(a) Finds a forward resultant of 3600 N. Obtains acceleration 3.0 m s⁻².

Finds a forward resultant of 3600 N. Obtains acceleration 3.0 m s⁻².

(b) Calculates net work 3600 × 30 = 1.08 × 10⁵ J. Equates this work to the increase in kinetic energy from rest. Obtains speed √180 = 13.4 m s⁻¹.

Calculates net work 3600 × 30 = 1.08 × 10⁵ J. Equates this work to the increase in kinetic energy from rest. Obtains speed √180 = 13.4 m s⁻¹.

(c) Uses inward force mv²/r. Obtains required sideways force 6750 N, greater than 6000 N. Concludes that the intended bend cannot be followed at this speed; the initial ground-frame motion on losing grip is tangential.

Uses inward force mv²/r. Obtains required sideways force 6750 N, greater than 6000 N. Concludes that the intended bend cannot be followed at this speed; the initial ground-frame motion on losing grip is tangential.

(d) Obtains √(6000 × 32/1200) = 12.6 m s⁻¹. Identifies an applicable limitation such as uneven ice, a changing grip limit or simultaneous braking using some of the available tyre force.

Obtains √(6000 × 32/1200) = 12.6 m s⁻¹. Identifies an applicable limitation such as uneven ice, a changing grip limit or simultaneous braking using some of the available tyre force.

(a) Obtains t = 45/18 = 2.50 s from horizontal motion.

Obtains t = 45/18 = 2.50 s from horizontal motion.

(b) Uses zero initial vertical velocity and drop = ½gt². Obtains drop = 0.5 × 9.8 × 2.50² = 30.6 m.

Uses zero initial vertical velocity and drop = ½gt². Obtains drop = 0.5 × 9.8 × 2.50² = 30.6 m.

(a) Obtains impulse magnitude 0.40 × 14 = 5.6 N s. Obtains rigid-stop average force 5.6/0.012 = 467 N. Obtains liner average force 5.6/0.080 = 70 N.

Obtains impulse magnitude 0.40 × 14 = 5.6 N s. Obtains rigid-stop average force 5.6/0.012 = 467 N. Obtains liner average force 5.6/0.080 = 70 N.

(b) Both stops remove the same momentum, so spreading the same impulse over a longer time reduces average force.

Both stops remove the same momentum, so spreading the same impulse over a longer time reduces average force.

(c) Obtains rigid-stop distance ½ × 14 × 0.012 = 0.084 m. Obtains liner stopping distance ½ × 14 × 0.080 = 0.56 m. Obtains kinetic-energy decrease ½ × 0.40 × 14² = 39.2 J.

Obtains rigid-stop distance ½ × 14 × 0.012 = 0.084 m. Obtains liner stopping distance ½ × 14 × 0.080 = 0.56 m. Obtains kinetic-energy decrease ½ × 0.40 × 14² = 39.2 J.

(a) Uses g = GM/r². Obtains g ≈ 7.68 N kg⁻¹ directed toward Earth.

Uses g = GM/r². Obtains g ≈ 7.68 N kg⁻¹ directed toward Earth.

(b) Obtains v = √(GM/r) ≈ 7.44 × 10³ m s⁻¹.

Obtains v = √(GM/r) ≈ 7.44 × 10³ m s⁻¹.

(a) Calculates E = V/d = 480/0.012 = 4.0 × 10⁴ V m⁻¹. Calculates F = qE = 1.2 × 10⁻⁴ N toward the negative plate.

Calculates E = V/d = 480/0.012 = 4.0 × 10⁴ V m⁻¹. Calculates F = qE = 1.2 × 10⁻⁴ N toward the negative plate.

(b) Obtains electric work qV = 1.44 × 10⁻⁶ J. Obtains electric-potential-energy change −1.44 × 10⁻⁶ J. Obtains kinetic-energy increase +1.44 × 10⁻⁶ J.

Obtains electric work qV = 1.44 × 10⁻⁶ J. Obtains electric-potential-energy change −1.44 × 10⁻⁶ J. Obtains kinetic-energy increase +1.44 × 10⁻⁶ J.

(c) E = V/d doubles, so force doubles to 2.4 × 10⁻⁴ N. The full-transit work qV remains 1.44 × 10⁻⁶ J because the potential difference is unchanged.

E = V/d doubles, so force doubles to 2.4 × 10⁻⁴ N. The full-transit work qV remains 1.44 × 10⁻⁶ J because the potential difference is unchanged.

(a) Uses F = qvB for perpendicular velocity and field. Obtains F = 6.91 × 10⁻¹⁵ N.

Uses F = qvB for perpendicular velocity and field. Obtains F = 6.91 × 10⁻¹⁵ N.

