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A Level H1 Physics Practice Paper 4

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TuitionGoWhere Exam Practice (AI)

Subject: Physics H1 (8867)
Level: A-Level
Paper: Practice Paper 1 (Mechanics Focus) - Version 4 of 5
Duration: 1 hour
Total Marks: 40

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates:

Answer all questions.
Write your answers in the spaces provided.
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.
This paper focuses on Section II: Newtonian Mechanics of the H1 Physics syllabus.

Section A: Structured Questions

Answer all questions in this section.

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

Answer space

2. A car of mass $1200 \text{ kg}$ travels at a constant speed of $25 \text{ m s}^{-1}$ along a straight horizontal road. The total resistive force acting on the car is $800 \text{ N}$ .
(a) Calculate the driving force provided by the engine.
[1]

Answer space

(b) Calculate the power developed by the engine.
[2]

Answer space

3. Fig. 3.1 shows a uniform plank $AB$ of length $4.0 \text{ m}$ and weight $200 \text{ N}$ . The plank rests on two supports $X$ and $Y$ . Support $X$ is at end $A$ , and support $Y$ is $1.0 \text{ m}$ from end $B$ . A student of weight $600 \text{ N}$ stands on the plank at a distance $x$ from $A$ .

(Diagram Description: Horizontal plank. Support X at left end (A). Support Y at 3.0m from A. Weight of plank acts at center (2.0m from A). Student stands at distance x.)

(a) On Fig. 3.1, draw and label arrows to represent all the vertical forces acting on the plank.
[2]

Answer space

(b) Determine the maximum value of $x$ such that the plank does not tip over about support $Y$ .
[3]

Answer space

4. A ball is thrown vertically upwards with an initial speed of $15 \text{ m s}^{-1}$ . Air resistance is negligible.
(a) Calculate the maximum height reached by the ball.
[2]

Answer space

(b) Sketch the graph of velocity $v$ against time $t$ for the motion of the ball from the instant it is thrown until it returns to the starting height. Take upward velocity as positive.
[2]

Answer space

5. Two trolleys, $A$ and $B$ , move along a smooth horizontal track. Trolley $A$ has mass $2.0 \text{ kg}$ and moves with velocity $4.0 \text{ m s}^{-1}$ to the right. Trolley $B$ has mass $3.0 \text{ kg}$ and is initially at rest. They collide and stick together.
(a) Calculate the common velocity of the trolleys after the collision.
[3]

Answer space

(b) Determine whether the collision is elastic or inelastic. Show your working.
[2]

Answer space

6. Fig. 6.1 shows a block of mass $5.0 \text{ kg}$ resting on a rough inclined plane at an angle of $30^\circ$ to the horizontal. The block is held in equilibrium by a horizontal force $F$ .

(Diagram Description: Inclined plane at 30 degrees. Block on plane. Horizontal force F pushing into the slope.)

(a) Draw a free-body diagram for the block, showing all forces acting on it.
[2]

Answer space

(b) Calculate the magnitude of the normal contact force exerted by the plane on the block if $F = 20 \text{ N}$ .
[3]

Answer space

7. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards from rest with a constant acceleration of $0.5 \text{ m s}^{-2}$ .
(a) Calculate the tension in the cable during the acceleration phase.
[2]

Answer space

(b) The load reaches a speed of $2.0 \text{ m s}^{-1}$ . Calculate the useful power delivered by the crane at this instant.
[2]

Answer space

8. Explain, in terms of forces, why a skydiver falling through the air eventually reaches a constant terminal velocity.
[3]

Answer space

9. A spring obeys Hooke’s Law. When a force of $10 \text{ N}$ is applied, the extension is $4.0 \text{ cm}$ .
(a) Calculate the spring constant $k$ .
[1]

Answer space

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

Answer space

10. A projectile is fired horizontally from the top of a cliff with a speed of $20 \text{ m s}^{-1}$ . It hits the ground $3.0 \text{ s}$ later. ( $g = 9.81 \text{ m s}^{-2}$ )
(a) Calculate the height of the cliff.
[2]

Answer space

(b) Calculate the horizontal distance from the base of the cliff to the point where the projectile lands.
[1]

Answer space

Section B: Data and Context Questions

Answer all questions in this section.

11. Fig. 11.1 shows the displacement-time graph for a particle undergoing simple harmonic motion.

(Diagram Description: Sinusoidal wave. Peak at t=0.5s, displacement=0.2m. Trough at t=1.5s, displacement=-0.2m. Period is 2.0s.)

