A Level H1 Physics Practice Paper 3

TuitionGoWhere Practice Paper - Physics H1 A-Level

ANSWER KEY AND MARKING SCHEME

Paper: Practice Paper 3 (Version 3 of 5) Total Marks: 80

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Section A: Structured Questions [50 marks]

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Question 1: Momentum and Kinetic Energy [4 marks]

(a) Calculate the velocity of the trolley. [1 mark]

• p = mv
• v = p/m = 15.0 / 2.5 = 6.0 m s⁻¹ [A1]

(b) Calculate the kinetic energy of the trolley. [1 mark]

• KE = ½mv² = ½ × 2.5 × (6.0)² = 45 J [A1]
• Accept: KE = p²/2m = (15.0)²/(2 × 2.5) = 45 J

(c) Calculate the velocity of the combined trolleys after collision. [2 marks]

• Conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v [M1]
• 2.5 × 6.0 + 1.5 × 0 = (2.5 + 1.5)v
• 15.0 = 4.0v
• v = 3.75 m s⁻¹ [A1]

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Question 2: Conservation of Linear Momentum [3 marks]

(a) State the principle of conservation of linear momentum. [2 marks]

• The total momentum of a closed/isolated system remains constant [B1]
• provided no external forces act on the system / in the absence of external forces [B1]
• Accept: "In a closed system, total momentum before = total momentum after, when no external forces act"

(b) Explain why the principle applies to the collision in Question 1(c). [1 mark]

• The track is frictionless, so no external horizontal forces act on the system during the collision [B1]
• Accept: "The system is isolated from external horizontal forces" / "Only internal forces act between the trolleys"

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Question 3: Free-Body Diagram and Equilibrium [5 marks]

(a) Draw and label all forces acting on the plank. [2 marks]

• Weight of plank (120 N) acting downwards at centre (2.0 m from A) [B1]
• Weight of person (650 N) acting downwards at distance x from A
• Reaction force at P (R_P) acting upwards at A
• Reaction force at Q (R_Q) acting upwards at 3.0 m from A
• All four forces correctly drawn and labelled [B1]

(b) Write equation for sum of vertical forces in equilibrium. [1 mark]

• R_P + R_Q = 120 + 650 = 770 N [B1]
• Accept: ΣF_y = 0 → R_P + R_Q - 120 - 650 = 0

(c) Derive expression for reaction force at Q in terms of x. [2 marks]

• Taking moments about P (clockwise positive): [M1]
• (120 × 2.0) + (650 × x) = R_Q × 3.0
• 240 + 650x = 3.0 R_Q [M1]
• R_Q = (240 + 650x) / 3.0 [A1]
• Accept equivalent algebraic manipulation

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Question 4: Equilibrium Calculation [4 marks]

(a) Calculate reaction force at Q when x = 2.5 m. [2 marks]

• R_Q = (240 + 650 × 2.5) / 3.0 [M1]
• R_Q = (240 + 1625) / 3.0 = 1865 / 3.0
• R_Q = 622 N (or 620 N to 2 s.f.) [A1]

(b) Determine maximum distance x before plank tips. [2 marks]

• Plank tips when R_P = 0 (plank loses contact with support P) [M1]
• Taking moments about Q: 120 × 1.0 = 650 × (x - 3.0)
• 120 = 650(x - 3.0)
• x - 3.0 = 120/650 = 0.185
• x = 3.185 m ≈ 3.2 m [A1]
• Accept: x = 3.18 m or 3.2 m

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Question 5: Motion with Air Resistance [4 marks]

(a) Sketch velocity-time graph with terminal velocity labelled. [2 marks]

• Graph starts at origin, curves upward with decreasing gradient [B1]
• Approaches horizontal asymptote (terminal velocity) clearly labelled [B1]
• Graph should be concave down (gradient decreases with time)

(b) Explain shape of graph in terms of forces. [2 marks]

• Initially, only weight acts, acceleration = g (steep gradient) [B1]
• As speed increases, air resistance increases, reducing net downward force
• Net force = weight - air resistance decreases, so acceleration decreases
• Eventually, air resistance = weight, net force = 0, speed becomes constant (terminal velocity) [B1]