(b) Uses r = mv/(qB). Obtains r ≈ 7.59 × 10⁻⁴ m. Explains that the perpendicular magnetic force changes direction but does no work, so speed remains constant.

Uses r = mv/(qB). Obtains r ≈ 7.59 × 10⁻⁴ m. Explains that the perpendicular magnetic force changes direction but does no work, so speed remains constant.

(a) Identifies opposite magnetic forces on the two current-carrying sides of the coil. Explains that the force pair creates a torque about the axle. Explains that the split-ring commutator reverses coil current every half-turn so torque continues in the same rotational sense. States that increasing current, field strength or turn count increases torque within operating limits. Notes that added resistance or heating can limit current, so a higher supply voltage does not guarantee proportional torque indefinitely.

Identifies opposite magnetic forces on the two current-carrying sides of the coil. Explains that the force pair creates a torque about the axle. Explains that the split-ring commutator reverses coil current every half-turn so torque continues in the same rotational sense. States that increasing current, field strength or turn count increases torque within operating limits. Notes that added resistance or heating can limit current, so a higher supply voltage does not guarantee proportional torque indefinitely.

(a) Calculates the flux-change magnitude per turn as 3.0 × 10⁻³ Wb. Calculates |ε| = 350 × 0.0030/0.060 = 17.5 V.

Calculates the flux-change magnitude per turn as 3.0 × 10⁻³ Wb. Calculates |ε| = 350 × 0.0030/0.060 = 17.5 V.

(b) States that changing flux induces an emf/current. Explains that the induced magnetic effect opposes the flux change rather than opposing the existing flux necessarily. Connects this opposition to energy conservation.

States that changing flux induces an emf/current. Explains that the induced magnetic effect opposes the flux change rather than opposing the existing flux necessarily. Connects this opposition to energy conservation.

(a) Calculates Vrms = 460/√2 ≈ 325 V. Calculates T = 1/f = 0.020 s.

Calculates Vrms = 460/√2 ≈ 325 V. Calculates T = 1/f = 0.020 s.

(b) States that rotation changes the magnetic flux through the coil. Links induced emf to the rate of change of flux. Explains that slip rings preserve connection while the emf reverses every half-turn.

States that rotation changes the magnetic flux through the coil. Links induced emf to the rate of change of flux. Explains that slip rings preserve connection while the emf reverses every half-turn.

(a) Uses V₂/V₁ = N₂/N₁ = 90/1800. Obtains V₂ = 12 V rms. Obtains I₂ = 360/12 = 30 A.

Uses V₂/V₁ = N₂/N₁ = 90/1800. Obtains V₂ = 12 V rms. Obtains I₂ = 360/12 = 30 A.

(a) Obtains 100 A at 20 kV. Obtains 10 A at 200 kV.

Obtains 100 A at 20 kV. Obtains 10 A at 200 kV.

(b) Obtains 100² × 3.0 = 30 kW at 20 kV. Obtains 10² × 3.0 = 0.30 kW at 200 kV.

Obtains 100² × 3.0 = 30 kW at 20 kV. Obtains 10² × 3.0 = 0.30 kW at 200 kV.

(c) Subtracts line heating to obtain delivered powers 1.970 MW and 1.99970 MW. Obtains line efficiencies 98.5% and 99.985% respectively. Selects 200 kV because only its line efficiency reaches 99.9%.

Subtracts line heating to obtain delivered powers 1.970 MW and 1.99970 MW. Obtains line efficiencies 98.5% and 99.985% respectively. Selects 200 kV because only its line efficiency reaches 99.9%.

(a) Uses Δx = λL/d. Obtains Δx = 3.12 × 10⁻³ m or 3.12 mm.

Uses Δx = λL/d. Obtains Δx = 3.12 × 10⁻³ m or 3.12 mm.

(b) Obtains spacing ratio 650/520 = 1.25. Obtains new spacing 3.90 mm.

Obtains spacing ratio 650/520 = 1.25. Obtains new spacing 3.90 mm.

(c) Obtains spacing 650 × 10⁻⁹ × 1.80/(0.60 × 10⁻³) = 1.95 mm. Bright fringes require an integer multiple of wavelength as path difference. Dark fringes require an odd multiple of half a wavelength as path difference.

Obtains spacing 650 × 10⁻⁹ × 1.80/(0.60 × 10⁻³) = 1.95 mm. Bright fringes require an integer multiple of wavelength as path difference. Dark fringes require an odd multiple of half a wavelength as path difference.

(a) Calculates Ek,max = 3.10 - 2.20 = 0.90 eV. Obtains a stopping-potential magnitude of 0.90 V.

Calculates Ek,max = 3.10 - 2.20 = 0.90 eV. Obtains a stopping-potential magnitude of 0.90 V.

