HSC Exam Practice

Sample response. Uniform acceleration is a constant rate of change of velocity — equal increases (or decreases) in velocity occur in equal time intervals. On a v–t graph, uniform acceleration appears as a straight line (either through the origin if starting from rest, or from an initial velocity).

Marking notes. 1 mark for defining uniform acceleration as constant rate of change of velocity; 1 mark for correctly describing the v–t graph appearance (straight line). Do not award full marks for “constant speed” — must reference velocity.

Sample response. (a) The gradient of a v–t graph represents acceleration, with units of m s⁻². (b) The area under a v–t graph represents displacement, with units of metres (m).

Marking notes. 1 mark for gradient = acceleration with correct units; 1 mark for area = displacement with correct units. Both are needed for full credit.

Sample response. Newton’s Second Law states F_net = ma. This can be rearranged to F = m × a, which has the same form as the equation of a straight line y = mx. When F is plotted on the y-axis and a on the x-axis, the equation F = m × a means the gradient of the straight line equals m — the mass of the object. Therefore, the gradient of an F vs a graph gives the mass of the object, in kg (since N ÷ m s⁻² = kg).

Marking notes. 1 mark for stating F_net = ma and rearranging to F = m × a; 1 mark for explicitly comparing to y = mx and identifying gradient = m (mass); 1 mark for confirming the units: N ÷ (m s⁻²) = kg.

Sample response. Step 1: Draw and label axes with the variable name, symbol, and unit in brackets (e.g. “Net Force, F (N)”). Step 2: Plot all data points accurately and draw a line of best fit (straight or curved to represent the trend). Step 3: Calculate the gradient using two widely-spaced points ON the line of best fit — not from the data table. Show rise/run working. Step 4: Interpret the gradient in physics terms using NESA language — state what physical quantity the gradient represents, link it to the relevant equation (e.g. “The gradient represents mass because F = ma has the form y = mx where gradient = m”), and include units.

Marking notes. 1 mark per step. For Step 4: must specify that the gradient is stated in physics terms with a link to an equation — not just a number.

Sample response. Data points contain random experimental error (measurement uncertainty from timing devices, sensors, or friction). Using only two specific data points directly imports this error into the gradient, making it unreliable and possibly unrepresentative of the true trend. Instead, the student should draw a line of best fit through all data points (averaging out the random errors) and then use two widely-spaced points on that line to calculate the gradient.

Marking notes. 1 mark for explaining that data points contain experimental error that makes the gradient unreliable; 1 mark for stating the correct method (use points on the line of best fit). Accept “the line of best fit averages out errors”.

Sample response. Gradient of v–t graph: represents acceleration (m s⁻²); calculated as Δv / Δt; tells us the rate of change of velocity. Gradient of F vs a graph: represents mass (kg); calculated as ΔF / Δa; tells us the inertia of the object and confirms Newton’s Second Law (F = ma in y = mx form). They differ in both the physical quantity they represent and their units.

Marking notes. 1 mark for v–t gradient = acceleration with units m s⁻²; 1 mark for F vs a gradient = mass with units kg. Must distinguish both the quantity and units to receive full credit.

Sample response (a) — gradient calculation. Using P1(1.0 m s⁻², 3.0 N) and P2(5.0 m s⁻², 15.0 N) on the line of best fit [1]: rise = ΔF = 15.0 − 3.0 = 12.0 N; run = Δa = 5.0 − 1.0 = 4.0 m s⁻² [1]. Gradient = 12.0 / 4.0 = 3.0 kg [1]. Units: N ÷ (m s⁻²) = kg.

Sample response (b) — NESA interpretation. The gradient of the F vs a graph represents the mass of the cart (3.0 kg) [1], because F_net = ma can be rearranged to F = m × a, which has the form y = mx where the gradient m equals the mass of the object [1]. Therefore the mass of the dynamics cart is 3.0 kg [1].

Sample response (c) — origin and y-intercept. The line passing through the origin confirms that net force is directly proportional to acceleration at constant mass (F ∝ a) [1], consistent with Newton’s Second Law. A non-zero positive y-intercept would indicate a constant friction force acting on the cart that was not accounted for in the net force measurement; this systematic error would mean the applied force is always greater than the net force by a constant amount equal to the friction force [1].

