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A Level H2 Physics Practice Paper 2

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Questions

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Subject: Physics
Level: A-Level H2
Paper: Practice Paper (Version 2 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend approximately 1 hour 30 minutes on this paper.
Use $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Structured Questions

Answer all questions in this section.

1. State the Principle of Conservation of Linear Momentum.
...................................................................................................................................................
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[2]

2. A ball of mass $0.15 \text{ kg}$ undergoes simple harmonic motion with an amplitude of $4.0 \text{ cm}$ and a frequency of $2.5 \text{ Hz}$ .
Calculate the maximum acceleration of the ball.

Maximum acceleration = __________________________ $\text{m s}^{-2}$ [3]

3. In an experiment to determine the acceleration due to gravity $g$ , a student drops a steel ball from rest and measures the time $t$ it takes to fall a distance $h$ .
State two precautions the student should take to improve the accuracy of the measurement of $t$ .

...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
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[2]

4. A proton moves with a speed of $2.0 \times 10^6 \text{ m s}^{-1}$ into a uniform magnetic field of flux density $0.50 \text{ T}$ . The velocity of the proton is perpendicular to the magnetic field.
State the direction of the magnetic force acting on the proton relative to its velocity and the magnetic field direction.

5. Define the term binding energy of a nucleus.

...................................................................................................................................................
...................................................................................................................................................
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[2]

6. A block of mass $2.0 \text{ kg}$ slides down a rough inclined plane at a constant velocity. The angle of inclination is $30^\circ$ .
Calculate the magnitude of the frictional force acting on the block.

Frictional force = __________________________ N [3]

7. An electron is accelerated from rest through a potential difference of $5.0 \text{ kV}$ .
Calculate the final speed of the electron. (Charge of electron $e = 1.60 \times 10^{-19} \text{ C}$ , Mass of electron $m_e = 9.11 \times 10^{-31} \text{ kg}$ )

Speed = __________________________ $\text{m s}^{-1}$ [3]

8. State Faraday’s Law of electromagnetic induction.

9. A satellite orbits the Earth in a circular orbit of radius $r$ .
Explain why the satellite is accelerating even though its speed is constant.

10. A spring obeys Hooke’s Law. When a load of $10 \text{ N}$ is applied, the extension is $5.0 \text{ cm}$ .
Calculate the elastic potential energy stored in the spring.

Energy = __________________________ J [3]

Section B: Data Analysis and Application

Answer all questions in this section.

11. A student investigates the relationship between the period $T$ of a simple pendulum and its length $L$ . The student plots a graph of $T^2$ against $L$ and obtains a straight line passing through the origin with a gradient of $4.0 \text{ s}^2 \text{ m}^{-1}$ .
(a) Determine the value of the acceleration due to gravity $g$ from this gradient.

$g$ = __________________________ $\text{m s}^{-2}$ [3]

(b) Suggest one reason why the graph might not pass exactly through the origin in a real experiment.

12. Two trolleys, A and B, move on a frictionless horizontal track. Trolley A has mass $0.50 \text{ kg}$ and moves with velocity $2.0 \text{ m s}^{-1}$ to the right. Trolley B has mass $0.30 \text{ kg}$ and is initially at rest. They collide and stick together.
(a) Calculate the common velocity of the trolleys after the collision.

Velocity = __________________________ $\text{m s}^{-1}$ [3]

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

Conclusion: __________________________ [3]

13. The diagram below shows a uniform beam of length $2.0 \text{ m}$ and weight $50 \text{ N}$ pivoted at one end. A vertical force $F$ is applied at the other end to keep the beam horizontal. A load of $100 \text{ N}$ is placed $0.50 \text{ m}$ from the pivot.
(a) Calculate the magnitude of the force $F$ .

$F$ = __________________________ N [3]

(b) State the condition for rotational equilibrium used in your calculation.

...................................................................................................................................................
[1]

14. A car of mass $1200 \text{ kg}$ travels around a horizontal circular bend of radius $50 \text{ m}$ . The coefficient of static friction between the tires and the road is $0.80$ .
(a) Calculate the maximum speed at which the car can travel without skidding.

Maximum speed = __________________________ $\text{m s}^{-1}$ [3]

(b) Explain what happens to the required centripetal force if the radius of the bend is doubled while the speed remains constant.

