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A Level H1 Physics Practice Paper 2

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A-Level Physics H1 Practice Paper - Mechanics (Version 2 of 5)

Subject: Physics H1
Level: A-Level
Paper: Practice Paper 2 (Mechanics Focus)
Duration: 1 hour 15 minutes
Total Marks: 60

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 booklet.
All working must be clearly shown. Numerical answers should be given to 3 significant figures unless otherwise stated.
The use of an approved scientific calculator is expected.
A Data and Formulae booklet is provided.

Section A: Structured Questions

Answer all questions in this section.

1. State the principle of conservation of linear momentum.
[2]

2. A ball of mass $0.15 \text{ kg}$ is dropped from rest. It hits the ground with a speed of $8.0 \text{ m s}^{-1}$ and rebounds vertically with a speed of $6.0 \text{ m s}^{-1}$ .
(a) Calculate the magnitude of the change in momentum of the ball during the impact.
[3]

(b) The contact time with the ground is $0.050 \text{ s}$ . Calculate the average resultant force acting on the ball during the impact.
[2]

3. A uniform plank $AB$ of length $4.0 \text{ m}$ and weight $200 \text{ N}$ rests horizontally on two supports $P$ and $Q$ . Support $P$ is at end $A$ , and support $Q$ is $1.0 \text{ m}$ from end $B$ . A student of weight $600 \text{ N}$ stands on the plank at a distance $x$ from $A$ .
(a) Draw a free-body diagram for the plank, showing all forces acting on it. Label the forces clearly.
[3]

(b) Determine the maximum value of $x$ such that the plank does not tip over.
[4]

4. A car of mass $1200 \text{ kg}$ travels along a straight horizontal road. The engine provides a constant driving force of $2400 \text{ N}$ . The total resistive force is proportional to the square of the speed, given by $F_R = kv^2$ , where $k$ is a constant.
(a) Explain why the acceleration of the car decreases as its speed increases.
[2]

(b) The maximum speed of the car is $40 \text{ m s}^{-1}$ . Calculate the value of the constant $k$ .
[3]

5. Two trolleys, $X$ and $Y$ , move on a smooth horizontal track. Trolley $X$ has mass $2.0 \text{ kg}$ and moves with velocity $3.0 \text{ m s}^{-1}$ to the right. Trolley $Y$ has mass $1.0 \text{ kg}$ and moves with velocity $1.0 \text{ m s}^{-1}$ to the left. They collide and stick together.
(a) Calculate the common velocity of the trolleys after the collision.
[3]

(b) Show that the collision is inelastic by comparing the total kinetic energy before and after the collision.
[3]

Section B: Data Interpretation and Application

Answer all questions in this section.

6. Figure 6.1 shows the velocity-time graph for a skydiver falling vertically from rest. Air resistance is significant.

(Imagine a graph where velocity increases rapidly at first, then the gradient decreases, eventually becoming a horizontal line at $v = 50 \text{ m s}^{-1}$ after $t = 20 \text{ s}$ .)

(a) Describe the motion of the skydiver during the first 20 seconds, referring to the forces acting on him.
[3]

(b) Estimate the distance fallen by the skydiver in the first 20 seconds.
[2]

(c) At $t = 25 \text{ s}$ , the skydiver opens his parachute. Explain, in terms of forces, why his speed decreases rapidly immediately after opening the parachute.
[2]

7. A block of mass $5.0 \text{ kg}$ is pulled up a rough inclined plane by a constant force $F$ parallel to the slope. The plane is inclined at $30^\circ$ to the horizontal. The block moves at a constant speed of $2.0 \text{ m s}^{-1}$ . The frictional force acting on the block is $10 \text{ N}$ .
(a) Draw a free-body diagram showing the four forces acting on the block.
[2]

(b) Calculate the magnitude of the force $F$ .
[3]

8. A spring obeys Hooke's Law. When a load of $10 \text{ N}$ is applied, the extension is $4.0 \text{ cm}$ .
(a) Calculate the spring constant $k$ .
[2]

(b) Calculate the elastic potential energy stored in the spring when the extension is $4.0 \text{ cm}$ .
[2]