Marking notes. Part (a): 1 mark for identifying P1 and P2 on the line (not data points); 1 mark for correct rise and run values with working; 1 mark for correct gradient = 3.0 kg. Part (b): 1 mark for stating mass = 3.0 kg; 1 mark for referencing F = m × a and y = mx; 1 mark for explicit NESA-standard interpretation sentence. Part (c): 1 mark for direct proportionality from straight line through origin; 1 mark for explaining that a positive y-intercept indicates constant friction/systematic error.

Sample response (a) — acceleration. Positive direction: forward (direction of motion). a = Δv / Δt = (20.0 − 8.0) / 6.0 = 12.0 / 6.0 = 2.0 m s⁻² (forward) [1]. The car undergoes uniform acceleration because the v–t relationship is linear [1].

Sample response (b) — net force. F_net = ma = 1200 × 2.0 = 2400 N (forward) [1]. Both mass and acceleration must be included with correct multiplication [1].

Sample response (c) — v–t sketch. Straight line from (0 s, 8.0 m s⁻¹) to (6.0 s, 20.0 m s⁻¹) [1]. Labels: initial velocity = 8.0 m s⁻¹; final velocity = 20.0 m s⁻¹; gradient = acceleration = 2.0 m s⁻²; shaded triangular area = displacement. [1]

Sample response (d) — assumption. Assumption: the acceleration is uniform (constant) throughout the 6.0 s interval [1]. If the acceleration is not uniform (varies with time), the v–t graph would be curved rather than straight, and the formula Δv / Δt would give an average acceleration rather than the instantaneous acceleration at any moment.

Marking notes. (a) 1 mark for correct formula and substitution; 1 mark for correct answer 2.0 m s⁻² with direction stated and positive direction defined. (b) 1 mark for F = ma correctly written; 1 mark for F = 2400 N with units and direction. (c) 1 mark for correct straight line with correct end-values; 1 mark for at least three of four required labels. (d) 1 mark for identifying uniform acceleration as the assumption, with a valid consequence if violated. Accept “frictionless” or “constant net force” as alternative assumptions with appropriate consequences.

Sample response. Graphical analysis of the trolley experiment allows Newton’s Second Law to be derived from data rather than imposed as an assumption. When a student applies different known net forces to a trolley of constant mass and measures the resulting acceleration, they generate a set of (a, F) pairs. Plotting F on the y-axis and a on the x-axis and drawing a line of best fit produces a graph of the form y = mx. The key insight is that this is not a check that F = ma is true — it is the experimental discovery of it. The student measures the gradient of their best-fit line and finds it has units of kg; by identifying this as the mass of the trolley, they have experimentally derived the form of Newton’s Second Law for themselves. The shape of the graph provides further information: a straight line through the origin is the graphical signature of direct proportionality (F ∝ a). If the line did not pass through the origin, it would indicate a systematic error — for example, a constant friction force on the trolley would shift the line upward by a constant amount equal to the friction, producing a positive y-intercept. The existence of such a y-intercept is therefore physically meaningful, not just a graphing artefact. In the trolley experiment conducted in many NSW HSC laboratories, a student who draws the line of best fit and correctly interprets the gradient as mass (confirmed by units: N ÷ m s⁻² = kg) and the origin-passing as direct proportionality has done something richer than verification: they have followed the same logical path as Newton, inferring a law from systematic measurement. A steeper line on the F vs a graph would indicate a greater mass; this can be confirmed by adding known mass blocks to the trolley and checking that the gradient increases by the same amount as the added mass. The requirement to use two widely-spaced points on the line of best fit (not data points) for the gradient calculation is also significant: it forces the student to distinguish between the theoretical trend and the noisy individual measurements, reinforcing the role of the line of best fit as a statistical model of the data rather than a connection of individual dots.

Marking criteria (7 marks). 1 = correctly identifies the gradient of the F vs a graph as mass with units kg (N ÷ m s⁻²). 1 = correctly links gradient to Newton’s Second Law in the form F = ma (y = mx). 1 = distinguishes between deriving the law (the graph shape provides the evidence) versus simply verifying a pre-assumed equation. 1 = correctly explains the significance of the straight line through the origin as evidence of direct proportionality (F ∝ a). 1 = correctly explains what a non-zero y-intercept would indicate (systematic error, e.g. constant friction). 1 = references a specific, realistic experimental context (trolley experiment, hanging mass + pulley, motion sensor, or comparable). 1 = reaches an explicit evaluative judgement: graphical analysis is a tool for deriving laws from data, not just checking them; discusses the requirement for line of best fit (averaging errors) or the significance of gradient interpretation step in NESA responses.