(a) State the amplitude of the motion.
[1]

(b) Determine the maximum speed of the particle.
[2]

Answer space

12. In an experiment to determine the acceleration of free fall $g$ , a steel ball is dropped from rest. The time $t$ taken to fall a distance $h$ is measured.
(a) State the equation relating $h$ , $g$ , and $t$ .
[1]

(b) Explain why plotting a graph of $h$ against $t^2$ allows for a more accurate determination of $g$ than calculating $g$ from a single pair of $h$ and $t$ values.
[2]

Answer space

13. A car of mass $1000 \text{ kg}$ is traveling at $20 \text{ m s}^{-1}$ . The driver applies the brakes, and the car comes to rest in $5.0 \text{ s}$ .
(a) Calculate the average braking force.
[2]

Answer space

(b) Calculate the work done by the braking force.
[2]

Answer space

14. Fig. 14.1 shows a ladder leaning against a smooth vertical wall. The foot of the ladder rests on rough horizontal ground.

(Diagram Description: Ladder at angle theta to ground. Wall is smooth. Ground is rough.)

(a) Explain why there is no vertical frictional force exerted by the wall on the ladder.
[1]

Answer space

(b) State the direction of the frictional force exerted by the ground on the ladder.
[1]

15. A satellite orbits the Earth in a circular path.
(a) State the direction of the resultant force acting on the satellite.
[1]

(b) Explain why the speed of the satellite remains constant even though it is accelerating.
[2]

Answer space

Section C: Extended Response

Answer the question in this section.

16. A student investigates the motion of a trolley rolling down a slope. The trolley starts from rest. The student measures the distance $s$ traveled and the time $t$ taken.
(a) Describe how the student can use the data to determine the acceleration of the trolley. Include a sketch of the graph you would plot.
[3]

Answer space

(b) The student repeats the experiment with a heavier trolley on the same slope. Assuming friction is negligible, state and explain how the acceleration of the heavier trolley compares to the lighter one.
[2]

Answer space

17. Define the term impulse.
[1]

18. A golf club strikes a stationary golf ball of mass $0.045 \text{ kg}$ . The club is in contact with the ball for $0.50 \text{ ms}$ . The ball leaves the club with a speed of $50 \text{ m s}^{-1}$ .
(a) Calculate the impulse imparted to the ball.
[2]

Answer space

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

Answer space

19. Fig. 19.1 shows a velocity-time graph for a car moving in a straight line.

(Diagram Description: Graph starts at v=0, t=0. Linear increase to v=20 at t=10. Constant v=20 from t=10 to t=30. Linear decrease to v=0 at t=40.)

(a) Calculate the total distance traveled by the car.
[3]

Answer space

(b) Calculate the average speed of the car for the entire journey.
[1]

Answer space

20. A box of mass $10 \text{ kg}$ is pushed across a horizontal floor with a constant horizontal force of $50 \text{ N}$ . The box moves at a constant velocity.
(a) State the magnitude of the frictional force acting on the box.
[1]

(b) The pushing force is increased to $70 \text{ N}$ . Assuming the frictional force remains constant, calculate the acceleration of the box.
[2]

Answer space

[END OF PAPER]

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Answers

TuitionGoWhere Exam Practice (AI) - Answer Key

Subject: Physics H1 (8867)
Paper: Practice Paper 1 (Mechanics Focus) - Version 4 of 5

Section A: Structured Questions

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

Answer: In a closed system (or isolated system) [B1], the total linear momentum remains constant (or is conserved) provided no external forces act [B1].
Note: Must mention "closed/isolated system" or "no external forces".

2. Car motion. [3]

(a) Driving force: Since speed is constant, acceleration is zero. By Newton's First Law, driving force equals resistive force.
- $F_{\text{drive}} = 800 \text{ N}$ [B1]
(b) Power:
- $P = Fv$ [M1]
- $P = 800 \times 25 = 20,000 \text{ W}$ (or $20 \text{ kW}$ ) [A1]

