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A Level H1 Physics Practice Paper 1
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TuitionGoWhere Practice Paper - Physics H1 A-Level (Marking Scheme)
Total Marks: 80
Section A: Mechanics [35 marks]
1. Momentum conservation investigation
(a) State the principle of conservation of linear momentum. [2]
Answer: In a closed system [1] / when no external forces act, the total momentum remains constant [1].
Alternative: Total momentum before collision = total momentum after collision, provided no external forces act [2].
(b)(i) Calculate velocity after inelastic collision [4]
Working: Taking right as positive direction: Initial momentum of P = 1.5 × 2.4 = 3.6 kg m s⁻¹ [1] Initial momentum of Q = 2.0 × (-1.8) = -3.6 kg m s⁻¹ [1] Total initial momentum = 3.6 + (-3.6) = 0 kg m s⁻¹ [1]
After collision: (1.5 + 2.0)v = 0 Therefore v = 0 m s⁻¹ [1]
(b)(ii) Show collision is inelastic [4]
Working: Initial KE = ½(1.5)(2.4)² + ½(2.0)(1.8)² [1] = ½(1.5)(5.76) + ½(2.0)(3.24) = 4.32 + 3.24 = 7.56 J [1]
Final KE = ½(3.5)(0)² = 0 J [1]
Since initial KE ≠ final KE, the collision is inelastic [1]
(c) Elastic collision calculation [5]
Working: For elastic collision, both momentum and kinetic energy are conserved.
Conservation of momentum: 1.5 × 3.0 + 2.0 × 0 = 1.5v₁ + 2.0v₂ 4.5 = 1.5v₁ + 2.0v₂ ... (1) [1]
Conservation of kinetic energy: ½(1.5)(3.0)² + 0 = ½(1.5)v₁² + ½(2.0)v₂² 6.75 = 0.75v₁² + v₂² ... (2) [1]
Using elastic collision formula or solving simultaneously: v₁ = (m₁ - m₂)u₁/(m₁ + m₂) = (1.5 - 2.0)(3.0)/(1.5 + 2.0) = -0.43 m s⁻¹ [1] v₂ = 2m₁u₁/(m₁ + m₂) = 2(1.5)(3.0)/(1.5 + 2.0) = 2.6 m s⁻¹ [1]
Answer: Trolley P: 0.43 m s⁻¹ to the left, Trolley Q: 2.6 m s⁻¹ to the right [1]
2. Beam equilibrium problem
(a) Force diagram [3] Diagram should show:
- Weight of beam (400 N downward) at 3.0 m from A [1]
- Weight W (downward) at 6.0 m from A [1]
- Tension T (900 N upward) at 4.0 m from A [1]
- Reaction at hinge A (components or resultant) [1]
(b)(i) Calculate weight W [4]
Working: Taking moments about A (clockwise positive): 400 × 3.0 + W × 6.0 - 900 × 4.0 = 0 [1] 1200 + 6W - 3600 = 0 [1] 6W = 2400 [1] W = 400 N [1]
(b)(ii) Reaction at hinge A [4]
Working: Vertical equilibrium: Rᵧ + 900 = 400 + 400 Rᵧ = 800 - 900 = -100 N (downward) [1]
Horizontal equilibrium: Rₓ = 0 [1]
Magnitude = 100 N [1] Direction = vertically downward [1]
3. Projectile motion
(a) Initial velocity components [2]
Answer: Horizontal: 30 cos 45° = 30 × 0.707 = 21.2 m s⁻¹ [1] Vertical: 30 sin 45° = 30 × 0.707 = 21.2 m s⁻¹ [1]
(b)(i) Maximum height [3]
Working: At maximum height, vᵧ = 0 Using v² = u² + 2as: 0 = (21.2)² + 2(-9.8)h [1] 0 = 449.44 - 19.6h [1] h = 449.44/19.6 = 22.9 m [1]
(b)(ii) Horizontal range [3]
Working: Time of flight: using s = ut + ½at² for vertical motion 0 = 21.2t - ½(9.8)t² [1] t = 2 × 21.2/9.8 = 4.33 s [1] Range = horizontal velocity × time = 21.2 × 4.33 = 91.8 m [1]