(c) The load is removed and replaced with a mass of $0.50 \text{ kg}$ . The mass is pulled down slightly and released. Describe the energy changes that occur during one complete oscillation, assuming no air resistance.
[3]

9. A projectile is launched from ground level with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $30^\circ$ to the horizontal. Air resistance is negligible.
(a) Calculate the horizontal component of the initial velocity.
[1]

(b) Calculate the maximum height reached by the projectile.
[3]

10. A satellite orbits the Earth in a circular path.
(a) State the force that provides the centripetal force for the satellite.
[1]

(b) Explain why the satellite is considered to be in a state of "free fall" despite maintaining a constant altitude.
[2]

Section C: Extended Response and Synthesis

Answer all questions in this section.

11. A car of mass $1000 \text{ kg}$ is traveling at $20 \text{ m s}^{-1}$ when the driver applies the brakes. The car comes to a stop in a distance of $40 \text{ m}$ .
(a) Calculate the average braking force.
[3]

(b) If the initial speed had been $40 \text{ m s}^{-1}$ (double the speed), calculate the new stopping distance, assuming the same average braking force.
[2]

12. Two spheres, $A$ and $B$ , are suspended by light inextensible strings. Sphere $A$ (mass $m$ ) is pulled back to a height $h$ and released. It strikes sphere $B$ (mass $2m$ ), which is initially at rest. The collision is elastic.
(a) Derive an expression for the speed of sphere $A$ just before impact in terms of $g$ and $h$ .
[2]

(b) State two conditions that must be satisfied for a collision to be perfectly elastic.
[2]

(c) After the collision, sphere $A$ rebounds. Explain, using the principle of conservation of momentum, why sphere $A$ must rebound rather than continue forward or stop, given that $m_A < m_B$ .
[3]

13. A lift of mass $500 \text{ kg}$ carries a passenger of mass $70 \text{ kg}$ . The lift accelerates upwards from rest at $1.5 \text{ m s}^{-2}$ .
(a) Calculate the tension in the cable supporting the lift.
[3]

(b) Calculate the normal reaction force exerted by the floor of the lift on the passenger.
[2]

(c) The lift then moves at constant velocity. State and explain how the normal reaction force on the passenger changes.
[2]

14. A river flows at $3.0 \text{ m s}^{-1}$ due East. A boat can travel at $5.0 \text{ m s}^{-1}$ in still water. The boat heads due North across the river.
(a) Calculate the resultant velocity of the boat relative to the bank (magnitude and direction).
[3]

(b) If the river is $100 \text{ m}$ wide, calculate the time taken to cross.
[2]

15. A student investigates the relationship between the force applied to a trolley and its acceleration. The trolley moves on a friction-compensated track.
(a) Sketch the expected graph of acceleration ( $a$ ) against force ( $F$ ).
[1]

(b) Explain what is meant by "friction-compensated" and why it is necessary for this experiment.
[2]

(c) If the track is not properly compensated (friction is present), sketch how the graph of $a$ against $F$ would differ from the ideal case.
[2]

16. A block of ice of mass $2.0 \text{ kg}$ slides down a smooth curved track from a height of $5.0 \text{ m}$ . It reaches the bottom and slides onto a rough horizontal surface where it comes to rest after traveling $10 \text{ m}$ .
(a) Calculate the speed of the ice at the bottom of the curved track.
[2]

(b) Calculate the average frictional force acting on the ice on the horizontal surface.
[3]

17. Define the term impulse.
[1]

18. A tennis racket strikes a ball. The force-time graph for the impact is triangular, with a peak force of $200 \text{ N}$ and a duration of $0.01 \text{ s}$ .
(a) Calculate the impulse delivered to the ball.
[2]

(b) If the ball has a mass of $0.06 \text{ kg}$ and was initially at rest, calculate its final speed.
[2]

19. A uniform ladder of weight $W$ leans against a smooth vertical wall and rests on a rough horizontal ground.
(a) Explain why there is no frictional force at the wall.
[1]

(b) State the direction of the frictional force at the ground.
[1]

(c) As a person climbs up the ladder, explain whether the frictional force at the ground increases, decreases, or stays the same.
[2]

