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A Level H1 Physics Practice Paper 1

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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

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

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

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

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

(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 = _________________°

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

(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

END OF PAPER

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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%)