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Question 6: Work, Energy, and Power [5 marks]

(a) Calculate tension in the cable. [1 mark]

• Constant speed → net force = 0
• T = mg = 850 × 9.81 = 8340 N (or 8.34 × 10³ N) [A1]

(b) Calculate work done lifting through 12.0 m. [2 marks]

• W = Fd = T × d [M1]
• W = 8340 × 12.0 = 100,080 J ≈ 1.00 × 10⁵ J [A1]
• Accept: W = mgh = 850 × 9.81 × 12.0 = 100,062 J

(c) Calculate power output during lift. [2 marks]

• P = W/t or P = Fv [M1]
• t = d/v = 12.0/0.40 = 30 s
• P = 100,080/30 = 3336 W ≈ 3.34 × 10³ W [A1]
• Alternative: P = Fv = 8340 × 0.40 = 3336 W

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Question 7: Projectile Motion [6 marks]

(a) Calculate time to reach ground. [2 marks]

• Vertical motion: s = ut + ½at², u_y = 0 [M1]
• 45.0 = 0 + ½ × 9.81 × t²
• t² = 90.0/9.81 = 9.174
• t = 3.03 s [A1]

(b) Calculate horizontal distance travelled. [1 mark]

• Horizontal: s_x = u_x × t = 15.0 × 3.03 = 45.5 m [A1]

(c) Calculate magnitude and direction of velocity just before impact. [3 marks]

• v_x = 15.0 m s⁻¹ (constant) [M1]
• v_y = u_y + at = 0 + 9.81 × 3.03 = 29.7 m s⁻¹ [M1]
• Magnitude: v = √(15.0² + 29.7²) = √(225 + 882) = √1107 = 33.3 m s⁻¹ [A1]
• Direction: θ = tan⁻¹(29.7/15.0) = tan⁻¹(1.98) = 63.2° below horizontal [A1]
• Accept: 63° below horizontal

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Question 8: Circuit Analysis [6 marks]

(a) Draw circuit diagram. [1 mark]

• Battery symbol with EMF labelled (12.0 V) [B1]
• Internal resistance (0.80 Ω) shown in series with battery
• External resistor (5.2 Ω) connected across terminals
• Correct circuit symbols and connections

(b) Calculate current in circuit. [2 marks]

• Total resistance: R_total = r + R = 0.80 + 5.2 = 6.0 Ω [M1]
• I = EMF / R_total = 12.0 / 6.0 = 2.0 A [A1]

(c) Calculate terminal potential difference. [1 mark]

• V = EMF - Ir = 12.0 - (2.0 × 0.80) = 12.0 - 1.6 = 10.4 V [A1]
• Alternative: V = IR = 2.0 × 5.2 = 10.4 V

(d) Calculate power dissipated in external resistor. [2 marks]

• P = I²R = (2.0)² × 5.2 [M1]
• P = 4.0 × 5.2 = 20.8 W [A1]
• Alternative: P = V²/R = (10.4)²/5.2 = 20.8 W or P = VI = 10.4 × 2.0 = 20.8 W

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Question 9: Potential Divider [5 marks]

(a) Draw circuit diagram. [1 mark]

• Two resistors in series across 9.0 V supply [B1]
• R₁ (3.0 kΩ) and R₂ (6.0 kΩ) clearly labelled
• Output voltage labelled across R₂

(b) Calculate current through resistors. [2 marks]

• R_total = 3.0 + 6.0 = 9.0 kΩ = 9000 Ω [M1]
• I = V / R_total = 9.0 / 9000 = 1.0 × 10⁻³ A = 1.0 mA [A1]

(c) Calculate output voltage across R₂. [2 marks]

• V_out = I × R₂ = (1.0 × 10⁻³) × 6000 [M1]
• V_out = 6.0 V [A1]
• Alternative: V_out = V_supply × [R₂/(R₁ + R₂)] = 9.0 × (6.0/9.0) = 6.0 V

