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Physics  ·  Year 12  ·  Module 5  ·  Lesson 5

HSC Exam Practice

Practical Investigation: Validating Projectile Motion

9 questions / 3 sections / 34 marks total
Section 1

Short answer

1.Short answer

1.1

State the theoretical relationship between horizontal range R and launch height h for a projectile launched horizontally at constant speed vx. Explain why plotting R against √h, rather than against h, produces a straight-line graph that passes through the origin.

3marks Band 3
1.2

Identify the independent variable, dependent variable, and one controlled variable in the horizontal-launch projectile practical described in the lesson. For each controlled variable you state, explain how it is controlled in the procedure.

4marks Band 3
1.3

Describe how using carbon paper placed over white paper on the floor improves the accuracy of the range measurement compared with trying to observe where the ball hits the floor directly. Identify the type of error (random or systematic) that carbon paper primarily addresses.

3marks Band 3–4
1.4

Explain why the ball bearing must be released from rest at the same marked position on the ramp for every trial. In your answer, name the variable that would be confounded if the release position changed, and use the equation R = vx√(2h/g) to justify your reasoning.

4marks Band 4
1.5

Outline why measuring time of flight with a handheld stopwatch introduces significant random error in this experiment. Propose one alternative measurement technique that would reduce this error and briefly explain why it is superior.

3marks Band 4
1.6

Describe what the physical meaning of the gradient of the R vs √h graph is in terms of measurable quantities. A student obtains a gradient of 0.890 m½. Use vx = mexp√(g/2) to find the launch speed implied by this gradient. Take g = 9.80 m s−2.

3marks Band 4
Section 2

Data response

2.Data response — horizontal-launch experiment

2.1

A student conducts the horizontal-launch ball-bearing practical. The data recorded are shown in the table below. Use g = 9.80 m s−2.

h (m) h (m½) R1 (m) R2 (m) R3 (m) Mean (m)
0.1000.3160.300.320.340.32
0.2000.4470.440.460.450.45
0.3000.5480.540.550.560.55
0.4000.6320.620.640.660.64
0.5000.7070.690.710.730.71
Table 2.1. Primary data from horizontal-launch ball-bearing experiment. Launch speed vx held constant by fixed release position. g = 9.80 m s−2. Illustrative data.

(a) Using the data in the table, calculate the theoretical time of flight at h = 0.400 m using t = √(2h/g). Show your working and give the answer to 3 significant figures. (2 marks)

(b) For the h = 0.400 m row, use the mean range R̄ = 0.64 m to calculate the experimental launch speed vx using the formula vx = R̄ / t, where t is your answer from part (a). Compare this to the value obtained from rearranging R = vx√(2h/g). (3 marks)

(c) The student calculates the experimental gradient from the R̄ vs √h data and obtains mexp = 1.00 m½. Calculate the theoretical gradient mtheory = vx√(2/g) using vx from part (b), then determine the percentage difference. State whether the model is validated using the <10% criterion from the lesson. (3 marks)

8marks Band 4–5
Section 3

Extended response

3.Extended response

3.1

Evaluate the design of the horizontal-launch ball-bearing practical described in the lesson as a method for validating the projectile motion model R = vx√(2h/g). In your response you must:

  • explain what specific prediction of the theoretical model the experiment tests, and justify why a linearised R vs √h graph is used instead of R vs h,
  • assess how well the design controls the three key controlled variables (launch speed, launch angle, and landing measurement precision),
  • distinguish between random and systematic errors present in the design, citing a specific example of each and explaining how each is addressed,
  • evaluate whether a percentage difference of less than 10% between the experimental gradient and the theoretical gradient constitutes valid support for the model, identifying one assumption of the model that cannot be verified by this experiment alone.
9marks Band 5–6

Physics · Year 12 · Module 5 · Lesson 5

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 3 marks · Band 3

Sample response. For a horizontal launch from height h, the vertical free-fall gives time of flight t = √(2h/g). The horizontal range is then R = vx·t = vx√(2h/g). This is a square-root function of h, not a linear one, so a direct R vs h plot is a curve. Writing R = [vx√(2/g)] × √h puts it in the form y = m√h, which is linear in √h with gradient m = vx√(2/g) and zero intercept. Plotting R vs √h therefore gives a straight line through the origin, which is easier to confirm visually and numerically as evidence for the model.

