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A Level H1 Physics Practice Paper 1
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Questions
TuitionGoWhere Practice Paper - Physics H1 A-Level
TuitionGoWhere Secondary School (AI)
Subject: Physics H1
Level: A-Level
Paper: PRACTICE Paper 2
Duration: 2 hours
Total Marks: 80
Name: _________________ Class: _________________ Date: _________________
Instructions to Candidates
- Answer ALL questions
- Write your answers in the spaces provided
- Show all working clearly for calculation questions
- Use appropriate units in your final answers
- The number of marks is given in brackets [ ] at the end of each question or part question
- Take g = 9.8 m s⁻²
Physical Constants:
- Speed of light in vacuum, c = 3.00 × 10⁸ m s⁻¹
- Planck constant, h = 6.63 × 10⁻³⁴ J s
- Electronic charge, e = 1.60 × 10⁻¹⁹ C
Section A [40 marks]
1. A projectile is launched horizontally from the top of a cliff with an initial speed of 25 m s⁻¹. The cliff is 45 m high.
(a) Calculate the time taken for the projectile to reach the ground. [3]
Time = _________________ s
(b) Calculate the horizontal distance travelled by the projectile. [2]
Horizontal distance = _________________ m
(c) Determine the speed of the projectile just before it hits the ground. [4]
Speed = _________________ m s⁻¹
2. Two ice hockey pucks A and B move towards each other on a frictionless horizontal surface. Puck A has mass 0.50 kg and velocity 8.0 m s⁻¹ to the right. Puck B has mass 0.30 kg and velocity 6.0 m s⁻¹ to the left.
The pucks collide and stick together.
(a) State the principle of conservation of momentum. [2]
(b) Calculate the velocity of the combined pucks immediately after collision. [4]
Velocity = _________________ m s⁻¹
(c) Calculate the kinetic energy lost in the collision. [5]
Kinetic energy lost = _________________ J
3. A uniform rod AB of length 2.4 m and weight 80 N is pivoted at point P, which is 0.8 m from end A. A force F is applied vertically downward at end B to keep the rod in horizontal equilibrium.
A ←—— 0.8 m ——→ P ←—— 1.6 m ——→ B
| | ↓
↑ F
Pivot
(a) Draw a diagram showing all forces acting on the rod. Label each force clearly. [3]
(b) Calculate the magnitude of force F. [3]
Force F = _________________ N
(c) Calculate the magnitude and direction of the reaction force at the pivot P. [4]
Reaction force = _________________ N, direction: _________________
4. In a potential divider circuit, a 12 V battery with internal resistance 2.0 Ω is connected to two resistors of 8.0 Ω and 4.0 Ω in series. A voltmeter is connected across the 4.0 Ω resistor.
┌─── 8.0 Ω ───┬─── 4.0 Ω ───┐
│ │ │
┌───┴───┐ │ │
│ 12 V │ │ │
│ 2.0 Ω │ V │
└───┬───┘ │ │
└──────────────┴──────────────┘
(a) Calculate the current in the circuit. [3]
Current = _________________ A
(b) Calculate the reading on the voltmeter. [2]
Voltmeter reading = _________________ V
(c) Explain why the voltmeter reading is less than the expected value from a simple potential divider calculation. [3]
Section B [40 marks]
5. A student investigates the photoelectric effect using a zinc plate and ultraviolet light of different wavelengths. The work function of zinc is 4.3 eV.
(a) Define work function. [2]
(b) Calculate the threshold wavelength for zinc. [4]
Threshold wavelength = _________________ m
(c) Light of wavelength 250 nm is incident on the zinc plate. Calculate:
(i) The maximum kinetic energy of the emitted photoelectrons [3]
Maximum kinetic energy = _________________ eV
(ii) The stopping potential [2]
Stopping potential = _________________ V
(d) The intensity of the incident light is now doubled while keeping the wavelength constant at 250 nm. State and explain the effect this has on:
(i) The maximum kinetic energy of the photoelectrons [2]
(ii) The number of photoelectrons emitted per second [2]
6. A conducting rod of mass 15 g and length 20 cm rests on two smooth parallel horizontal rails in a uniform magnetic field of flux density 0.25 T. The magnetic field is perpendicular to the plane of the rails.
Rail ═══════════════════════════════
┃ ┃
┃ Rod ┃ ← B (into page)
┃ ┃
Rail ═══════════════════════════════
A current of 0.80 A flows through the rod.
(a) Calculate the magnetic force on the rod. [2]
Magnetic force = _________________ N
(b) The rails are now tilted at an angle θ to the horizontal. The rod remains in equilibrium under the action of its weight and the magnetic force.