15. In a photoelectric effect experiment, light of wavelength $400 \text{ nm}$ is incident on a metal surface. The work function of the metal is $2.0 \text{ eV}$ .
(a) Calculate the energy of a single photon of this light in Joules. ( $h = 6.63 \times 10^{-34} \text{ J s}$ , $c = 3.00 \times 10^8 \text{ m s}^{-1}$ )

Energy = __________________________ J [3]

(b) Determine the maximum kinetic energy of the emitted photoelectrons in eV. ( $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$ )

Maximum KE = __________________________ eV [3]

Section C: Extended Response

Answer all questions in this section.

16. A projectile is launched from ground level with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $60^\circ$ to the horizontal. Air resistance is negligible.
(a) Calculate the maximum height reached by the projectile.

Maximum height = __________________________ m [4]

(b) Calculate the horizontal range of the projectile.

Range = __________________________ m [4]

17. A block of mass $5.0 \text{ kg}$ is pulled up a rough inclined plane by a constant force of $40 \text{ N}$ parallel to the plane. The plane is inclined at $30^\circ$ to the horizontal. The block moves a distance of $10 \text{ m}$ up the plane at a constant speed.
(a) Calculate the work done by the applied force.

Work done = __________________________ J [2]

(b) Calculate the gain in gravitational potential energy of the block.

Gain in GPE = __________________________ J [3]

18. A particle of mass $m$ moves in a vertical circle of radius $r$ on the end of a string.
(a) Derive an expression for the minimum speed $v_{min}$ the particle must have at the top of the circle to maintain circular motion.

$v_{min}$ = __________________________ [4]

(b) If the speed at the bottom of the circle is $\sqrt{5gr}$ , show that the tension in the string at the bottom is $6mg$ .

[4]

19. Two stars, each of mass $M$ , orbit their common center of mass in circular orbits of radius $R$ .
(a) Show that the gravitational force between the stars is $\frac{GM^2}{4R^2}$ .

[2]

(b) Derive an expression for the orbital period $T$ of the stars in terms of $G$ , $M$ , and $R$ .

$T$ = __________________________ [4]

20. A rocket of initial mass $M_0$ is launched vertically from rest. It ejects gas at a constant speed $u$ relative to the rocket at a constant rate $\frac{dm}{dt}$ .
(a) State Newton’s Second Law of Motion in terms of momentum.

(b) Explain why the acceleration of the rocket increases with time, assuming the thrust is constant and air resistance is negligible.

(c) If the thrust is $F$ and the mass at time $t$ is $M(t)$ , write an equation for the acceleration $a(t)$ of the rocket.

$a(t)$ = __________________________ [2]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Answer Key and Marking Scheme (Version 2)

Total Marks: 60

Section A: Structured Questions

1. State the Principle of Conservation of Linear Momentum.
Answer:
In a closed system (or isolated system) [1], the total momentum before an interaction (collision/explosion) is equal to the total momentum after the interaction, provided no external forces act [1].
(Accept: "Total momentum of a system remains constant if the resultant external force is zero.")
[2]

2. Calculate the maximum acceleration of the ball.
Answer:
$\omega = 2\pi f = 2\pi(2.5) = 5\pi \text{ rad s}^{-1}$ [1]
$a_{max} = \omega^2 x_0$ [1]
$a_{max} = (5\pi)^2 \times 0.04 = 25\pi^2 \times 0.04 = \pi^2 \approx 9.87 \text{ m s}^{-2}$ [1]
Answer: $9.9 \text{ m s}^{-2}$ (2 s.f.)
[3]

3. State two precautions to improve accuracy of $t$ .
Answer:

Use an electronic timer/light gate to eliminate human reaction time error [1].
Repeat the measurement several times and take the average to reduce random error [1].
(Other valid answers: Ensure the ball is dropped from rest; Use a large height $h$ to reduce percentage uncertainty in time.)
[2]

4. Direction of magnetic force.
Answer:
Perpendicular to both the velocity of the proton and the magnetic field direction [1].
(Accept: "Perpendicular to the plane containing v and B")
[1]

5. Define binding energy.
Answer:
The energy required to completely separate a nucleus into its constituent protons and neutrons [1] (to infinity) [1].
(Alternatively: The energy released when protons and neutrons combine to form a nucleus.)
[2]

6. Calculate frictional force.
Answer:
Since velocity is constant, acceleration is zero, so resultant force is zero [1].
Component of weight down the slope $= mg \sin \theta$ [1]
$F_f = 2.0 \times 9.81 \times \sin 30^\circ = 9.81 \text{ N}$ [1]
Answer: $9.8 \text{ N}$
[3]