(b)(iii) Speed on return to ground [2]
Working: By symmetry or energy conservation, speed = initial speed [1] Speed = 30 m s⁻¹ [1]
(c) Velocity-time graphs [3]
(i) Horizontal velocity vs time: horizontal line at 21.2 m s⁻¹ [1] (ii) Vertical velocity vs time: straight line from +21.2 m s⁻¹ to -21.2 m s⁻¹ [2]
Section B: Electricity and Magnetism [25 marks]
4. Battery circuits
(a)(i) Current calculation [2]
Working: Total resistance = 1.5 + 10.0 = 11.5 Ω [1] Current = 12.0/11.5 = 1.04 A [1]
(a)(ii) Terminal voltage [2]
Working: Terminal voltage = EMF - I × r = 12.0 - 1.04 × 1.5 [1] = 12.0 - 1.56 = 10.4 V [1]
(a)(iii) Power in external resistor [2]
Working: P = I²R = (1.04)² × 10.0 [1] = 10.8 W [1]
(b) Batteries in parallel [4]
Working: Two identical batteries in parallel: EMF = 12.0 V, internal resistance = 1.5/2 = 0.75 Ω [1] Total resistance = 0.75 + 10.0 = 10.75 Ω [1] Current = 12.0/10.75 = 1.12 A [1] Answer: 1.12 A [1]
(c) Advantages of parallel batteries [2]
Answer: Lower internal resistance [1], can supply more current / longer battery life [1]
5. Magnetic forces
(a) Force magnitude [2]
Working: F = BIL = 0.15 × 4.0 × 0.25 [1] = 0.15 N [1]
(b) Force direction [2]
Answer: Using right-hand rule: current east, field up, force is north [2]
(c) Force at 30° angle [3]
Working: F = BIL sin θ = 0.15 × 4.0 × 0.25 × sin 30° [1] = 0.15 × 0.5 [1] = 0.075 N [1]
(d) Force between parallel wires [3]
Working: F/L = μ₀I₁I₂/(2πd) [1] = (4π × 10⁻⁷ × 2.0 × 2.0)/(2π × 0.05) [1] = 1.6 × 10⁻⁵ N m⁻¹ [1]
Section C: Waves and Modern Physics [20 marks]
6. Double-slit interference
(a) Fringe spacing [3]
Working: λ = 650 × 10⁻⁹ m, a = 0.80 × 10⁻³ m, D = 2.5 m [1] Fringe spacing = λD/a = (650 × 10⁻⁹ × 2.5)/(0.80 × 10⁻³) [1] = 2.0 × 10⁻³ m = 2.0 mm [1]
(b) Why slits must be narrow and close [3]
Answer: Narrow slits ensure good diffraction [1], close slits give reasonable fringe spacing [1], both needed for clear interference pattern [1]
(c) Effect of blue light [3]
Answer: Fringe spacing decreases [1] because fringe spacing is proportional to wavelength [1] and blue light has shorter wavelength than red [1]
7. Photoelectric effect
(a) Photon energy [2]
Working: E = hf = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(400 × 10⁻⁹) [1] = 4.97 × 10⁻¹⁹ J [1]
(b) Show photoelectrons are emitted [2]
Working: Photon energy = 4.97 × 10⁻¹⁹ J = 3.11 eV [1] Since 3.11 eV > 2.8 eV (work function), photoelectrons are emitted [1]
(c)(i) Maximum kinetic energy [3]
Working: KEₘₐₓ = hf - Φ = 3.11 - 2.8 = 0.31 eV [1] = 0.31 × 1.60 × 10⁻¹⁹ [1] = 5.0 × 10⁻²⁰ J [1]
(c)(ii) Stopping potential [2]
Working: eVₛ = KEₘₐₓ [1] Vₛ = 0.31 V [1]
(d) Effect of intensity [3]
Answer: Intensity affects number of photons, not energy per photon [1]. Maximum KE depends only on photon energy (frequency) [1]. Higher intensity gives more photoelectrons but same maximum KE [1].