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A Level H2 Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H2 Level: A-Level Paper: Practice Paper 2 (Mechanics Focus) Version: 2 of 5 Duration: 1 hour 30 minutes Total Marks: 60

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of 20 questions on the topic of Mechanics.
Answer all questions in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend about 1 hour 30 minutes on this paper.
You may use a calculator.
Take g = 9.81 m s⁻² unless otherwise stated.

Section A: Structured Questions (20 marks)

Answer all questions in this section.

1. A car of mass 1200 kg accelerates uniformly from rest to 25 m s⁻¹ in 8.0 s along a straight, level road.

(a) Calculate the acceleration of the car. [2]

(b) Calculate the resultant force acting on the car during the acceleration. [2]

(c) The car then maintains a constant speed of 25 m s⁻¹. The total resistive force acting on the car at this speed is 800 N. Calculate the power output of the engine. [2]

2. A ball of mass 0.50 kg is dropped from a height of 20 m above the ground. Air resistance is negligible.

(a) Calculate the speed of the ball just before it hits the ground. [2]

(b) On impact with the ground, the ball rebounds vertically with a speed of 12 m s⁻¹. The time of contact with the ground is 0.080 s. Calculate the average force exerted by the ground on the ball. [3]

3. A particle moves along a straight line with velocity v given by v = 4t² − 6t + 2, where v is in m s⁻¹ and t is in seconds.

(a) Find the acceleration of the particle at t = 2.0 s. [2]

(b) Determine the displacement of the particle between t = 0 and t = 3.0 s. [3]

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

Section B: Calculation and Application (20 marks)

Answer all questions in this section.

5. A projectile is launched from ground level with an initial speed of 30 m s⁻¹ at an angle of 40° above the horizontal.

(a) Calculate the horizontal and vertical components of the initial velocity. [2]

(b) Determine the time taken for the projectile to reach its maximum height. [2]

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

6. A block of mass 3.0 kg is pulled up a rough incline of angle 25° to the horizontal by a force of 40 N applied parallel to the incline. The coefficient of kinetic friction between the block and the incline is 0.25.

(a) Draw a free-body diagram showing all forces acting on the block. [3]

(b) Calculate the normal reaction force on the block. [2]

7. Two objects, A of mass 2.0 kg and B of mass 3.0 kg, are connected by a light inextensible string passing over a smooth, frictionless pulley. Object A rests on a smooth horizontal table, while object B hangs freely.

(a) Calculate the acceleration of the system. [2]

(b) Calculate the tension in the string. [2]

Section C: Analysis and Extended Response (20 marks)

Answer all questions in this section.

8. A particle of mass 0.20 kg performs simple harmonic motion with amplitude 0.15 m and period 2.0 s.

(a) Calculate the angular frequency of the motion. [1]

(b) Write an expression for the displacement x as a function of time t, assuming x = 0 at t = 0 and the particle is moving in the positive direction. [2]

(d) Calculate the maximum acceleration of the particle. [2]

(e) Determine the total energy of the oscillating system. [2]

9. A satellite of mass 500 kg orbits the Earth in a circular orbit at a height of 400 km above the Earth's surface.

Given: Mass of Earth = 5.97 × 10²⁴ kg, Radius of Earth = 6.37 × 10⁶ m, Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².

(a) Calculate the orbital radius of the satellite. [1]

(b) Calculate the gravitational force acting on the satellite. [2]

(d) Calculate the period of the orbit in hours. [2]

10. A roller coaster car of mass 500 kg starts from rest at point A, which is 40 m above the ground. It travels down a frictionless track to point B at ground level, then enters a vertical circular loop of radius 10 m.

(a) Calculate the speed of the car at point B. [2]

(b) Calculate the speed of the car at the top of the loop (point C). [2]

(d) Explain whether the car will successfully complete the loop without falling off. [2]

END OF PAPER

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Answers

TuitionGoWhere Practice Paper - Physics H2 A-Level (Answers)

TuitionGoWhere Practice Paper (AI) - Answer Key and Marking Scheme

Subject: Physics H2 Level: A-Level Paper: Practice Paper 2 (Mechanics Focus) Version: 2 of 5

Section A: Structured Questions (20 marks)

1. Car acceleration and power

(a) Acceleration of the car [2 marks]

a = (v − u)/t = (25 − 0)/8.0 = 3.125 m s⁻² [1 mark for correct substitution, 1 mark for correct answer]
Answer: 3.13 m s⁻² (3 s.f.)

