HSC Exam Practice

Sample response. Momentum (p) is the product of an object’s mass and velocity: p = mv. It is measured in kilogram metres per second (kg m/s) and is a vector quantity in the direction of the velocity. Impulse (J) is the product of the average net force and the contact time over which it acts: J = FΔt. It is measured in newton seconds (N s) and is a vector in the direction of the net force. They have the same unit because N s = (kg m/s²) × s = kg m/s — and because impulse equals the change in momentum (the impulse-momentum theorem).

Marking notes. 1 = momentum definition (p = mv, unit, vector); 1 = impulse definition (J = FΔt, unit, vector); 1 = units are equivalent (N s = kg m/s with algebraic reasoning); 1 = both are vectors and impulse equals change in momentum.

Sample response. The impulse-momentum theorem states that the net impulse acting on an object equals its change in momentum: J = Δp = mv_f − mv_i. Derivation: Newton’s Second Law gives F = ma = mΔv/Δt. Multiplying both sides by Δt: FΔt = mΔv = mv_f − mv_i = Δp. Therefore J = FΔt = Δp.

Marking notes. 1 = states J = Δp (impulse equals change in momentum); 1 = starts derivation from F = ma or F = mΔv/Δt; 1 = correctly arrives at FΔt = mΔv = Δp.

Sample response. Step 1 — Positive direction: toward the bat (initial direction of ball) = positive. v_i = +38 m/s; v_f = −32 m/s; Δt = 0.0025 s.

(a) Δp = mv_f − mv_i = 0.16(−32) − 0.16(+38) = −5.12 − 6.08 = −11.2 N s.

(b) J = Δp = −11.2 N s (bat exerts impulse in the negative direction — pushing ball back).

(c) F = J/Δt = −11.2/0.0025 = −4480 N (in the negative direction; the bat exerts 4480 N on the ball away from the bat).

Marking notes. 1 = positive direction defined AND signed values used (v_f negative); 1 = Δp = −11.2 N s correct (accept ±0.1 N s); 1 = J = Δp stated explicitly; 1 = F = −4480 N with correct sign or direction stated. Deduct 1 mark if positive direction not defined and student uses magnitudes only.

Sample response. The student made the error of subtracting the magnitudes of the speeds (38 − 32 = 6 m/s) instead of subtracting the signed velocities. Because the ball reverses direction, the final velocity is negative. Correct: Δv = v_f − v_i = (−32) − (+38) = −70 m/s. The student’s value of 6 m/s underestimates Δv by more than 11 times and gives a physically incorrect result that ignores the direction reversal.

Marking notes. 1 = identifies error as using magnitudes (speeds) rather than signed velocities; 1 = correct Δv = −70 m/s shown with reasoning; 1 = explains that direction reversal requires a sign change, not subtraction of magnitudes.

Sample response. An airbag inflates rapidly between the occupant and the car interior during a crash. It extends the contact time from approximately 2 ms (hitting the dashboard) to approximately 50 ms. By the impulse-momentum theorem, J = FΔt = Δp. The change in momentum (Δp) is the same regardless of whether the airbag deploys, because the occupant undergoes the same change in velocity. However, because Δt is increased by a factor of ~25, the average force F must decrease by the same factor (F = Δp/Δt). The reduced force is less likely to cause fatal head or chest injuries. The quantity that changes is Δt (and therefore F); the quantity that remains constant is Δp. A second example is a crumple zone: the front of the car deforms progressively, extending the stopping time from ~100 ms to ~200 ms and halving the average deceleration force on occupants.

Marking notes. 1 = explains airbag extends Δt; 1 = J = FΔt = Δp used correctly, identifying Δp as constant and F as decreasing; 1 = states the factor by which F reduces (quantitative or relative); 1 = names a second safety device (crumple zone, helmet, gymnastics mat, or equivalent) with same-principle explanation.

Sample response (a). Energy conservation: mgh = ½mv² ⇒ v = √(2gh) = √(2 × 9.8 × 3.2) = √(62.72) = 7.92 m/s (downward).