20. A particle moves in a straight line. Its displacement $s$ (in meters) from a fixed point is given by $s = 2t^2 - 8t + 5$ , where $t$ is time in seconds.
(a) Calculate the velocity of the particle at $t = 3.0 \text{ s}$ .
[2]

(b) Determine the time at which the particle is instantaneously at rest.
[2]

End of Paper

Answers

A-Level Physics H1 Practice Paper - Mechanics (Version 2 of 5)

Marking Scheme and Answer Key

Total Marks: 60

Section A

1. Principle of Conservation of Linear Momentum

Answer: In a closed/isolated system [B1], the total linear momentum remains constant (or is conserved) provided no external resultant force acts on the system [B1].
Marks: 2

2. Ball Impact

(a) Change in Momentum
- Take upward as positive.
- Initial velocity $u = -8.0 \text{ m s}^{-1}$ , Final velocity $v = +6.0 \text{ m s}^{-1}$ .
- $\Delta p = m(v - u)$ [M1]
- $\Delta p = 0.15(6.0 - (-8.0)) = 0.15(14.0)$ [M1]
- $\Delta p = 2.1 \text{ N s}$ (or $\text{kg m s}^{-1}$ ) [A1]
- Marks: 3
(b) Average Resultant Force
- $F_{avg} = \frac{\Delta p}{\Delta t}$ [M1]
- $F_{avg} = \frac{2.1}{0.050} = 42 \text{ N}$ [A1]
- Marks: 2

3. Plank Equilibrium

(a) Free-Body Diagram
- Weight of plank ( $200 \text{ N}$ ) acting downwards at center ( $2.0 \text{ m}$ from A). [B1]
- Weight of student ( $600 \text{ N}$ ) acting downwards at distance $x$ . [B1]
- Reaction force at P ( $R_P$ ) acting upwards at A. [B1]
- Reaction force at Q ( $R_Q$ ) acting upwards at $3.0 \text{ m}$ from A. [B1] (Max 3 marks, accept clear labels)
- Marks: 3
(b) Maximum Distance $x$
- Condition for tipping: The plank is on the verge of tipping about support Q. Reaction at P becomes zero ( $R_P = 0$ ). [M1]
- Take moments about Q.
- Clockwise moment = Anticlockwise moment.
- Weight of student creates clockwise moment: $600 \times (x - 3.0)$ ? No, student is at $x$ from A. Q is at $3.0$ m from A.
- Distance of student from Q = $(3.0 - x)$ if $x < 3$ , or $(x - 3.0)$ if $x > 3$ .
- Wait, let's look at the geometry. A is at 0. Q is at 3.0 m (1m from B, length 4m). Center of mass is at 2.0 m.
- Moments about Q:
 - Plank weight ( $200 \text{ N}$ ) acts at 2.0 m. Distance from Q = $3.0 - 2.0 = 1.0 \text{ m}$ . This creates an anticlockwise moment (tending to rotate A down). Moment = $200 \times 1.0 = 200 \text{ Nm}$ .
 - Student weight ( $600 \text{ N}$ ) acts at $x$ . To tip, student must be to the right of Q? No, if student moves right of Q, they create a clockwise moment. The plank weight creates an anticlockwise moment.
 - For equilibrium limit: Moment of Student = Moment of Plank Weight.
 - $600 \times (x - 3.0) = 200 \times 1.0$ [M1]
 - $600(x - 3.0) = 200$
 - $x - 3.0 = \frac{200}{600} = \frac{1}{3}$
 - $x = 3.0 + 0.333 = 3.33 \text{ m}$ [A1]
- Alternatively, if the question implies tipping about P (student moves left), but student starts at A? "Maximum value of x" usually implies moving towards the overhang or far support. Here Q is the pivot for tipping if he goes too far right.
- Answer: $3.33 \text{ m}$ [A1]
- Marks: 4