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Question 10: Photoelectric Effect [4 marks]

(a) Calculate photon energy in joules. [2 marks]

• E = hf = hc/λ [M1]
• E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹)
• E = (1.989 × 10⁻²⁵) / (4.50 × 10⁻⁷)
• E = 4.42 × 10⁻¹⁹ J [A1]

(b) Determine if electrons will be emitted. [2 marks]

• Work function in joules: Φ = 2.0 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ J [M1]
• Photon energy (4.42 × 10⁻¹⁹ J) > work function (3.20 × 10⁻¹⁹ J)
• Therefore, electrons WILL be emitted [A1]
• Accept: Convert photon energy to eV: 4.42 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.76 eV > 2.0 eV, so emission occurs

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Question 11: Photoelectric Effect Analysis [4 marks]

(a) Explain why maximum kinetic energy increases as wavelength decreases. [2 marks]

• Shorter wavelength means higher frequency (f = c/λ) [B1]
• Higher frequency photons have greater energy (E = hf)
• From Eₘₐₓ = hf - Φ, greater photon energy means greater maximum kinetic energy (for same work function) [B1]

(b) Estimate threshold wavelength. [2 marks]

• At threshold, Eₘₐₓ = 0, so hf₀ = Φ [M1]
• Plot Eₘₐₓ vs 1/λ or use data to extrapolate
• Using data: when Eₘₐₓ = 0, 1/λ₀ ≈ 1.67 × 10⁶ m⁻¹
• λ₀ ≈ 600 nm (accept 580-620 nm) [A1]
• Accept: Linear extrapolation from given data points

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Section B: Free Response Questions [30 marks]

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Question 12: Mechanics – Collisions and Energy [15 marks]

(a)(i) Calculate velocity after collision. [2 marks]

• Conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v [M1]
• 1200 × 20.0 + 800 × 0 = (1200 + 800)v
• 24000 = 2000v
• v = 12.0 m s⁻¹ [A1]

(a)(ii) Calculate total kinetic energy before and after. [3 marks]

• KE_before = ½ × 1200 × (20.0)² = 240,000 J [M1]
• KE_after = ½ × 2000 × (12.0)² = 144,000 J [M1]
• KE_before = 2.40 × 10⁵ J, KE_after = 1.44 × 10⁵ J [A1]

(a)(iii) Explain whether collision is elastic or inelastic. [2 marks]

• In an elastic collision, kinetic energy is conserved [B1]
• Here, KE_after < KE_before (loss of 96,000 J)
• Therefore, the collision is inelastic [B1]
• Accept: "Kinetic energy is not conserved, so collision is inelastic"

(b)(i) Calculate speed just before hitting floor. [2 marks]

• v² = u² + 2as, u = 0 [M1]
• v² = 0 + 2 × 9.81 × 2.0 = 39.24
• v = 6.26 m s⁻¹ (downward) [A1]

(b)(ii) Calculate speed just after leaving floor. [2 marks]

• Using rebound height: v² = u² + 2as, v = 0 at max height [M1]
• 0 = u² + 2(-9.81)(1.6)
• u² = 31.392
• u = 5.60 m s⁻¹ (upward) [A1]

(b)(iii) Calculate impulse exerted by floor on ball. [2 marks]

• Impulse = Δp = m(v_final - v_initial) [M1]
• Taking upward as positive: v_initial = -6.26, v_final = +5.60
• Impulse = 0.15 × (5.60 - (-6.26)) = 0.15 × 11.86 = 1.78 N s [A1]

(b)(iv) Calculate average force exerted by floor. [2 marks]

• F_avg = Impulse / Δt [M1]
• F_avg = 1.78 / 0.050 = 35.6 N [A1]
• Direction: upward

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Question 13: Electricity – Circuits and Power [15 marks]

(a)(i) Calculate resistance of one lamp at rated voltage. [2 marks]

• P = V²/R → R = V²/P [M1]
• R = (6.0)²/12.0 = 36.0/12.0 = 3.0 Ω [A1]