Marking notes. 1 mark for stating the relationship R = vx√(2h/g) or equivalent; 1 mark for explaining that R is proportional to √h (not h), making R vs h a curve; 1 mark for explaining that plotting R against √h linearises the relationship, producing a straight line through the origin with gradient vx√(2/g).

1.2

Section 1 · Short answer · 4 marks · Band 3

Sample response. Independent variable: launch height h (the exit height of the ramp above the floor), varied deliberately across at least five values. Dependent variable: horizontal range R (measured from the point directly below the ramp exit to the centre of the landing mark). Controlled variable (one from): launch speed vx — controlled by releasing the ball from the same marked position on the ramp from rest every trial, so the same gravitational PE is converted to the same KE at the exit. OR: launch angle — controlled by checking the ramp exit is horizontal with a spirit level before each set of trials, so θ = 0° for every launch. OR: the ball used — controlled by using the same dense steel ball bearing for every trial, keeping mass, diameter and air-resistance properties constant.

Marking notes. 1 mark IV (launch height h); 1 mark DV (horizontal range R); 1 mark for correctly naming one controlled variable; 1 mark for a correct method of control for that variable. Accept any one well-described controlled variable from the lesson.

1.3

Section 1 · Short answer · 3 marks · Band 3–4

Sample response. Carbon paper transfers the ink of the ball's first contact point to the white paper below, providing a clear permanent mark of the exact landing location. This is more accurate than naked-eye observation because the ball can bounce or roll after landing, making it impossible to identify the first-contact point by eye. Carbon paper marks the first impact position, not a post-bounce position. This primarily addresses random error in the range measurement: without carbon paper, different observers (or the same observer at different heights) might read slightly different landing positions, introducing scatter into R values at each height. The carbon paper mark is objective and does not vary with observer reaction time.

Marking notes. 1 mark for explaining that carbon paper marks the first contact point, preventing ambiguity from bouncing/rolling; 1 mark for stating that this is not dependent on observer reaction time or judgment; 1 mark for correctly identifying the error type as random (variation in observed landing position across trials).

1.4

Section 1 · Short answer · 4 marks · Band 4

Sample response. The ball is released from rest at the same marked position to ensure the same gravitational potential energy (mghramp) is converted to kinetic energy (½mvx2) at the exit, giving a constant horizontal launch speed vx. If the release position changed, vx would change. From R = vx√(2h/g), R depends on both vx and h: if vx varies between trials at different heights, the change in R cannot be attributed solely to the change in h. The variable confounded is the launch speed vx. Since the experiment aims to measure how R depends on h (with vx constant), an uncontrolled vx makes the gradient of the R vs √h graph meaningless as a test of the model.

Marking notes. 1 mark for explaining that release position determines vx via energy conversion; 1 mark for naming the confounded variable (launch speed vx); 1 mark for using the equation R = vx√(2h/g) to show that both vx and h appear in R, so varying vx prevents isolating the h dependence; 1 mark for stating that the R vs √h gradient would be invalid as a test of the model if vx is uncontrolled.

1.5

Section 1 · Short answer · 3 marks · Band 4

Sample response. A handheld stopwatch introduces large random error because human reaction time (≈0.2 s) is comparable to the actual time of flight (e.g. t = √(2 × 0.40 / 9.80) ≈ 0.29 s at h = 0.40 m). Starting and stopping the stopwatch by hand adds an unpredictable timing error of up to ±0.2 s — almost as large as the quantity being measured, making the result unreliable. An improvement: use a slow-motion video camera or a motion sensor to record the moment of launch and the moment of landing. These methods can resolve events to within a few milliseconds, reducing random timing uncertainty by two orders of magnitude compared with manual stopwatch operation.

Marking notes. 1 mark for quantifying or explaining that human reaction time (≈0.2 s) is comparable to or significant relative to the time of flight; 1 mark for identifying this as random error (varies each trial); 1 mark for naming a superior technique (slow-motion video, motion sensor, or photogate pair) and giving a correct reason why it reduces timing uncertainty.