4. Car Dynamics

(a) Acceleration Decrease
- Resultant force $F_{res} = F_{drive} - F_R = F_{drive} - kv^2$ . [M1]
- As $v$ increases, $F_R$ increases, so $F_{res}$ decreases. Since $a = F_{res}/m$ , acceleration decreases. [A1]
- Marks: 2
(b) Constant $k$
- At max speed, $a = 0$ , so $F_{drive} = F_R$ . [M1]
- $2400 = k(40)^2$
- $k = \frac{2400}{1600} = 1.5 \text{ kg m}^{-1}$ (or $\text{N s}^2 \text{m}^{-2}$ ) [A1]
- Marks: 3 (1 for condition, 1 for substitution, 1 for answer)
(c) Power
- $P = F v$ [M1]
- $P = 2400 \times 40 = 96,000 \text{ W}$ or $96 \text{ kW}$ [A1]
- Marks: 2

5. Inelastic Collision

(a) Common Velocity
- Conservation of Momentum: $m_X u_X + m_Y u_Y = (m_X + m_Y)v$ [M1]
- Taking right as positive: $(2.0)(3.0) + (1.0)(-1.0) = (2.0 + 1.0)v$
- $6.0 - 1.0 = 3.0v$
- $5.0 = 3.0v \Rightarrow v = 1.67 \text{ m s}^{-1}$ (to the right) [A1]
- Marks: 3
(b) Inelastic Proof
- $KE_{initial} = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(1.0)(-1.0)^2 = 9.0 + 0.5 = 9.5 \text{ J}$ [M1]
- $KE_{final} = \frac{1}{2}(3.0)(1.667)^2 = 4.17 \text{ J}$ [M1]
- $KE_{initial} \neq KE_{final}$ (KE is lost), so collision is inelastic. [A1]
- Marks: 3

Section B

6. Skydiver Graph

(a) Motion Description
- Initially, weight > air resistance, so there is a resultant downward force and acceleration. [B1]
- As speed increases, air resistance increases, reducing the resultant force and thus acceleration (gradient of graph decreases). [B1]
- Eventually, air resistance equals weight. Resultant force is zero, acceleration is zero, and terminal velocity is reached. [B1]
- Marks: 3
(b) Distance Fallen
- Distance = Area under v-t graph. [M1]
- Approximate area (counting squares or trapezium estimate). Graph goes to 50 m/s in 20s. If linear, area = $0.5 \times 20 \times 50 = 500$ . But it's curved (concave down). Area is greater than triangle.
- Let's assume standard shape. Area $\approx 0.7 \times 20 \times 50$ ? Or count squares.
- Accept reasonable estimate between $600 \text{ m}$ and $800 \text{ m}$ depending on curve shape provided in visual. Let's assume approx $700 \text{ m}$ . [A1]
- Marks: 2
(c) Parachute Opening
- Opening parachute greatly increases surface area, causing a large increase in air resistance. [B1]
- Air resistance becomes much larger than weight, creating a large upward resultant force (deceleration). [B1]
- Marks: 2

7. Inclined Plane

(a) Free-Body Diagram
- Weight ( $mg$ ) vertically down. [B1]
- Normal reaction ( $N$ ) perpendicular to slope. [B1]
- Friction ( $F_f$ ) down the slope (opposing motion up). [B1] (Wait, motion is UP, so friction is DOWN).
- Applied Force ( $F$ ) up the slope. [B1]
- Marks: 2 (Accept 2 correct forces for 1 mark, all 4 for 2)
(b) Force F
- Constant speed $\Rightarrow$ Equilibrium. Forces up slope = Forces down slope.
- $F = mg \sin(30^\circ) + F_f$ [M1]
- $F = (5.0)(9.81)(0.5) + 10$
- $F = 24.525 + 10 = 34.5 \text{ N}$ [A1]
- Marks: 3
(c) Power
- $P = F v$ [M1]
- $P = 34.525 \times 2.0 = 69.0 \text{ W}$ (or $69.1 \text{ W}$ ) [A1]
- Marks: 2

8. Spring

(a) Spring Constant
- $F = kx \Rightarrow k = F/x$ [M1]
- $k = 10 / 0.04 = 250 \text{ N m}^{-1}$ [A1]
- Marks: 2
(b) Elastic Potential Energy
- $E = \frac{1}{2}kx^2$ or $\frac{1}{2}Fx$ [M1]
- $E = 0.5 \times 10 \times 0.04 = 0.20 \text{ J}$ [A1]
- Marks: 2
(c) Energy Changes
- At lowest point: Max Elastic PE, Min KE, Min Gravitational PE (relative to equilibrium).
- Moving up: Elastic PE converts to KE and Gravitational PE.
- At equilibrium: Max KE.
- At highest point: Max Gravitational PE, Min Elastic PE (depending on reference), Zero KE.
- Key point: Continuous interchange between Kinetic Energy, Gravitational Potential Energy, and Elastic Potential Energy. Total energy is conserved. [B1 for each valid transition mention, max 3]
- Marks: 3