(b) Resultant force [2 marks]

F = ma = 1200 × 3.125 = 3750 N [1 mark for formula, 1 mark for correct answer]
Answer: 3750 N (or 3.75 × 10³ N)

At constant speed, driving force = resistive force = 800 N [1 mark for recognising equilibrium]
P = Fv = 800 × 25 = 20,000 W [1 mark for correct calculation]
Answer: 2.0 × 10⁴ W (or 20 kW)

2. Ball drop and rebound

(a) Speed just before impact [2 marks]

v² = u² + 2as = 0 + 2 × 9.81 × 20 = 392.4 [1 mark for correct substitution]
v = √392.4 = 19.8 m s⁻¹ [1 mark for correct answer]
Answer: 19.8 m s⁻¹

(b) Average force during impact [3 marks]

Change in velocity: Δv = 12 − (−19.8) = 31.8 m s⁻¹ (taking upward as positive) [1 mark for correct sign convention]
Acceleration: a = Δv/Δt = 31.8/0.080 = 397.5 m s⁻² [1 mark for correct calculation]
F = ma = 0.50 × 397.5 = 198.75 N [1 mark for correct force]
Answer: 199 N (upward) (3 s.f.)

3. Particle kinematics

(a) Acceleration at t = 2.0 s [2 marks]

a = dv/dt = 8t − 6 [1 mark for correct differentiation]
At t = 2.0 s: a = 8(2.0) − 6 = 10 m s⁻² [1 mark for correct substitution and answer]
Answer: 10 m s⁻²

(b) Displacement between t = 0 and t = 3.0 s [3 marks]

s = ∫v dt = ∫(4t² − 6t + 2) dt = (4/3)t³ − 3t² + 2t + C [1 mark for correct integration]
s(3) = (4/3)(27) − 3(9) + 2(3) = 36 − 27 + 6 = 15 m [1 mark for correct evaluation at t = 3]
s(0) = 0 [1 mark for correct displacement]
Answer: 15 m

4. Principle of conservation of linear momentum [2 marks]

The total momentum of a system remains constant [1 mark]
provided no net external force acts on the system [1 mark]
Accept: "In a closed/isolated system, the total momentum before an interaction equals the total momentum after the interaction, in the absence of external forces."

Section B: Calculation and Application (20 marks)

5. Projectile motion

(a) Velocity components [2 marks]

u_x = 30 cos 40° = 22.98 m s⁻¹ [1 mark]
u_y = 30 sin 40° = 19.28 m s⁻¹ [1 mark]
Answer: u_x = 23.0 m s⁻¹, u_y = 19.3 m s⁻¹

(b) Time to maximum height [2 marks]

At max height, v_y = 0: 0 = u_y − gt [1 mark for correct equation]
t = u_y/g = 19.28/9.81 = 1.97 s [1 mark for correct answer]
Answer: 1.97 s

h = u_y²/(2g) = (19.28)²/(2 × 9.81) [1 mark for correct formula]
h = 371.7/19.62 = 18.94 m [1 mark for correct answer]
Answer: 18.9 m

(d) Horizontal range [2 marks]

Total time of flight = 2 × 1.97 = 3.94 s [1 mark for recognising symmetry]
Range = u_x × total time = 22.98 × 3.94 = 90.5 m [1 mark for correct answer]
Answer: 90.5 m

6. Block on incline

(a) Free-body diagram [3 marks]

Weight (mg) acting vertically downward [1 mark]
Normal reaction (N) perpendicular to incline [1 mark]
Applied force (40 N) parallel to incline upward; friction (f) parallel to incline downward [1 mark]
Award marks for correctly labelled forces with directions

(b) Normal reaction force [2 marks]

N = mg cos 25° = 3.0 × 9.81 × cos 25° [1 mark for correct formula]
N = 29.43 × 0.9063 = 26.67 N [1 mark for correct answer]
Answer: 26.7 N

Frictional force: f = μN = 0.25 × 26.67 = 6.67 N [1 mark]
Component of weight down incline: mg sin 25° = 3.0 × 9.81 × sin 25° = 12.44 N [1 mark]
Net force up incline: F_net = 40 − 6.67 − 12.44 = 20.89 N
a = F_net/m = 20.89/3.0 = 6.96 m s⁻² [1 mark for correct acceleration]
Answer: 6.96 m s⁻² (or 7.0 m s⁻² to 2 s.f.)