Calculate the angle θ. [4]
Angle θ = _________________°
(c) Suggest why the rails need to be smooth for this experiment. [2]
7. In a double-slit experiment, coherent light of wavelength 650 nm passes through two parallel slits separated by 0.40 mm. The interference pattern is observed on a screen 2.5 m away from the slits.
(a) Calculate the fringe spacing on the screen. [3]
Fringe spacing = _________________ m
(b) The experiment is repeated using light of a different wavelength. The fringe spacing is now measured to be 3.2 mm.
Calculate the wavelength of this light. [3]
Wavelength = _________________ m
(c) State two conditions necessary for observable interference fringes to be formed. [2]
(i) ___________________________________________________________________
(ii) ___________________________________________________________________
(d) Explain why interference fringes are not normally observed when light from two separate lamps illuminates the double slit. [3]
8. A lamp is rated at 12 V, 36 W when operating normally.
(a) Calculate the resistance of the lamp when operating normally. [2]
Resistance = _________________ Ω
(b) Three identical lamps are connected in series across a 36 V supply.
(i) Calculate the current through each lamp. [3]
Current = _________________ A
(ii) Calculate the power dissipated in each lamp. [2]
Power = _________________ W
(c) Comment on the brightness of each lamp compared to normal operation. [2]
END OF PAPER
Answers
TuitionGoWhere Practice Paper - Physics H1 A-Level (Answer Key)
Section A [40 marks]
1. Projectile motion [9 marks]
(a) Time to reach ground [3] Working: Using s = ut + ½at² for vertical motion 45 = 0 + ½ × 9.8 × t² [M1] t² = 90/9.8 = 9.18 [M1] t = 3.03 s [A1]
Answer: Time = 3.0 s
(b) Horizontal distance [2] Working: Horizontal velocity remains constant at 25 m s⁻¹ Distance = speed × time = 25 × 3.03 = 75.8 m [M1][A1]
Answer: Horizontal distance = 76 m
(c) Speed just before impact [4] Working: Horizontal component: vₓ = 25 m s⁻¹ [M1] Vertical component: vᵧ = u + at = 0 + 9.8 × 3.03 = 29.7 m s⁻¹ [M1] Resultant speed = √(vₓ² + vᵧ²) = √(25² + 29.7²) = √(625 + 882) [M1] = √1507 = 38.8 m s⁻¹ [A1]
Answer: Speed = 39 m s⁻¹
2. Collision of pucks [11 marks]
(a) Conservation of momentum [2] Answer: In a closed/isolated system, the total momentum remains constant [B1] provided no external forces act on the system [B1]
(b) Velocity after collision [4] Working: Taking rightward as positive: Initial momentum of A = 0.50 × 8.0 = +4.0 kg m s⁻¹ [M1] Initial momentum of B = 0.30 × (-6.0) = -1.8 kg m s⁻¹ [M1] Total initial momentum = 4.0 - 1.8 = 2.2 kg m s⁻¹ Total mass after collision = 0.50 + 0.30 = 0.80 kg [M1] Final velocity = 2.2/0.80 = 2.75 m s⁻¹ (rightward) [A1]
Answer: Velocity = 2.8 m s⁻¹ to the right
(c) Kinetic energy lost [5] Working: Initial KE of A = ½ × 0.50 × 8.0² = 16.0 J [M1] Initial KE of B = ½ × 0.30 × 6.0² = 5.4 J [M1] Total initial KE = 16.0 + 5.4 = 21.4 J [M1] Final KE = ½ × 0.80 × 2.75² = 3.03 J [M1] KE lost = 21.4 - 3.03 = 18.4 J [A1]
Answer: Kinetic energy lost = 18 J
3. Rod equilibrium [10 marks]
(a) Force diagram [3] Answer: Should show:
- Weight of rod (80 N downward at center, 1.2 m from A) [B1]
- Applied force F (downward at B) [B1]
- Reaction force at P (upward) [B1]
(b) Magnitude of force F [3] Working: Taking moments about P: Clockwise: Weight of rod × 0.4 = 80 × 0.4 = 32 N m [M1] Anticlockwise: F × 1.6 [M1] For equilibrium: F × 1.6 = 32 F = 32/1.6 = 20 N [A1]
Answer: Force F = 20 N
(c) Reaction at pivot [4] Working: Vertical equilibrium: R = Weight + F = 80 + 20 = 100 N [M2] Direction: upward [M1] (No horizontal forces, so reaction is purely vertical) [M1]
Answer: Reaction force = 100 N, direction: upward
4. Potential divider circuit [10 marks]