9. Projectile

(a) Horizontal Component
- $u_x = 20 \cos(30^\circ) = 17.3 \text{ m s}^{-1}$ [A1]
- Marks: 1
(b) Max Height
- $u_y = 20 \sin(30^\circ) = 10 \text{ m s}^{-1}$ . At max height, $v_y = 0$ .
- $v^2 = u^2 + 2as \Rightarrow 0 = 10^2 + 2(-9.81)h$ [M1]
- $19.62 h = 100 \Rightarrow h = 5.10 \text{ m}$ [A1]
- Marks: 3
(c) Time of Flight
- Time to max height: $v = u + at \Rightarrow 0 = 10 - 9.81t \Rightarrow t = 1.02 \text{ s}$ .
- Total time = $2 \times 1.02 = 2.04 \text{ s}$ [A1]
- Marks: 2

10. Satellite

(a) Centripetal Force
- Gravitational force (or Weight) [B1]
- Marks: 1
(b) Free Fall
- The only force acting on the satellite is gravity. [B1]
- It is constantly accelerating towards the Earth (changing direction), which is the definition of free fall, even though its tangential velocity keeps it in orbit. [B1]
- Marks: 2

Section C

11. Car Braking

(a) Braking Force
- Work-Energy Principle: Work done by brakes = Change in KE.
- $F d = \frac{1}{2}mv^2$ [M1]
- $F(40) = 0.5(1000)(20)^2 = 200,000$
- $F = \frac{200,000}{40} = 5000 \text{ N}$ [A1]
- Marks: 3
(b) New Stopping Distance
- $F d' = \frac{1}{2}m(v')^2$
- $5000 d' = 0.5(1000)(40)^2 = 800,000$
- $d' = \frac{800,000}{5000} = 160 \text{ m}$ [A1]
- Marks: 2
(c) Relationship
- KE is proportional to $v^2$ . [B1]
- Since stopping force is constant, stopping distance is proportional to KE, and thus proportional to $v^2$ . Doubling speed quadruples the distance. [B1]
- Marks: 2

12. Elastic Collision Spheres

(a) Speed before impact
- Conservation of Energy: $mgh = \frac{1}{2}mv^2$ [M1]
- $v = \sqrt{2gh}$ [A1]
- Marks: 2
(b) Elastic Conditions
- Total kinetic energy is conserved. [B1]
- Relative speed of approach equals relative speed of separation. [B1]
- Marks: 2
(c) Rebound Explanation
- Conservation of Momentum: $m u_A = m v_A + 2m v_B$ .
- Conservation of KE (or relative speed): $u_A = v_B - v_A$ (since B is at rest).
- Solving these shows $v_A$ is negative (rebound) because $m_A < m_B$ . Specifically, for elastic collision with target at rest, $v_A = u_A \frac{m_A - m_B}{m_A + m_B}$ . Since $m_A < m_B$ , the numerator is negative. [M1 for logic, A1 for conclusion]
- Marks: 3

13. Lift

(a) Tension
- Total mass $M = 500 + 70 = 570 \text{ kg}$ .
- $T - Mg = Ma \Rightarrow T = M(g + a)$ [M1]
- $T = 570(9.81 + 1.5) = 570(11.31) = 6447 \text{ N}$ [A1]
- Marks: 3
(b) Normal Reaction
- For passenger: $R - mg = ma \Rightarrow R = m(g + a)$ [M1]
- $R = 70(11.31) = 792 \text{ N}$ [A1]
- Marks: 2
(c) Constant Velocity
- Acceleration is zero. [B1]
- $R = mg$ . The normal reaction decreases to equal the passenger's weight ( $70 \times 9.81 = 687 \text{ N}$ ). [B1]
- Marks: 2