3. Plank equilibrium. [5]

(a) Forces:
- 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 at X ( $R_X$ ) acting upwards at A. [B1 - if drawn]
- Reaction at Y ( $R_Y$ ) acting upwards at $3.0 \text{ m}$ from A. [B1 - if drawn]
- Note: Award marks for correct labels and directions.
(b) Maximum $x$ $x$ (tipping point):
- At tipping point, reaction at X becomes zero ( $R_X = 0$ ). The plank pivots about Y. [M1]
- Take moments about Y:
  - Clockwise moment (Student): $600 \times (x - 3.0)$ ? No, distance from Y is $(3.0 - x)$ if $x < 3$ . Wait, student is between A and B. If student moves towards B, moment increases. Let's define distances from Y.
  - Center of mass is at $2.0 \text{ m}$ from A. Y is at $3.0 \text{ m}$ from A. Distance CM to Y = $1.0 \text{ m}$ .
  - Moment of Plank Weight about Y (Counter-Clockwise): $200 \text{ N} \times 1.0 \text{ m} = 200 \text{ N m}$ .
  - Moment of Student about Y (Clockwise): $600 \text{ N} \times d$ , where $d$ is distance from Y towards B.
  - For equilibrium limit: $200 = 600 \times d \Rightarrow d = 200/600 = 0.333 \text{ m}$ .
  - Position $x$ from A: $x = 3.0 + 0.333 = 3.33 \text{ m}$ .
- Alternative interpretation: If student is to the left of Y, they help balance. Tipping happens if student goes too far right.
- Sum of moments about Y = 0.
- $200(1.0) = 600(x - 3.0)$ assuming $x > 3$ .
- $200 = 600(x - 3)$ .
- $0.333 = x - 3$ .
- $x = 3.33 \text{ m}$ . [A1]
- Check: Is $x \le 4.0$ ? Yes.

4. Vertical projectile. [4]

(a) Max height:
- $v^2 = u^2 + 2as$ [M1]
- $0 = 15^2 + 2(-9.81)s$
- $s = 225 / 19.62 = 11.47 \text{ m}$ [A1] (Accept $11.5 \text{ m}$ )
(b) Graph:
- Straight line with negative gradient. [B1]
- Starts at $+15$ , crosses t-axis, ends at $-15$ (symmetry). [B1]

5. Collision. [5]

(a) Common velocity:
- Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ [M1]
- $(2.0)(4.0) + 0 = (2.0 + 3.0)v$
- $8.0 = 5.0v$
- $v = 1.6 \text{ m s}^{-1}$ [A1] Direction: To the right. [B1]
(b) Elastic/Inelastic:
- KE before: $\frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$ [M1]
- KE after: $\frac{1}{2}(5.0)(1.6)^2 = 6.4 \text{ J}$
- KE is not conserved ( $16 \neq 6.4$ ). Therefore, inelastic. [A1]

6. Inclined plane with horizontal force. [5]

(a) Free-body diagram:
- Weight ( $mg$ ) vertically down. [B1]
- Normal contact force ( $N$ ) perpendicular to plane. [B1]
- Applied force ( $F$ ) horizontal. [B1]
- Friction ( $f$ ) parallel to plane (up or down depending on tendency, usually up if F is small, but here F pushes in. Let's assume equilibrium). Note: Question asks for forces, not calculation of friction direction yet, but typically friction opposes motion tendency.
(b) Normal contact force:
- Resolve forces perpendicular to the plane.
- $N = mg \cos(30^\circ) + F \sin(30^\circ)$ [M1]
- $N = (5.0)(9.81)\cos(30^\circ) + 20 \sin(30^\circ)$ [M1]
- $N = 42.48 + 10 = 52.48 \text{ N}$ [A1] (Accept $52.5 \text{ N}$ )

7. Crane lifting load. [4]

(a) Tension:
- $T - mg = ma$ [M1]
- $T = m(g + a) = 500(9.81 + 0.5) = 500(10.31)$
- $T = 5155 \text{ N}$ [A1] (Accept $5160 \text{ N}$ or $5.2 \text{ kN}$ )
(b) Power:
- $P = Tv$ [M1] (Use Tension, not just weight, as it's accelerating)
- $P = 5155 \times 2.0 = 10,310 \text{ W}$ [A1] (Accept $10.3 \text{ kW}$ )

8. Terminal velocity explanation. [3]

Initially, weight > air resistance, so there is a resultant downward force and acceleration. [B1]
As speed increases, air resistance increases. [B1]
Eventually, air resistance equals weight. Resultant force is zero, so acceleration is zero and velocity becomes constant (terminal velocity). [B1]

9. Spring. [3]

(a) Spring constant:
- $F = kx \Rightarrow k = F/x$ [M1]
- $k = 10 / 0.04 = 250 \text{ N m}^{-1}$ [A1]
(b) Elastic Potential Energy:
- $E = \frac{1}{2}kx^2$ or $\frac{1}{2}Fx$ [M1]
- $E = 0.5 \times 10 \times 0.04 = 0.2 \text{ J}$ [A1]

10. Horizontal projectile. [3]

(a) Height:
- $s = ut + \frac{1}{2}at^2$ (vertical)
- $u_y = 0, a = 9.81, t = 3.0$
- $h = 0 + 0.5(9.81)(3.0)^2 = 44.145 \text{ m}$ [A1] (Accept $44.1 \text{ m}$ )
(b) Horizontal distance:
- $d = v_x t = 20 \times 3.0 = 60 \text{ m}$ [B1]

