HSC Exam Practice

Sample response. Static friction acts on a stationary object, opposing its tendency to slide; it adjusts in magnitude up to a maximum of μ_sF_N before the object moves. Kinetic friction acts on an already-sliding object and has a constant value of μ_kF_N that does not depend on speed. The key difference in magnitude is that μ_s > μ_k for any surface pair, so the maximum static friction is always greater than kinetic friction — it takes more force to start an object moving than to keep it sliding.

Marking notes. 1 mark for correct definition of static friction (stationary object, adjusts to maximum); 1 mark for correct definition of kinetic friction (sliding object, constant value); 1 mark for explicitly stating μ_s > μ_k and what this means practically.

Sample response. Newton’s First Law: an object will remain at rest, or continue moving in a straight line at constant velocity, unless acted on by a net external force. Example: packages in the back of a braking van. The packages are the object. No horizontal force acts on the packages (there is no seatbelt or restraint), so by Newton’s First Law they continue moving forward at the van’s original velocity while the van decelerates around them.

Marking notes. 1 mark for correct formal statement of Newton’s First Law (no net force → no change in motion); 1 mark for naming a specific object in a valid example; 1 mark for correctly explaining why the law applies (zero net force on that specific object).

Sample response. A constant velocity means the net force is zero (Newton’s First Law), not that there are no forces. When a car cruises at constant speed on a flat road, two horizontal forces act: the driving force from the engine (forward) and friction/air resistance (backward). These forces are equal in magnitude and opposite in direction, so they cancel to give F_net = 0, resulting in zero acceleration (constant velocity). Forces are present — they are simply balanced.

Marking notes. 1 mark for stating that constant velocity means F_net = 0, not no forces; 1 mark for naming both forces present (driving force forward and friction/drag backward); 1 mark for explaining they are equal and opposite, producing F_net = 0 and zero acceleration.

Sample response (Vector Protocol). Step 1 — positive direction: forward = positive. Step 2 — FBD: W = 12×9.8 = 117.6 N down; F_N = 117.6 N up; F_applied = 90 N forward; f_k backward. Step 3 — friction: F_N = mg = 117.6 N. f_k = μ_kF_N = 0.40×117.6 = 47.0 N (backward, negative). Step 4: F_net = 90 − 47.0 = 43.0 N (forward). a = F_net/m = 43.0/12 = 3.6 m s⁻² (forward).

Marking notes. 1 mark for correct F_N = 117.6 N; 1 mark for correct f_k = 47.0 N; 1 mark for correct F_net = 43.0 N; 1 mark for correct a = 3.6 m s⁻² with direction. Penalise missing direction on the final answer.

Sample response. Newton’s First Law applies when F_net = 0: the object is at rest or moving at constant velocity. Newton’s Second Law applies when F_net ≠ 0: the object is accelerating. Braking scenario example for N1: the unsecured packages in the back of the van experience no horizontal force and continue moving forward at constant velocity — Newton’s First Law (F_net = 0). Braking scenario example for N2: the van itself experiences a backward net friction force from the road, causing it to decelerate — Newton’s Second Law (F_net = ma, acceleration directed backward).

Marking notes. 1 mark for correct condition for N1 (F_net = 0, constant velocity or rest); 1 mark for correct condition for N2 (F_net ≠ 0, acceleration); 1 mark for valid N1 example from braking scenario; 1 mark for valid N2 example from braking scenario.

Sample response. The packages are not “thrown forward” — no force acts on them horizontally. When the van brakes, friction from the road acts backward on the van’s tyres, decelerating the van (Newton’s Second Law). The packages, however, have no horizontal force acting on them (no restraint), so by Newton’s First Law they continue moving forward at the original velocity while the van decelerates and “comes back” to hit them. The seatbelted driver does not move forward relative to the van because the seatbelt exerts a backward force on the driver, providing the net force (F_net = ma) needed to decelerate the driver at the same rate as the van — Newton’s Second Law.

Marking notes. 1 mark for correctly explaining the packages’ motion (no horizontal force, continue at original velocity — Newton’s First Law); 1 mark for explaining the van’s braking using Newton’s Second Law and friction; 1 mark for explaining the driver (seatbelt force provides F_net to decelerate driver with van).