(a)(ii) Draw circuit. [2 marks]

• Two lamps in parallel, this combination in series with third lamp [B1]
• Connected across 12.0 V battery
• Correct circuit symbols and connections [B1]

(a)(iii) Calculate total resistance. [3 marks]

• Resistance of parallel pair: 1/R_parallel = 1/3.0 + 1/3.0 = 2/3.0 [M1]
• R_parallel = 1.5 Ω [M1]
• R_total = R_parallel + R_series = 1.5 + 3.0 = 4.5 Ω [A1]

(a)(iv) Determine which lamp(s) will be brightest. [3 marks]

• Total current: I = 12.0/4.5 = 2.67 A [M1]
• Current through series lamp = 2.67 A
• Current through each parallel lamp = 2.67/2 = 1.33 A [M1]
• Power in series lamp: P = I²R = (2.67)² × 3.0 = 21.3 W
• Power in each parallel lamp: P = (1.33)² × 3.0 = 5.33 W
• The series lamp is brightest because it carries the largest current [A1]
• Accept: Series lamp dissipates most power, therefore brightest

(b)(i) Derive expression for power in R. [2 marks]

• Circuit current: I = EMF/(R + r) = 9.0/(R + 1.5) [M1]
• Power in R: P = I²R = [9.0/(R + 1.5)]² × R
• P = 81R/(R + 1.5)² [A1]

(b)(ii) Determine R for maximum power. [3 marks]

• Maximum power when R = r (maximum power transfer theorem) [M1]
• R = 1.5 Ω [A1]
• Check: dP/dR = 0 gives R = r
• P_max = 81 × 1.5/(1.5 + 1.5)² = 121.5/9 = 13.5 W [A1]
• Accept: R = 1.5 Ω with or without verification

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Question 14: Waves and Quantum Physics [15 marks]

(a)(i) Calculate fringe spacing. [2 marks]

• β = λD/a [M1]
• β = (600 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³)
• β = (1.20 × 10⁻⁶) / (5.0 × 10⁻⁴) = 2.4 × 10⁻³ m = 2.4 mm [A1]

(a)(ii) Describe how pattern changes with increased slit separation. [2 marks]

• Fringe spacing decreases (β ∝ 1/a) [B1]
• Pattern becomes more compressed/closer together
• Brightness/intensity of fringes remains unchanged (same amount of light) [B1]

(b)(i) Calculate maximum kinetic energy in eV. [3 marks]

• Photon energy: E = hc/λ [M1]
• E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(200 × 10⁻⁹) = 9.945 × 10⁻¹⁹ J
• E in eV = 9.945 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 6.22 eV [M1]
• K.E._max = E - Φ = 6.22 - 4.5 = 1.72 eV [A1]

(b)(ii) Calculate stopping potential. [2 marks]

• eV_s = K.E._max [M1]
• V_s = 1.72 V [A1]

(b)(iii) Explain immediate emission at low intensity. [3 marks]

• Wave theory predicts energy is spread over wavefront, requiring time to accumulate sufficient energy [B1]
• Photon model: light consists of photons, each with energy E = hf [B1]
• If photon energy > work function, a single photon can eject an electron immediately
• Low intensity means fewer photons, but each still has sufficient energy for immediate emission [B1]

(b)(iv) State one observation unexplained by wave theory. [1 mark]

• Any one of: [B1]
  • Existence of threshold frequency (no emission below certain frequency regardless of intensity)
  • Immediate emission of electrons (no time delay)
  • Maximum kinetic energy depends on frequency, not intensity
  • Kinetic energy of electrons independent of intensity

(b)(v) Explain how photon model accounts for observation. [2 marks]

• Light consists of discrete photons, each with energy E = hf [B1]
• One photon interacts with one electron
• If hf > Φ, electron is emitted with K.E._max = hf - Φ
• This explains [chosen observation] because... [B1]
• Example for threshold frequency: "Photons with frequency below threshold have energy less than work function, so cannot eject electrons regardless of how many photons arrive (intensity)"

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END OF ANSWER KEY