Section B: Data and Context Questions

11. SHM Graph. [3]

(a) Amplitude: $0.2 \text{ m}$ [B1]
(b) Max speed:
- $\omega = 2\pi / T = 2\pi / 2.0 = \pi \text{ rad s}^{-1}$ [M1]
- $v_{\max} = \omega A = \pi \times 0.2 = 0.628 \text{ m s}^{-1}$ [A1]

12. Free fall experiment. [3]

(a) Equation: $h = \frac{1}{2}gt^2$ [B1]
(b) Graph advantage:
- Plotting $h$ vs $t^2$ gives a straight line through the origin. [B1]
- The gradient is $\frac{1}{2}g$ . Using a line of best fit reduces the effect of random errors in individual time/distance measurements. [B1]

13. Braking car. [4]

(a) Average braking force:
- $a = (v - u) / t = (0 - 20) / 5.0 = -4.0 \text{ m s}^{-2}$ [M1]
- $F = ma = 1000 \times (-4.0) = -4000 \text{ N}$ [A1]
- Magnitude is $4000 \text{ N}$ .
(b) Work done:
- Work done = Change in KE [M1]
- $W = \frac{1}{2}mv^2 - 0 = 0.5 \times 1000 \times 20^2 = 200,000 \text{ J}$ [A1]
- (Or $W = Fs = 4000 \times (avg\_speed \times t) = 4000 \times 10 \times 5 = 200,000 \text{ J}$ )

14. Ladder. [2]

(a) No vertical friction at wall: The wall is smooth, so it cannot exert a frictional force parallel to its surface. Friction acts parallel to the contact surface. [B1]
(b) Friction at ground: The ladder tends to slip outwards (away from the wall) at the base. Therefore, friction acts horizontally towards the wall. [B1]

15. Satellite. [3]

(a) Direction of force: Towards the center of the Earth. [B1]
(b) Constant speed:
- The force (gravity) is perpendicular to the velocity vector. [B1]
- Therefore, the force does no work on the satellite, and only changes the direction of velocity, not its magnitude (speed). [B1]

Section C: Extended Response

16. Trolley on slope. [5]

(a) Method:
- Plot a graph of distance $s$ (y-axis) against $t^2$ (x-axis). [B1]
- Since $s = \frac{1}{2}at^2$ (starting from rest), the graph should be a straight line through the origin. [B1]
- The gradient of the line is equal to $\frac{1}{2}a$ . Therefore, $a = 2 \times \text{gradient}$ . [B1]
(b) Heavier trolley:
- Acceleration remains the same. [B1]
- Explanation: The component of weight down the slope is $mg \sin \theta$ . By Newton's 2nd Law, $mg \sin \theta = ma$ . Mass $m$ cancels out, so $a = g \sin \theta$ , which is independent of mass. [B1]

17. Impulse. [1]

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

18. Golf ball. [4]

(a) Impulse:
- $I = \Delta p = m(v - u)$ [M1]
- $I = 0.045(50 - 0) = 2.25 \text{ N s}$ [A1]
(b) Average Force:
- $F_{\text{avg}} = I / \Delta t$ [M1]
- $\Delta t = 0.50 \text{ ms} = 0.50 \times 10^{-3} \text{ s}$
- $F_{\text{avg}} = 2.25 / (0.50 \times 10^{-3}) = 4500 \text{ N}$ [A1]

19. Velocity-time graph area. [4]

(a) Total distance:
- Area under graph = Distance.
- Area 1 (Triangle, 0-10s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$ [M1]
- Area 2 (Rectangle, 10-30s): $20 \times 20 = 400 \text{ m}$ [M1]
- Area 3 (Triangle, 30-40s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$ [M1]
- Total Distance = $100 + 400 + 100 = 600 \text{ m}$ [A1]
(b) Average speed:
- $\text{Avg Speed} = \text{Total Distance} / \text{Total Time}$ [M1]
- $600 / 40 = 15 \text{ m s}^{-1}$ [A1]

20. Box on floor. [3]

(a) Frictional force:
- Since velocity is constant, forces are balanced.
- $f = F_{\text{push}} = 50 \text{ N}$ [B1]
(b) New acceleration:
- New resultant force $F_{\text{res}} = 70 - 50 = 20 \text{ N}$ [M1]
- $a = F_{\text{res}} / m = 20 / 10 = 2.0 \text{ m s}^{-2}$ [A1]