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A Level H2 Physics Practice Paper 2

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A Level H2 Physics From Real Exams Generated by Qwen3.6 Plus Updated 2026-08-17
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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:
ω=2πf=2π(2.5)=5π rad s−1\omega = 2\pi f = 2\pi(2.5) = 5\pi \text{ rad s}^{-1} [1]
amax=ω2x0a_{max} = \omega^2 x_0 [1]
amax=(5π)2×0.04=25π2×0.04=π2≈9.87 m s−2a_{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 m s−29.9 \text{ m s}^{-2} (2 s.f.)
[3]

3. State two precautions to improve accuracy of tt.
Answer:

  1. Use an electronic timer/light gate to eliminate human reaction time error [1].
  2. 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 hh 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 =mgsin⁡θ= mg \sin \theta [1]
Ff=2.0×9.81×sin⁡30∘=9.81 NF_f = 2.0 \times 9.81 \times \sin 30^\circ = 9.81 \text{ N} [1]
Answer: 9.8 N9.8 \text{ N}
[3]

7. Calculate final speed of electron.
Answer:
Gain in KE = Loss in EPE
12mv2=eV\frac{1}{2}mv^2 = eV [1]
v=2eVmv = \sqrt{\frac{2eV}{m}}
v=2×1.60×10−19×50009.11×10−31v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 5000}{9.11 \times 10^{-31}}} [1]
v=1.756×1015=4.19×107 m s−1v = \sqrt{1.756 \times 10^{15}} = 4.19 \times 10^7 \text{ m s}^{-1} [1]
Answer: 4.2×107 m s−14.2 \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: ε=−d(NΦ)dt\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=Fx=100.05=200 N m−1k = \frac{F}{x} = \frac{10}{0.05} = 200 \text{ N m}^{-1} [1]
E=12kx2E = \frac{1}{2}kx^2 [1]
E=12(200)(0.05)2=100×0.0025=0.25 JE = \frac{1}{2}(200)(0.05)^2 = 100 \times 0.0025 = 0.25 \text{ J} [1]
Answer: 0.25 J0.25 \text{ J}
[3]

Section B: Data Analysis and Application

11. (a) Determine gg.
Answer:
Formula: T=2πLg⇒T2=4π2gLT = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g} L [1]
Gradient =4π2g= \frac{4\pi^2}{g} [1]
4.0=4π2g⇒g=4π24.0=π2≈9.87 m s−24.0 = \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 m s−29.9 \text{ m s}^{-2}
[3]

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

12. (a) Common velocity.
Answer:
Conservation of momentum: mAuA+mBuB=(mA+mB)vm_A u_A + m_B u_B = (m_A + m_B)v [1]
(0.50)(2.0)+0=(0.50+0.30)v(0.50)(2.0) + 0 = (0.50 + 0.30)v
1.0=0.80v1.0 = 0.80 v
v=1.25 m s−1v = 1.25 \text{ m s}^{-1} [1]
Answer: 1.3 m s−11.3 \text{ m s}^{-1} (2 s.f.) [1]
[3]

(b) Elastic or inelastic?
Answer:
Initial KE =12(0.50)(2.0)2=1.0 J= \frac{1}{2}(0.50)(2.0)^2 = 1.0 \text{ J} [1]
Final KE =12(0.80)(1.25)2=0.625 J= \frac{1}{2}(0.80)(1.25)^2 = 0.625 \text{ J} [1]
KE is not conserved (1.0≠0.6251.0 \neq 0.625), so the collision is inelastic [1].
[3]

13. (a) Calculate force FF.
Answer:
Taking moments about the pivot:
Clockwise moments = Anticlockwise moments [1]
(100×0.50)+(50×1.0)=F×2.0(100 \times 0.50) + (50 \times 1.0) = F \times 2.0 (Weight acts at center, 1.0m from pivot) [1]
50+50=2F50 + 50 = 2F
100=2F⇒F=50 N100 = 2F \Rightarrow F = 50 \text{ N} [1]
Answer: 50 N50 \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: Fc=mv2rF_c = \frac{mv^2}{r} [1]
Max friction Fmax=μmgF_{max} = \mu mg
μmg=mv2r⇒v=μgr\mu mg = \frac{mv^2}{r} \Rightarrow v = \sqrt{\mu gr} [1]
v=0.80×9.81×50=392.4=19.8 m s−1v = \sqrt{0.80 \times 9.81 \times 50} = \sqrt{392.4} = 19.8 \text{ m s}^{-1} [1]
Answer: 20 m s−120 \text{ m s}^{-1} (2 s.f.)
[3]

(b) Effect of doubling radius.
Answer:
Fc=mv2rF_c = \frac{mv^2}{r}. If rr doubles and vv is constant, FcF_c is halved [1].
Therefore, the required centripetal force decreases [1].
[2]