7. Calculate final speed of electron.
Answer:
Gain in KE = Loss in EPE
$\frac{1}{2}mv^2 = eV$ [1]
$v = \sqrt{\frac{2eV}{m}}$
$v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 5000}{9.11 \times 10^{-31}}}$ [1]
$v = \sqrt{1.756 \times 10^{15}} = 4.19 \times 10^7 \text{ m s}^{-1}$ [1]
Answer: $4.2 \times 10^7 \text{ m s}^{-1}$
[3]

8. State Faraday’s Law.
Answer:
The induced e.m.f. is proportional to the rate of change of magnetic flux linkage [1] (or magnetic flux) [1].
(Accept: $\varepsilon = -\frac{d(N\Phi)}{dt}$ with explanation)
[2]

9. Why is the satellite accelerating?
Answer:
Velocity is a vector quantity (has direction) [1].
The direction of the velocity is constantly changing as it moves in a circle, so there is a change in velocity, which means acceleration [1].
[2]

10. Calculate elastic potential energy.
Answer:
Spring constant $k = \frac{F}{x} = \frac{10}{0.05} = 200 \text{ N m}^{-1}$ [1]
$E = \frac{1}{2}kx^2$ [1]
$E = \frac{1}{2}(200)(0.05)^2 = 100 \times 0.0025 = 0.25 \text{ J}$ [1]
Answer: $0.25 \text{ J}$
[3]

Section B: Data Analysis and Application

11. (a) Determine $g$ .
Answer:
Formula: $T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g} L$ [1]
Gradient $= \frac{4\pi^2}{g}$ [1]
$4.0 = \frac{4\pi^2}{g} \Rightarrow g = \frac{4\pi^2}{4.0} = \pi^2 \approx 9.87 \text{ m s}^{-2}$ [1]
Answer: $9.9 \text{ m s}^{-2}$
[3]

(b) Reason for non-zero intercept.
Answer:
Systematic error in measuring length $L$ (e.g., measured to top of bob instead of center of mass) [1].
[1]

12. (a) Common velocity.
Answer:
Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ [1]
$(0.50)(2.0) + 0 = (0.50 + 0.30)v$
$1.0 = 0.80 v$
$v = 1.25 \text{ m s}^{-1}$ [1]
Answer: $1.3 \text{ m s}^{-1}$ (2 s.f.) [1]
[3]

(b) Elastic or inelastic?
Answer:
Initial KE $= \frac{1}{2}(0.50)(2.0)^2 = 1.0 \text{ J}$ [1]
Final KE $= \frac{1}{2}(0.80)(1.25)^2 = 0.625 \text{ J}$ [1]
KE is not conserved ( $1.0 \neq 0.625$ ), so the collision is inelastic [1].
[3]

13. (a) Calculate force $F$ .
Answer:
Taking moments about the pivot:
Clockwise moments = Anticlockwise moments [1]
$(100 \times 0.50) + (50 \times 1.0) = F \times 2.0$ (Weight acts at center, 1.0m from pivot) [1]
$50 + 50 = 2F$
$100 = 2F \Rightarrow F = 50 \text{ N}$ [1]
Answer: $50 \text{ N}$
[3]

(b) Condition for rotational equilibrium.
Answer:
The sum of clockwise moments equals the sum of anticlockwise moments about any point [1].
[1]

14. (a) Maximum speed.
Answer:
Centripetal force provided by friction: $F_c = \frac{mv^2}{r}$ [1]
Max friction $F_{max} = \mu mg$
$\mu mg = \frac{mv^2}{r} \Rightarrow v = \sqrt{\mu gr}$ [1]
$v = \sqrt{0.80 \times 9.81 \times 50} = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$ [1]
Answer: $20 \text{ m s}^{-1}$ (2 s.f.)
[3]

(b) Effect of doubling radius.
Answer:
$F_c = \frac{mv^2}{r}$ . If $r$ doubles and $v$ is constant, $F_c$ is halved [1].
Therefore, the required centripetal force decreases [1].
[2]

15. (a) Energy of photon.
Answer:
$E = \frac{hc}{\lambda}$ [1]
$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{400 \times 10^{-9}}$ [1]
$E = 4.97 \times 10^{-19} \text{ J}$ [1]
Answer: $5.0 \times 10^{-19} \text{ J}$
[3]