Sample response (b). Positive direction = downward. v_i = +7.92 m/s; v_f = 0. Δp = m(v_f − v_i) = 60(0 − 7.92) = −475 N s (upward impulse stops gymnast). Accept −475 to −476 N s.

Sample response (c). Δp is the same in both cases (−475 N s). Mat: F = −475/0.50 = −950 N. Concrete: F = −475/0.012 = −39 600 N. Gymnast weight W = 60 × 9.8 = 588 N. Mat: 950/588 = 1.6× W. Concrete: 39 600/588 = 67× W. Award 1 mark for both correct forces, 1 mark for correct weight comparison, 1 mark for clear working with positive direction defined.

Sample response (d). Equal areas mean equal impulse: the total impulse delivered by each surface is identical. Since both bring the gymnast from the same speed to rest (same Δp), the impulse-momentum theorem (J = Δp) requires that both F-t curves enclose the same area. This is consistent with the theorem — a larger peak force with shorter contact time (concrete spike) and a smaller force with longer contact time (mat broad curve) can deliver the same impulse. The theorem does not constrain the shape of the force, only the area.

Marking notes (a): 1 = v = √(2gh) with correct substitution; 1 = v ≈ 7.9 m/s. (b): 1 = defines positive direction AND signs v correctly; 1 = Δp ≈ −475 N s. (c): 1 = both forces correct; 1 = correct W calculation; 1 = forces expressed as multiples of W. (d): 1 = equal areas = equal impulse; 1 = links to J = Δp and explains different force shapes can give same area.

Sample response. The strategy of extending contact time in collisions is grounded in the impulse-momentum theorem: J = FΔt = Δp. For any collision where an object decelerates from speed v to rest, the change in momentum (Δp = mv) is fixed by the mass and initial speed. The theorem shows that F = Δp/Δt; therefore, increasing Δt while holding Δp constant proportionally decreases F. This is the physical mechanism underlying all safety devices that protect against collision forces. A vehicle safety application is the crumple zone. The front section of a car is engineered to deform progressively during a frontal collision, extending the stopping time from approximately 100 ms (rigid vehicle) to approximately 200 ms (with crumple zones). This halves the average deceleration force on occupants, reducing the likelihood of fatal chest and head injuries. The same Δp is delivered, but over twice the time. A sporting application is the gymnastics mat, which extends the stopping time from approximately 10 ms (concrete) to approximately 500 ms. This reduces the average force by a factor of 50 for the same landing, making the difference between a serious fracture or spinal injury and a routine landing. A limitation of the strategy is that Δt cannot be increased indefinitely: vehicle crumple zones require a certain amount of crush space, and if the zone is used up before the occupant is protected, the structure becomes rigid and the force spikes. Similarly, a mat cannot be made arbitrarily thick without affecting athletic performance. Furthermore, the strategy reduces average force but not peak force: real collision forces peak above the average (triangular or bell-shaped pulses), so even with extended contact time, a sharp peak can still cause injury if the force profile is not carefully designed. An engineer could improve safety device design by shaping the force-time curve, not just extending Δt: a device that delivers a flat (constant) force over the longest possible Δt minimises peak force. This can be achieved with energy-absorbing foam that compresses at a nearly constant resistance, or with programmable airbag materials that are tuned to absorb energy at a controlled rate.

Marking criteria (6 marks): 1 = clearly states J = FΔt = Δp and identifies that Δp is fixed while Δt and F vary; 1 = vehicle safety application with correct physics reasoning (Δp constant, Δt extended, F reduced quantitatively or relatively); 1 = sporting/recreational application with correct reasoning and numbers or ratios; 1 = limitation identified (finite crush space, or peak vs average force distinction, or structural constraint); 1 = second limitation or further nuance (e.g. effect on performance, or peak vs average force if not already used); 1 = specific engineering improvement with physical justification (constant-force profile, programmable materials, or equivalent).