(a) Calculates the net force as 3600 N forward. Uses a = Fnet/m to obtain 3.0 m s⁻² forward.

Calculates the net force as 3600 N forward. Uses a = Fnet/m to obtain 3.0 m s⁻² forward.

(b) Calculates Wnet = 3600 × 30 = 1.08 × 10⁵ J. States the work-energy theorem: net work equals the change in kinetic energy. Concludes that the groomer's kinetic energy increases by 1.08 × 10⁵ J.

Calculates Wnet = 3600 × 30 = 1.08 × 10⁵ J. States the work-energy theorem: net work equals the change in kinetic energy. Concludes that the groomer's kinetic energy increases by 1.08 × 10⁵ J.

(c) Calculates a = 12²/32 = 4.5 m s⁻². Calculates Fnet = 1600 × 4.5 = 7.2 × 10³ N toward the centre.

Calculates a = 12²/32 = 4.5 m s⁻². Calculates Fnet = 1600 × 4.5 = 7.2 × 10³ N toward the centre.

(d) Identifies friction as the inward force required for circular motion. States that insufficient inward force produces a larger turning radius than intended. Predicts that the shuttle initially moves outward relative to the planned circular path, tangentially in the ground frame.

Identifies friction as the inward force required for circular motion. States that insufficient inward force produces a larger turning radius than intended. Predicts that the shuttle initially moves outward relative to the planned circular path, tangentially in the ground frame.

(a) Uses horizontal motion t = x/vx. Obtains t = 45/18 = 2.50 s.

Uses horizontal motion t = x/vx. Obtains t = 45/18 = 2.50 s.

(b) Calculates the drop as 0.5 × 9.8 × 2.50² = 30.6 m. Calculates vertical impact speed as 9.8 × 2.50 = 24.5 m s⁻¹ downward. Combines components to obtain impact speed approximately 30.4 m s⁻¹.

Calculates the drop as 0.5 × 9.8 × 2.50² = 30.6 m. Calculates vertical impact speed as 9.8 × 2.50 = 24.5 m s⁻¹ downward. Combines components to obtain impact speed approximately 30.4 m s⁻¹.

(a) Calculates the momentum-change magnitude as 0.40 × 14 = 5.6 N s. Calculates the rigid-stop average force as about 467 N. Calculates the liner average force as 70 N. Explains that the longer stopping time lowers average force for the same impulse. Notes that energy transferred to deformation, heat and sound means the package-liner collision is not elastic.

Calculates the momentum-change magnitude as 0.40 × 14 = 5.6 N s. Calculates the rigid-stop average force as about 467 N. Calculates the liner average force as 70 N. Explains that the longer stopping time lowers average force for the same impulse. Notes that energy transferred to deformation, heat and sound means the package-liner collision is not elastic.

(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) Uses v = √(GM/r) to obtain approximately 7.44 × 10³ m s⁻¹. States that gravity supplies the centripetal resultant force. Explains that acceleration is non-zero because velocity direction changes even when speed is constant.

Uses v = √(GM/r) to obtain approximately 7.44 × 10³ m s⁻¹. States that gravity supplies the centripetal resultant force. Explains that acceleration is non-zero because velocity direction changes even when speed is constant.

(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) Uses W = qV to obtain 1.44 × 10⁻⁶ J. States that electric potential energy decreases for motion along the force on a positive charge. States that kinetic energy increases by the same amount if other energy transfers are negligible.

Uses W = qV to obtain 1.44 × 10⁻⁶ J. States that electric potential energy decreases for motion along the force on a positive charge. States that kinetic energy increases by the same amount if other energy transfers are negligible.

(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 to obtain V₂ = 12 V rms. Uses P = VI to obtain I₂ = 360/12 = 30 A.

Uses V₂/V₁ = N₂/N₁ to obtain V₂ = 12 V rms. Uses P = VI to obtain I₂ = 360/12 = 30 A.

(b) Uses ideal power equality to obtain I₁ = 360/240 = 1.5 A. States that AC produces changing magnetic flux in the core. Explains that a steady DC input would not provide continuing flux change and induced secondary emf.

Uses ideal power equality to obtain I₁ = 360/240 = 1.5 A. States that AC produces changing magnetic flux in the core. Explains that a steady DC input would not provide continuing flux change and induced secondary emf.

(a) Calculates currents of 100 A at 20 kV and 10 A at 200 kV. Calculates line losses of 30 kW and 0.30 kW respectively using I²R. Concludes that the tenfold voltage increase reduces line loss by a factor of 100 for fixed resistance and input power. Explains that a step-up transformer enables high-voltage low-current transmission. Explains that local step-down transformers restore safer useful distribution voltages, with real transformer losses acknowledged.

Calculates currents of 100 A at 20 kV and 10 A at 200 kV. Calculates line losses of 30 kW and 0.30 kW respectively using I²R. Concludes that the tenfold voltage increase reduces line loss by a factor of 100 for fixed resistance and input power. Explains that a step-up transformer enables high-voltage low-current transmission. Explains that local step-down transformers restore safer useful distribution voltages, with real transformer losses acknowledged.

(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) States that the fringe spacing increases. Uses proportionality Δx ∝ λ to obtain a factor 650/520 = 1.25. Predicts a new spacing of approximately 3.90 mm.

States that the fringe spacing increases. Uses proportionality Δx ∝ λ to obtain a factor 650/520 = 1.25. Predicts a new spacing of approximately 3.90 mm.

(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) States that bound electrons occupy quantised energy states. States that a photon energy equals the difference between an allowed pair of states. Explains that only particular energy differences, and therefore wavelengths, are emitted.

States that bound electrons occupy quantised energy states. States that a photon energy equals the difference between an allowed pair of states. Explains that only particular energy differences, and therefore wavelengths, are emitted.

(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.