(b) Maximum KE in eV.
Answer:
Work function $\Phi = 2.0 \text{ eV} = 2.0 \times 1.60 \times 10^{-19} = 3.20 \times 10^{-19} \text{ J}$ [1]
$KE_{max} = E - \Phi = 4.97 \times 10^{-19} - 3.20 \times 10^{-19} = 1.77 \times 10^{-19} \text{ J}$ [1]
In eV: $\frac{1.77 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.11 \text{ eV}$ [1]
Answer: $1.1 \text{ eV}$
[3]

Section C: Extended Response

16. (a) Maximum height.
Answer:
Vertical component of initial velocity $u_y = 20 \sin 60^\circ = 17.32 \text{ m s}^{-1}$ [1]
At max height, $v_y = 0$ . Using $v^2 = u^2 + 2as$ :
$0 = (17.32)^2 + 2(-9.81)h$ [1]
$h = \frac{300}{19.62} = 15.29 \text{ m}$ [2]
Answer: $15 \text{ m}$ (2 s.f.)
[4]

(b) Horizontal range.
Answer:
Time to reach max height: $v = u + at \Rightarrow 0 = 17.32 - 9.81t \Rightarrow t = 1.765 \text{ s}$ [1]
Total time of flight $T = 2t = 3.53 \text{ s}$ [1]
Horizontal velocity $u_x = 20 \cos 60^\circ = 10 \text{ m s}^{-1}$ [1]
Range $= u_x \times T = 10 \times 3.53 = 35.3 \text{ m}$ [1]
Answer: $35 \text{ m}$
[4]

17. (a) Work done by applied force.
Answer:
$W = F d \cos \theta$ . Force is parallel to displacement, so $\theta = 0$ .
$W = 40 \times 10 = 400 \text{ J}$ [2]
[2]

(b) Gain in GPE.
Answer:
Vertical height gained $h = d \sin 30^\circ = 10 \times 0.5 = 5.0 \text{ m}$ [1]
$\Delta GPE = mgh = 5.0 \times 9.81 \times 5.0$ [1]
$\Delta GPE = 245.25 \text{ J}$ [1]
Answer: $245 \text{ J}$
[3]

(c) Explanation.
Answer:
Work is done against friction as well as gravity [1].
The difference between work done and GPE gain is the energy dissipated as heat due to friction [1].
[2]

18. (a) Minimum speed at top.
Answer:
At the top, forces acting downwards are Tension $T$ and Weight $mg$ .
Resultant force provides centripetal acceleration: $T + mg = \frac{mv^2}{r}$ [1]
For minimum speed, tension $T \ge 0$ . Limiting case $T=0$ [1].
$mg = \frac{mv_{min}^2}{r}$ [1]
$v_{min}^2 = gr \Rightarrow v_{min} = \sqrt{gr}$ [1]
[4]

(b) Tension at bottom.
Answer:
At bottom, forces are Tension $T$ (up) and Weight $mg$ (down).
$T - mg = \frac{mv^2}{r}$ [1]
Given $v = \sqrt{5gr}$ , so $v^2 = 5gr$ [1]
$T - mg = \frac{m(5gr)}{r} = 5mg$ [1]
$T = 5mg + mg = 6mg$ [1]
[4]

19. (a) Gravitational force.
Answer:
Distance between centers of the two stars is $R + R = 2R$ [1].
$F = \frac{G M_1 M_2}{d^2} = \frac{G M M}{(2R)^2} = \frac{GM^2}{4R^2}$ [1]
[2]

(b) Orbital period.
Answer:
Gravitational force provides centripetal force for circular motion of radius $R$ :
$\frac{GM^2}{4R^2} = M R \omega^2$ [1]
$\frac{GM}{4R^2} = R \left(\frac{2\pi}{T}\right)^2$ [1]
$\frac{GM}{4R^3} = \frac{4\pi^2}{T^2}$ [1]
$T^2 = \frac{16\pi^2 R^3}{GM} \Rightarrow T = 4\pi \sqrt{\frac{R^3}{GM}}$ [1]
[4]

20. (a) Newton’s Second Law.
Answer:
The resultant force acting on an object is equal to the rate of change of its momentum [1].
[1]

(b) Why acceleration increases.
Answer:
Thrust is constant, so the upward force is constant [1].
As fuel is ejected, the mass $M(t)$ of the rocket decreases [1].
Since $a = \frac{F_{net}}{M}$ , as $M$ decreases, $a$ increases [1].
[3]

(c) Equation for acceleration.
Answer:
Resultant force $F_{net} = \text{Thrust} - \text{Weight} = F - M(t)g$ [1]
$a(t) = \frac{F - M(t)g}{M(t)}$ or $\frac{F}{M(t)} - g$ [1]
[2]