15. (a) Energy of photon.
Answer:
E=hcλE = \frac{hc}{\lambda} [1]
E=6.63×10−34×3.00×108400×10−9E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{400 \times 10^{-9}} [1]
E=4.97×10−19 JE = 4.97 \times 10^{-19} \text{ J} [1]
Answer: 5.0×10−19 J5.0 \times 10^{-19} \text{ J}
[3]

(b) Maximum KE in eV.
Answer:
Work function Φ=2.0 eV=2.0×1.60×10−19=3.20×10−19 J\Phi = 2.0 \text{ eV} = 2.0 \times 1.60 \times 10^{-19} = 3.20 \times 10^{-19} \text{ J} [1]
KEmax=E−Φ=4.97×10−19−3.20×10−19=1.77×10−19 JKE_{max} = E - \Phi = 4.97 \times 10^{-19} - 3.20 \times 10^{-19} = 1.77 \times 10^{-19} \text{ J} [1]
In eV: 1.77×10−191.60×10−19=1.11 eV\frac{1.77 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.11 \text{ eV} [1]
Answer: 1.1 eV1.1 \text{ eV}
[3]

Section C: Extended Response

16. (a) Maximum height.
Answer:
Vertical component of initial velocity uy=20sin⁡60∘=17.32 m s−1u_y = 20 \sin 60^\circ = 17.32 \text{ m s}^{-1} [1]
At max height, vy=0v_y = 0. Using v2=u2+2asv^2 = u^2 + 2as:
0=(17.32)2+2(−9.81)h0 = (17.32)^2 + 2(-9.81)h [1]
h=30019.62=15.29 mh = \frac{300}{19.62} = 15.29 \text{ m} [2]
Answer: 15 m15 \text{ m} (2 s.f.)
[4]

(b) Horizontal range.
Answer:
Time to reach max height: v=u+at⇒0=17.32−9.81t⇒t=1.765 sv = u + at \Rightarrow 0 = 17.32 - 9.81t \Rightarrow t = 1.765 \text{ s} [1]
Total time of flight T=2t=3.53 sT = 2t = 3.53 \text{ s} [1]
Horizontal velocity ux=20cos⁡60∘=10 m s−1u_x = 20 \cos 60^\circ = 10 \text{ m s}^{-1} [1]
Range =ux×T=10×3.53=35.3 m= u_x \times T = 10 \times 3.53 = 35.3 \text{ m} [1]
Answer: 35 m35 \text{ m}
[4]

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

(b) Gain in GPE.
Answer:
Vertical height gained h=dsin⁡30∘=10×0.5=5.0 mh = d \sin 30^\circ = 10 \times 0.5 = 5.0 \text{ m} [1]
ΔGPE=mgh=5.0×9.81×5.0\Delta GPE = mgh = 5.0 \times 9.81 \times 5.0 [1]
ΔGPE=245.25 J\Delta GPE = 245.25 \text{ J} [1]
Answer: 245 J245 \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 TT and Weight mgmg.
Resultant force provides centripetal acceleration: T+mg=mv2rT + mg = \frac{mv^2}{r} [1]
For minimum speed, tension T≥0T \ge 0. Limiting case T=0T=0 [1].
mg=mvmin2rmg = \frac{mv_{min}^2}{r} [1]
vmin2=gr⇒vmin=grv_{min}^2 = gr \Rightarrow v_{min} = \sqrt{gr} [1]
[4]

(b) Tension at bottom.
Answer:
At bottom, forces are Tension TT (up) and Weight mgmg (down).
T−mg=mv2rT - mg = \frac{mv^2}{r} [1]
Given v=5grv = \sqrt{5gr}, so v2=5grv^2 = 5gr [1]
T−mg=m(5gr)r=5mgT - mg = \frac{m(5gr)}{r} = 5mg [1]
T=5mg+mg=6mgT = 5mg + mg = 6mg [1]
[4]

19. (a) Gravitational force.
Answer:
Distance between centers of the two stars is R+R=2RR + R = 2R [1].
F=GM1M2d2=GMM(2R)2=GM24R2F = \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 RR:
GM24R2=MRω2\frac{GM^2}{4R^2} = M R \omega^2 [1]
GM4R2=R(2πT)2\frac{GM}{4R^2} = R \left(\frac{2\pi}{T}\right)^2 [1]
GM4R3=4π2T2\frac{GM}{4R^3} = \frac{4\pi^2}{T^2} [1]
T2=16π2R3GM⇒T=4πR3GMT^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)M(t) of the rocket decreases [1].
Since a=FnetMa = \frac{F_{net}}{M}, as MM decreases, aa increases [1].
[3]

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