.06(-30 - 20) = 0.06(-50) = -3.0 \text{ N s}$ . [M1, A1] Magnitude is $3.0 \text{ N s}$ . [B1] (b) $F_{avg} = \Delta p / \Delta t = 3.0 / 0.01 = 300 \text{ N}$ . [M1, A1]

12. Airbags increase the time of impact ( $\Delta t$ ) for the passenger to stop. [B1] Since Impulse ( $\Delta p$ ) is fixed (change in momentum is constant), and $F \Delta t = \Delta p$ . [B1] Increasing $\Delta t$ reduces the average force $F$ exerted on the passenger, reducing injury. [B1]

13. (a) Thrust = Rate of change of momentum of gas = $(dm/dt) \times v_{exhaust}$ . $F = 5.0 \times 2000 = 10,000 \text{ N}$ . [M1, A1] (b) Weight of rocket $W = mg = 1000 \times 9.81 = 9810 \text{ N}$ . [M1] Thrust ( $10,000 \text{ N}$ ) > Weight ( $9810 \text{ N}$ ). [B1] Therefore, there is a resultant upward force, and the rocket will lift off. [A1]

14. (a) The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded. [B1] (b) $EPE = \frac{1}{2}kx^2 = 0.5 \times 50 \times (0.2)^2 = 25 \times 0.04 = 1.0 \text{ J}$ . [M1, A1]

15. (a) $u_x = u \cos \theta = 20 \cos 45^\circ = 14.14 \text{ m s}^{-1}$ . [A1] (b) At max height, $v_y = 0$ . $v_y = u_y - gt \Rightarrow 0 = (20 \sin 45^\circ) - 9.81 t$ . $14.14 = 9.81 t \Rightarrow t = 1.44 \text{ s}$ . [M1, A1] (c) Total time of flight = $2 \times t_{max\_height} = 2.88 \text{ s}$ . Range = $u_x \times T = 14.14 \times 2.88 = 40.7 \text{ m}$ . [M1, A1]

16. (a) Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B) v$ . $1.0(2.0) + 0 = (1.0 + 2.0) v$ . $2.0 = 3.0 v \Rightarrow v = 0.67 \text{ m s}^{-1}$ . [M1, A1] (Exact: 2/3 m/s) (b) $KE_{initial} = \frac{1}{2}(1.0)(2.0)^2 = 2.0 \text{ J}$ . [M1] $KE_{final} = \frac{1}{2}(3.0)(2/3)^2 = 1.5 \times (4/9) = 6/9 = 0.67 \text{ J}$ . [M1] Loss = $2.0 - 0.67 = 1.33 \text{ J}$ . Fraction lost = $1.33 / 2.0 = 0.67$ (or $2/3$ ). [A1]

17. (a) Total mass $M = 800 + 200 = 1000 \text{ kg}$ . Resultant Force $F_{net} = Ma = 1000 \times 1.5 = 1500 \text{ N}$ . [M1] $T - Mg = F_{net} \Rightarrow T = Mg + F_{net}$ . $T = (1000 \times 9.81) + 1500 = 9810 + 1500 = 11,310 \text{ N}$ . [M1, A1] (b) Tension is equal to the weight. [B1]

18. (a) Kinetic energy decreases [B1] and gravitational potential energy increases. [B1] (b) Zero. [B1]

19. (a) $a_c = v^2 / r = 15^2 / 50 = 225 / 50 = 4.5 \text{ m s}^{-2}$ . [M1, A1] (b) $F_c = ma_c = 1000 \times 4.5 = 4500 \text{ N}$ . [M1, A1] (c) Friction between tires and road. [B1]

20. (a) Work done = Area under Force-distance graph. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 5 \times 10 = 25 \text{ J}$ . [M1, A1] (b) Work done = Gain in KE. $25 = \frac{1}{2}mv^2 = 0.5(2.0)v^2 = v^2$ . [M1] $v = \sqrt{25} = 5.0 \text{ m s}^{-1}$ . [A1]