14. River Boat

(a) Resultant Velocity
- $v_{res} = \sqrt{3.0^2 + 5.0^2} = \sqrt{9 + 25} = \sqrt{34} = 5.83 \text{ m s}^{-1}$ [M1]
- Direction: $\theta = \tan^{-1}(3/5) = 31.0^\circ$ East of North. [A1]
- Marks: 3
(b) Time to Cross
- Time depends on velocity component perpendicular to bank ( $5.0 \text{ m s}^{-1}$ ).
- $t = \frac{100}{5.0} = 20 \text{ s}$ [A1]
- Marks: 2
(c) Distance Downstream
- $d = v_{river} \times t = 3.0 \times 20 = 60 \text{ m}$ [A1]
- Marks: 1

15. Friction Experiment

(a) Graph Sketch
- Straight line through origin. Positive gradient. [B1]
- Marks: 1
(b) Friction-Compensated
- The track is tilted slightly so that the component of weight down the slope balances friction. [B1]
- This ensures that the resultant force on the trolley is equal to the applied force only, allowing $F=ma$ to be tested directly without subtracting friction. [B1]
- Marks: 2
(c) Uncompensated Graph
- Line does not pass through origin. [B1]
- It has an x-intercept (force required to overcome static friction before acceleration starts). Gradient remains same (mass unchanged). [B1]
- Marks: 2

16. Ice Block

(a) Speed at Bottom
- $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ [M1]
- $v = \sqrt{2(9.81)(5.0)} = \sqrt{98.1} = 9.90 \text{ m s}^{-1}$ [A1]
- Marks: 2
(b) Frictional Force
- Work done by friction = Loss in KE.
- $F_f d = \frac{1}{2}mv^2$ [M1]
- $F_f (10) = 0.5(2.0)(9.90)^2 = 98.1 \text{ J}$
- $F_f = 9.81 \text{ N}$ [A1]
- Marks: 3

17. Impulse Definition

Answer: Impulse is the product of the average force and the time interval over which it acts ( $F \Delta t$ ). Alternatively, it is the change in momentum. [B1]
Marks: 1

18. Tennis Ball

(a) Impulse
- Impulse = Area under Force-Time graph.
- Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$ [M1]
- $I = 0.5 \times 0.01 \times 200 = 1.0 \text{ N s}$ [A1]
- Marks: 2
(b) Final Speed
- $I = \Delta p = m(v - u)$
- $1.0 = 0.06(v - 0)$ [M1]
- $v = \frac{1.0}{0.06} = 16.7 \text{ m s}^{-1}$ [A1]
- Marks: 2

19. Ladder

(a) No Friction at Wall
- The wall is smooth. [B1]
- Marks: 1
(b) Friction Direction at Ground
- Towards the wall (to oppose the tendency of the ladder to slip outwards). [B1]
- Marks: 1
(c) Person Climbing
- As the person climbs up, the moment of their weight about the base increases. [M1]
- To maintain rotational equilibrium, the normal reaction from the wall must increase. Consequently, the horizontal frictional force at the ground (which balances the wall's normal reaction) must increase. [A1]
- Marks: 2

20. Kinematics Calculus

(a) Velocity at t=3
- $v = \frac{ds}{dt} = 4t - 8$ [M1]
- At $t=3$ , $v = 4(3) - 8 = 4 \text{ m s}^{-1}$ [A1]
- Marks: 2
(b) Instantaneously at Rest
- $v = 0 \Rightarrow 4t - 8 = 0$ [M1]
- $t = 2.0 \text{ s}$ [A1]
- Marks: 2
(c) Total Distance
- Particle changes direction at $t=2$ .
- Distance 1 ( $t=0$ to $2$ ): $s(0) = 5$ , $s(2) = 2(4) - 16 + 5 = -3$ . Distance = $|-3 - 5| = 8 \text{ m}$ . [M1]
- Distance 2 ( $t=2$ to $4$ ): $s(4) = 2(16) - 32 + 5 = 5$ . Distance = $|5 - (-3)| = 8 \text{ m}$ . [M1]
- Total Distance = $8 + 8 = 16 \text{ m}$ [A1]
- Marks: 3

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