7. Connected masses

(a) Acceleration of the system [2 marks]

For mass B (hanging): m_B g − T = m_B a [1 mark for correct equation]
For mass A (on table): T = m_A a
Combining: m_B g = (m_A + m_B)a → a = (3.0 × 9.81)/(2.0 + 3.0) = 29.43/5.0 = 5.886 m s⁻² [1 mark for correct answer]
Answer: 5.89 m s⁻²

(b) Tension in the string [2 marks]

T = m_A a = 2.0 × 5.886 = 11.77 N [1 mark for correct substitution]
Answer: 11.8 N [1 mark for correct answer]

Section C: Analysis and Extended Response (20 marks)

8. Simple harmonic motion

(a) Angular frequency [1 mark]

ω = 2π/T = 2π/2.0 = π rad s⁻¹
Answer: 3.14 rad s⁻¹ (or π rad s⁻¹)

(b) Displacement expression [2 marks]

x = A sin(ωt) since x = 0 at t = 0 and moving positive [1 mark for correct form]
x = 0.15 sin(πt) [1 mark for correct substitution]
Answer: x = 0.15 sin(πt) m

v_max = ωA = π × 0.15 [1 mark for correct formula]
v_max = 0.471 m s⁻¹ [1 mark for correct answer]
Answer: 0.471 m s⁻¹

(d) Maximum acceleration [2 marks]

a_max = ω²A = π² × 0.15 [1 mark for correct formula]
a_max = 9.87 × 0.15 = 1.48 m s⁻² [1 mark for correct answer]
Answer: 1.48 m s⁻²

(e) Total energy [2 marks]

E_total = ½mω²A² = ½ × 0.20 × π² × (0.15)² [1 mark for correct formula]
E_total = 0.10 × 9.87 × 0.0225 = 0.0222 J [1 mark for correct answer]
Answer: 0.0222 J (or 2.22 × 10⁻² J)

9. Satellite orbit

(a) Orbital radius [1 mark]

r = R_Earth + h = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m
Answer: 6.77 × 10⁶ m

(b) Gravitational force [2 marks]

F = GMm/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 500)/(6.77 × 10⁶)² [1 mark for correct substitution]
F = (1.991 × 10¹⁷)/(4.583 × 10¹³) = 4344 N [1 mark for correct answer]
Answer: 4.34 × 10³ N

Gravitational force provides centripetal force: GMm/r² = mv²/r → v = √(GM/r) [1 mark for correct derivation]
v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6.77 × 10⁶)] = √(5.88 × 10⁷) = 7668 m s⁻¹ [1 mark for correct answer]
Answer: 7.67 × 10³ m s⁻¹ (or 7.67 km s⁻¹)

(d) Orbital period [2 marks]

T = 2πr/v = 2π × 6.77 × 10⁶/7668 [1 mark for correct formula]
T = 5547 s = 5547/3600 = 1.54 hours [1 mark for correct conversion]
Answer: 1.54 hours

10. Roller coaster

(a) Speed at point B [2 marks]

Conservation of energy: mgh_A = ½mv_B² [1 mark for correct energy equation]
v_B = √(2gh_A) = √(2 × 9.81 × 40) = √784.8 = 28.0 m s⁻¹ [1 mark for correct answer]
Answer: 28.0 m s⁻¹

(b) Speed at top of loop (point C) [2 marks]

Height at C = 2 × 10 = 20 m above ground
Energy conservation: mgh_A = mgh_C + ½mv_C² [1 mark for correct equation]
½v_C² = g(h_A − h_C) = 9.81 × (40 − 20) = 196.2
v_C = √(2 × 196.2) = √392.4 = 19.8 m s⁻¹ [1 mark for correct answer]
Answer: 19.8 m s⁻¹

At top of loop: mg + N = mv_C²/r [1 mark for correct force equation]
N = mv_C²/r − mg = 500 × (19.8)²/10 − 500 × 9.81
N = 500 × 39.2 − 4905 = 19,600 − 4905 = 14,695 N [1 mark for correct answer]
Answer: 1.47 × 10⁴ N (downward)

(d) Will the car complete the loop? [2 marks]

For the car to stay on the track at the top, N ≥ 0, so v_C²/r ≥ g, i.e., v_C ≥ √(gr) = √(9.81 × 10) = 9.90 m s⁻¹ [1 mark for correct condition]
Since v_C = 19.8 m s⁻¹ > 9.90 m s⁻¹, the car will successfully complete the loop [1 mark for correct conclusion with reasoning]
Answer: Yes, because the speed at the top (19.8 m s⁻¹) exceeds the minimum required speed (9.90 m s⁻¹) to maintain contact with the track.

END OF ANSWER KEY

Marking Summary:

Section A: 20 marks (Questions 1–4)
Section B: 20 marks (Questions 5–7)
Section C: 20 marks (Questions 8–10)
Total: 60 marks