HSC Exam Practice

Sample response. Momentum is the product of an object’s mass and velocity: p = mv. Its SI unit is kg m/s (or equivalently N s). A positive direction must be defined because momentum is a vector — its sign carries directional information. Without a sign convention, objects moving in opposite directions would be assigned incorrect signs, leading to errors in Δp calculations and conservation-of-momentum equations.

Marking notes. 1 mark for the correct definition p = mv and SI unit. 1 mark for stating that momentum is a vector. 1 mark for explaining that a positive direction is needed to correctly assign signs to velocities (and therefore momenta) of objects moving in opposite directions.

Sample response. Define positive direction as toward the wall. For the ball that sticks: v_i = +v, v_f = 0. Δp = m(0 − v) = −mv. For the ball that bounces: v_i = +v, v_f = −v (reversed direction). Δp = m(−v − v) = −2mv. The bouncing ball undergoes a direction reversal, so Δv = 2v rather than v, giving a momentum change twice as large in magnitude.

Marking notes. 1 mark for correct Δp for each case (sticky: −mv; bounce: −2mv or equivalent with a defined sign convention). 1 mark for explaining the direction reversal. 1 mark for the conclusion that the bouncing ball has double the momentum change.

Sample response. Law: in a closed system, the total momentum before any interaction equals the total momentum after (Σp_before = Σp_after). Conditions: (1) no net external force acts on the system; (2) the system is closed — no mass enters or leaves. Example where it cannot be applied: a ball decelerating due to friction while rolling along a floor — friction is an external force that continuously reduces the ball’s momentum, violating the closed-system condition.

Marking notes. 1 mark for correct statement of the law (Σp_before = Σp_after). 1 mark for condition 1 (no net external force). 1 mark for condition 2 (closed system / no mass in or out). 1 mark for a valid counter-example (friction deceleration, rocket not in closed system, etc.) with brief justification.

Sample response. In an elastic collision, both momentum and kinetic energy are conserved, and the objects separate after the collision. In a perfectly inelastic collision, momentum is conserved but kinetic energy is not — the maximum possible kinetic energy is lost — and the objects stick together and move as one combined mass. To confirm elastic: measure velocities before and after; calculate KE before and after; if KE_after = KE_before, the collision is elastic. To confirm perfectly inelastic: verify that the objects move together (same velocity) after impact, and show that KE_after < KE_before.

Marking notes. 1 mark for correct definition of elastic (momentum and KE both conserved, objects separate). 1 mark for correct definition of perfectly inelastic (momentum conserved, max KE lost, objects stick). 1 mark for experimental method for elastic (calculate KE before and after). 1 mark for experimental method for perfectly inelastic (verify same velocity after; measure KE loss).

Sample response. The claim is incorrect. The seatbelt does not reduce the impulse — the impulse J = Δp = mΔv is fixed by the mass and the velocity change required to stop the passenger, which is the same regardless of whether a seatbelt is worn. What the seatbelt does is extend the contact time Δt over which the same impulse is delivered. From J = FΔt, for a fixed J, a larger Δt means a proportionally smaller average force F. It is the force that is reduced, not the impulse.

Marking notes. 1 mark for identifying the specific error (seatbelt does not reduce impulse). 1 mark for explaining that impulse is fixed by Δp = mΔv (same for both). 1 mark for correct explanation using J = FΔt: seatbelt increases Δt, which reduces F for the same J.

Sample response. Newton’s Third Law: the racquet exerts a force on the ball directed forward (the direction the ball will travel), and simultaneously the ball exerts an equal and opposite force on the racquet (directed backward, opposing the swing). Both forces have the same magnitude and act for the same contact time Δt, so both objects receive equal and opposite impulses. The ball accelerates far more than the racquet because the ball has far less mass (F = ma: same F, smaller m gives larger a). If the ball is 0.06 kg and the racquet is 0.3 kg, the racquet experiences 5 times less acceleration than the ball for the same force.

Marking notes. 1 mark for correctly stating that the two forces are equal in magnitude and opposite in direction, acting for the same time. 1 mark for stating that both objects receive equal and opposite impulses (Newton 3 in terms of impulse). 1 mark for explaining that the ball accelerates more because it has less mass (F = ma, same F, different m).

Sample response (a). As m₂ increases, the fraction of KE lost increases. When m₂ = 1 kg, only 0.33 (33%) of KE is lost; when m₂ = 18 kg, 0.90 (90%) is lost. The fraction increases non-linearly — the rate of increase slows as m₂ becomes very large relative to m₁.

Marking notes (a). 1 mark for correctly identifying the trend (increasing fraction as m₂ increases). 1 mark for supporting the description with at least two specific data values from the table.

Sample response (b). For m₂ = 6 kg: (m₁ + m₂)v_f = m₁u₁. (2 + 6)v_f = 2 × 5 = 10. v_f = 10/8 = 1.25 m/s. This matches the table. Momentum is conserved: p_before = 2×5 = 10 kg m/s; p_after = 8×1.25 = 10 kg m/s. ✓

Marking notes (b). 1 mark for correct conservation equation; 1 mark for v_f = 1.25 m/s with momentum check.