(b) States that photon energy and therefore maximum kinetic energy remain unchanged. States that more photons arrive per unit time. Predicts a greater emission rate and photocurrent when other conditions are unchanged.

States that photon energy and therefore maximum kinetic energy remain unchanged. States that more photons arrive per unit time. Predicts a greater emission rate and photocurrent when other conditions are unchanged.

(a) Uses ΔE = hc/λ. Obtains ΔE ≈ 4.09 × 10⁻¹⁹ J or 2.55 eV.

Uses ΔE = hc/λ. Obtains ΔE ≈ 4.09 × 10⁻¹⁹ J or 2.55 eV.

(b) Quantised energy-state differences permit only particular emitted photon energies and therefore particular wavelengths.

Quantised energy-state differences permit only particular emitted photon energies and therefore particular wavelengths.

(a) Calculates the Earth-frame dilated lifetime as about 11.1 μs. Calculates an Earth-frame mean travel distance of about 3.25 km and recognises survival is probabilistic rather than guaranteed. Calculates the 8.0 km atmospheric distance in the muon frame as about 1.59 km. Explains that time dilation in Earth's frame and length contraction in the muon frame are consistent descriptions of the same events. Avoids adding the relativistic effects twice or claiming the muons experience a dilated proper lifetime.

Calculates the Earth-frame dilated lifetime as about 11.1 μs. Calculates an Earth-frame mean travel distance of about 3.25 km and recognises survival is probabilistic rather than guaranteed. Calculates the 8.0 km atmospheric distance in the muon frame as about 1.59 km. Explains that time dilation in Earth's frame and length contraction in the muon frame are consistent descriptions of the same events. Avoids adding the relativistic effects twice or claiming the muons experience a dilated proper lifetime.

(a) States an aim to determine the relationship between applied force and acceleration for the cart. Identifies force as continuous independent variable, acceleration as dependent variable, and cart mass as controlled.

States an aim to determine the relationship between applied force and acceleration for the cart. Identifies force as continuous independent variable, acceleration as dependent variable, and cart mass as controlled.

(b) Finds a gradient close to 0.42 (m s⁻²) N⁻¹. Uses gradient = 1/m to estimate mass about 2.4 kg. Notes the near-linear pattern and requests uncertainty/repeats before judging agreement precisely.

Finds a gradient close to 0.42 (m s⁻²) N⁻¹. Uses gradient = 1/m to estimate mass about 2.4 kg. Notes the near-linear pattern and requests uncertainty/repeats before judging agreement precisely.

(a) Predicts that decreasing slit width will increase diffraction spread or central maximum width. Identifies wavelength and screen distance as controlled variables.

Predicts that decreasing slit width will increase diffraction spread or central maximum width. Identifies wavelength and screen distance as controlled variables.

(b) Requires the beam path to be below or above eye level with controlled access and no direct viewing. Uses several continuous slit widths with repeated width measurements. Keeps alignment and screen distance fixed and records uncertainty in fringe-width measurement.

Requires the beam path to be below or above eye level with controlled access and no direct viewing. Uses several continuous slit widths with repeated width measurements. Keeps alignment and screen distance fixed and records uncertainty in fringe-width measurement.

(a) Identifies an approximately linear proportional relationship. Estimates gradient approximately 6.0 mN A⁻¹.

Identifies an approximately linear proportional relationship. Estimates gradient approximately 6.0 mN A⁻¹.

(b) Uses gradient = lB for a perpendicular single wire. Calculates B ≈ 0.0060/0.080 = 0.075 T. Notes a limitation such as field non-uniformity, balance resolution or unreported repeats/uncertainty.

Uses gradient = lB for a perpendicular single wire. Calculates B ≈ 0.0060/0.080 = 0.075 T. Notes a limitation such as field non-uniformity, balance resolution or unreported repeats/uncertainty.

(a) Identifies that the absolute word 'always' exceeds the tested angle range and assumptions. Identifies launch speed as a confounding variable because it changed with angle. Requires measured launch speed, repeats and uncertainty at each angle while controlling launch height and apparatus. States that any conclusion must be restricted to the tested launcher conditions and supported range pattern. Proposes clear reporting of raw/processed data, uncertainty bars, the latch limitation and a revised conditional claim rather than causal overstatement.

Identifies that the absolute word 'always' exceeds the tested angle range and assumptions. Identifies launch speed as a confounding variable because it changed with angle. Requires measured launch speed, repeats and uncertainty at each angle while controlling launch height and apparatus. States that any conclusion must be restricted to the tested launcher conditions and supported range pattern. Proposes clear reporting of raw/processed data, uncertainty bars, the latch limitation and a revised conditional claim rather than causal overstatement.