(a) Current calculation [3] Working: Total resistance = 2.0 + 8.0 + 4.0 = 14.0 Ω [M1] Current = EMF/Total resistance = 12/14.0 [M1] = 0.857 A [A1]
Answer: Current = 0.86 A
(b) Voltmeter reading [2] Working: Voltage across 4.0 Ω resistor = IR = 0.857 × 4.0 [M1] = 3.43 V [A1]
Answer: Voltmeter reading = 3.4 V
(c) Explanation [3] Answer: The internal resistance of the battery acts as an additional series resistor [B1] This reduces the terminal voltage available to the external circuit [B1] Therefore less voltage appears across the 4.0 Ω resistor than in an ideal potential divider [B1]
Section B [40 marks]
5. Photoelectric effect [15 marks]
(a) Work function definition [2] Answer: The minimum energy required to remove an electron from the surface of a material [B1] (or the minimum photon energy needed to cause photoemission) [B1]
(b) Threshold wavelength [4] Working: At threshold: hf₀ = Φ, so hc/λ₀ = Φ [M1] λ₀ = hc/Φ [M1] Convert work function: Φ = 4.3 × 1.60 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J [M1] λ₀ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(6.88 × 10⁻¹⁹) = 2.89 × 10⁻⁷ m [A1]
Answer: Threshold wavelength = 2.9 × 10⁻⁷ m
(c)(i) Maximum kinetic energy [3] Working: Photon energy = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(250 × 10⁻⁹) [M1] = 7.96 × 10⁻¹⁹ J = 4.97 eV [M1] KEₘₐₓ = hf - Φ = 4.97 - 4.3 = 0.67 eV [A1]
Answer: Maximum kinetic energy = 0.67 eV
(c)(ii) Stopping potential [2] Working: eVₛ = KEₘₐₓ [M1] Vₛ = 0.67 V [A1]
Answer: Stopping potential = 0.67 V
(d)(i) Effect on maximum KE [2] Answer: No change in maximum kinetic energy [B1] Maximum KE depends only on photon frequency/wavelength, not intensity [B1]
(d)(ii) Effect on number of photoelectrons [2] Answer: Number of photoelectrons doubles [B1] Intensity determines the number of photons per second, hence photoelectrons per second [B1]
6. Conducting rod in magnetic field [8 marks]
(a) Magnetic force [2] Working: F = BIL = 0.25 × 0.80 × 0.20 [M1] = 0.040 N [A1]
Answer: Magnetic force = 0.040 N
(b) Angle calculation [4] Working: For equilibrium on inclined rails: Component of weight down the slope = mg sin θ [M1] = 0.015 × 9.8 × sin θ = 0.147 sin θ N [M1] This equals magnetic force: 0.147 sin θ = 0.040 [M1] sin θ = 0.040/0.147 = 0.272 θ = 15.8° [A1]
Answer: Angle θ = 16°
(c) Why rails must be smooth [2] Answer: To eliminate friction forces [B1] So that only weight and magnetic force act on the rod [B1]
7. Double-slit interference [11 marks]
(a) Fringe spacing [3] Working: β = λD/a [M1] = (650 × 10⁻⁹ × 2.5)/(0.40 × 10⁻³) [M1] = 4.06 × 10⁻³ m [A1]
Answer: Fringe spacing = 4.1 × 10⁻³ m
(b) New wavelength [3] Working: β = λD/a, so λ = βa/D [M1] λ = (3.2 × 10⁻³ × 0.40 × 10⁻³)/2.5 [M1] = 5.12 × 10⁻⁷ m [A1]
Answer: Wavelength = 5.1 × 10⁻⁷ m
(c) Conditions for interference [2] Answer: (i) Sources must be coherent (constant phase difference) [B1] (ii) Sources must have similar amplitudes/intensities [B1]
Accept: monochromatic light, sources close together
(d) Why separate lamps don't interfere [3] Answer: Light from separate lamps is incoherent [B1] The phase difference between the two sources changes randomly [B1] No stable interference pattern can be observed [B1]
8. Lamp circuits [6 marks]
(a) Lamp resistance [2] Working: R = V²/P = 12²/36 [M1] = 4.0 Ω [A1]
Answer: Resistance = 4.0 Ω
(b)(i) Current in series circuit [3] Working: Total resistance = 3 × 4.0 = 12.0 Ω [M1] Current = V/R = 36/12.0 [M1] = 3.0 A [A1]
Answer: Current = 3.0 A
(b)(ii) Power in each lamp [2] Working: P = I²R = 3.0² × 4.0 [M1] = 36 W [A1]
Answer: Power = 36 W
(c) Brightness comparison [2] Answer: Each lamp operates at normal power (36 W) [B1] so each lamp has normal brightness [B1]
Total: 80 marks
Grade Boundaries (Indicative):
- A: 68-80 marks (85-100%)
- B: 60-67 marks (75-84%)
- C: 52-59 marks (65-74%)
- D: 44-51 marks (55-64%)
- E: 36-43 marks (45-54%)