Part (a). Convert: 100 km h⁻¹ = 100×1000/3600 = 27.78 m s⁻¹. a = Δv/Δt = (27.78 − 0)/8.2 = 3.39 m s⁻² (forward). Marks: 1 for correct unit conversion; 1 for correct a = 3.39 m s⁻².

Part (b). Step 1: positive direction = forward. Step 2: F_net = ma = 1 500×3.39 = 5 083 N (forward). Step 3: F_net = F_drive − f. Therefore f = F_drive − F_net = 6 500 − 5 083 = 1 417 N (backward). Marks: 1 for positive direction stated; 1 for correct F_net = 5 083 N; 1 for correct friction f = 1 417 N.

Part (c). F_N = mg = 1 500×9.8 = 14 700 N. μ_k = f/F_N = 1 417/14 700 = 0.096. Marks: 1 for F_N = 14 700 N; 1 for correct μ_k ≈ 0.096.

Part (d). Wet road: f_wet = μ_k(wet)×mg = 0.06×14 700 = 882 N. F_net,wet = 6 500 − 882 = 5 618 N. a_wet = 5 618/1 500 = 3.75 m s⁻². The car accelerates faster on the wet road (3.75 m s⁻² vs 3.39 m s⁻²) because rolling resistance friction is lower on a wet surface, so the net forward force is larger. Note: in a real road safety context, wet roads reduce braking and cornering friction — the specific friction measured here is rolling resistance, not braking friction. Marks: 1 for correct f_wet = 882 N; 1 for correct a_wet = 3.75 m s⁻²; 1 for explaining the car accelerates faster with reference to reduced friction and increased net force.

Sample response. The claim is scientifically inaccurate as stated, for the following reason: heavier vehicles do experience a larger friction force during braking (f = μmg; larger m means larger f). However, they also have greater mass resisting deceleration. Applying Newton’s Second Law: a = F_net/m = μmg/m = μg. Mass cancels entirely. On the same surface with the same μ_k, all vehicles decelerate at the same rate regardless of mass. From kinematics (v² = u² + 2as), since deceleration a = μg and initial speed u are identical, the stopping distance s = u²/(2μg) is also the same for any mass. The claim is therefore wrong: more friction does not lead to a longer stopping distance — it leads to a proportionally greater deceleration that exactly offsets the greater inertia.

The claim partially holds under one specific condition: if the braking system has a fixed maximum braking force (for example, if brake fade limits the force a heavy truck can apply), then the deceleration a = F_brake/m is smaller for larger m (fixed force, larger denominator). Heavy trucks can experience this on long downhill grades where brake overheating reduces available braking force, increasing stopping distance relative to a lighter vehicle. Additionally, at very high speeds, heavy vehicles carry more kinetic energy (E_k = ½mv²), which must all be dissipated by the brakes; if the braking force is limited, stopping distance increases with mass.

One engineering strategy to reduce stopping distance is anti-lock braking (ABS). ABS prevents wheels from locking under heavy braking, maintaining rolling contact rather than sliding. Because μ_s > μ_k, rolling contact uses a higher coefficient than sliding, allowing a larger deceleration force and shorter stopping distance. ABS also preserves steering control during emergency braking, further improving safety.

Marking criteria (7 marks). 1 = correctly identifies the claim as scientifically inaccurate with a clear statement. 1 = applies f = μmg and a = F/m = μg to show mass cancels, so deceleration is independent of mass. 1 = uses kinematics to show stopping distance is also mass-independent under constant μ. 1 = identifies a valid condition under which the claim partially holds (e.g. limited braking force, brake fade). 1 = develops the partial condition with correct physics reasoning (fixed force / larger mass gives smaller a). 1 = describes one engineering strategy (ABS, crumple zones, regenerative braking) and explains the physics correctly. 1 = reaches an explicit evaluative judgement integrating when the claim is correct vs incorrect, with precise terminology (Newton’s Second Law, F_net = ma, μ_k, stopping distance).