Sample response (c). m₂/(m₁+m₂) = 18/(2+18) = 18/20 = 0.90. This matches the table. As m₂ → ∞ (or m₂ ≫ m₁), the fraction approaches 1 — meaning almost all of the initial kinetic energy is lost. A real-world example is a small car hitting a large concrete wall or an immovable cliff face: the car (small mass) retains almost none of its kinetic energy after the collision, and the wall (effectively infinite mass) does not move. Virtually all the car’s kinetic energy is dissipated as heat, sound, and structural deformation.

Marking notes (c). 1 mark for correct substitution verifying 0.90. 1 mark for correctly stating that the fraction approaches 1 as m₂ becomes very large. 1 mark for a valid real-world example with explanation (car into wall, ball hitting floor, etc.).

Sample response (a). Positive direction = direction of rocket travel (right). Initial total momentum = 0. 0 = m_rocket v_rocket + m_gas v_gas. 0 = (6 − 0.05) v_rocket + 0.05×(−420). 0 = 5.95 v_rocket − 21. v_rocket = 21/5.95 = +3.53 m/s.

Marking notes (a). 1 mark for correct conservation of momentum equation (initial p = 0; rocket mass after = 6 − 0.05 = 5.95 kg). 1 mark for correct v_rocket = +3.53 m/s.

Sample response (b). KE_rocket = ½×5.95×3.53² = ½×5.95×12.46 = 37.1 J. KE_gas = ½×0.05×420² = ½×0.05×176400 = 4410 J. Total KE = 4447 J. Before the explosion, the rocket was at rest and all KE = 0. The kinetic energy came from the chemical energy stored in the rocket propellant; combustion converted chemical potential energy into kinetic energy of both the rocket and the exhaust gas.

Marking notes (b). 1 mark for both KE values correct (accept rounding). 1 mark for correctly identifying the source as chemical (potential) energy stored in the propellant.

Sample response. The impulse-momentum theorem states that J = FΔt = Δp: the impulse delivered to an object equals its change in momentum, and the average force required equals the impulse divided by the contact time. In a collision, a person who goes from 100 km/h to rest must experience a fixed Δp = mΔv regardless of how the stop occurs. Safety devices reduce injury by maximising Δt, thereby minimising F for the same Δp.

Device 1 — Seatbelt: Without a seatbelt, an unrestrained passenger continues at the car’s pre-collision speed and hits the dashboard in ∼20–50 ms. A seatbelt extends the stopping time to ∼200–300 ms, reducing the average force by a factor of 6–15. For a 70 kg person stopping from 15 m/s: Δp = 70×15 = 1050 N s. With seatbelt (Δt = 0.25 s): F ≈ 4200 N (survivable). Without (Δt = 0.02 s): F ≈ 52 500 N — equivalent to over 5 tonnes pressing on the chest, causing fatal internal injuries.

Device 2 — Bicycle / sports helmet: A foam-lined helmet extends the contact time during a head impact by deforming progressively. The foam increases Δt from ∼2 ms (rigid surface) to ∼10–15 ms, reducing peak force to the skull by a factor of 5–7. For a head mass of 5 kg striking a road at 8 m/s: Δp = 40 N s. Without helmet: F ≈ 20 000 N — well above the skull fracture threshold of ∼6000 N. With helmet: F ≈ 2700–4000 N — below the injury threshold.

Newton’s Third Law: Collisions are dangerous precisely because of Newton’s Third Law. When a car strikes a wall, the wall exerts an equal and opposite force on the car. This force has the same magnitude as the car exerts on the wall. The car occupants are then subjected to the same impulse as the car itself. Since human tissue has limited tolerance for force (not impulse), the key variable is always the average force — which depends on how quickly the momentum change occurs. A rigid object imposes a very high force over a very short time; a deformable safety device spreads the same impulse over a longer time.

Limitations: Safety devices have physical limits. A seatbelt cannot extend Δt indefinitely — the available deformation distance is constrained by the vehicle cabin. In a high-speed collision (>100 km/h), even a seatbelt may not reduce peak force below the injury threshold. Similarly, helmets are designed for a specific energy range — a high-speed motorcycle crash exceeds the foam’s deformation capacity, and the helmet bottoms out, making it effectively rigid. A second limitation is that the vector direction of the impulse matters: oblique impacts deliver off-axis forces that can cause rotational injuries even when linear peak force is below threshold.

Marking criteria (8 marks): 1 = correctly states J = FΔt = Δp and explains how extending Δt reduces F for the same Δp. 1 = named Device 1 (seatbelt or airbag) with specific mechanism and quantitative example. 1 = named Device 2 (helmet, crumple zone, crash mat) with specific mechanism. 1 = discusses Newton’s Third Law and explains why it makes collisions dangerous (equal forces on occupant as on vehicle). 1 = quantitative comparison showing the force difference (with and without device) using J = FΔt. 1 = identifies at least one physical limitation of safety devices (energy limit, cabin space, oblique impacts). 1 = second limitation OR a second quantitative example from a different context (road vs sports). 1 = reaches an explicit evaluative judgement: safety devices reduce injury by extending Δt but cannot eliminate risk above certain energy thresholds.