1.6

Section 1 · Short answer · 3 marks · Band 4

Sample response. The gradient of the R vs √h graph equals vx√(2/g), where vx is the horizontal launch speed and g is the acceleration due to gravity. Physically, it represents the proportionality constant relating range to the square root of launch height — a larger gradient means the ball is launched faster (larger vx). For mexp = 0.890 m½: vx = mexp × √(g/2) = 0.890 × √(9.80/2) m s−1 = 0.890 × √(4.90) m s−1 = 0.890 × 2.214 m s−1 = 1.97 m s−1.

Marking notes. 1 mark for stating that the gradient = vx√(2/g) and explaining its physical meaning in terms of launch speed; 1 mark for correct substitution into vx = mexp√(g/2); 1 mark for the correct numerical answer 1.97 m s−1 (accept 1.96–1.98 m s−1). Units check: m½ × √(m s−2) = m½ × m½ s−1 = m s−1.

2.1

Section 2 · Data response · 8 marks · Band 4–5

Sample response (a) — time of flight at h = 0.400 m. t = √(2h/g) = √((2 × 0.400 m) / (9.80 m s−2)) = √(0.08163 s2 × 10 — recalculate: 2 × 0.400 = 0.800; 0.800 / 9.80 = 0.08163; √0.08163 = 0.2857 s) ⇒ t = √(0.8000 / 9.80) = √(0.08163) = 0.2857 s. t = 0.286 s (3 sig figs). Units: √(m / (m s−2)) = √(s2) = s ✓.

Sample response (b) — launch speed from R = 0.64 m, t = 0.286 s. vx = R / t = 0.64 m / 0.286 s = 2.238 m s−1. Cross-check using rearranged formula: vx = R × √(g / 2h) = 0.64 × √(9.80 / 0.800) m s−1 = 0.64 × √(12.25 s−2) = 0.64 × 3.500 s−1 = 2.240 m s−1. Both methods give vx ≈ 2.24 m s−1 (3 sig figs) — the close agreement confirms consistency of the two approaches. Units: m / s = m s−1 ✓.

Sample response (c) — gradient comparison. Theoretical gradient: mtheory = vx√(2/g) = 2.24 × √(2/9.80) m½ = 2.24 × 0.4518 m½ = 1.012 m½. % difference = |mexp − mtheory| / mtheory × 100% = |1.00 − 1.012| / 1.012 × 100% = 0.012 / 1.012 × 100% = 1.19%. Because 1.2% < 10%, the model is validated by this experiment.

Marking notes. Part (a): 1 mark for correct substitution; 1 mark for t = 0.286 s (accept 0.285–0.287 s). Part (b): 1 mark for applying vx = R/t correctly; 1 mark for the correct numerical answer ~2.24 m s−1; 1 mark for confirming the result using the rearranged formula or noting the two methods agree. Part (c): 1 mark for calculating mtheory using their vx; 1 mark for correct % difference calculation; 1 mark for correctly invoking the <10% criterion and concluding the model is validated.

3.1

Section 3 · Extended response · 9 marks · Band 5–6

Sample Band 6 response (annotated by mark).

[Mark 1 — testable prediction and linearisation] The experiment tests the specific prediction R ∝ √h: if horizontal launch speed vx is held constant, the theoretical model R = vx√(2h/g) predicts that range scales as the square root of launch height. A direct R vs h graph would be a curve, which is harder to assess statistically. Plotting R vs √h linearises this to y = mx, where gradient m = vx√(2/g) and the y-intercept should be zero; this provides two simultaneous tests: the linearity of the relationship, and the zero-intercept prediction (that range is zero when launch height is zero).

[Mark 2 — controls launch speed] The launch speed vx is controlled by releasing the ball from rest at the same marked position on the ramp for every trial. This uses energy conservation (mghramp = ½mvx2) to ensure reproducible exit speed. The control is effective as long as rolling friction on the ramp is negligible and the ramp geometry does not change between trials. A limitation is that rolling friction is not directly measured, so a small systematic reduction in vx due to friction cannot be ruled out.

[Mark 3 — controls launch angle] The launch angle is controlled by checking the ramp exit is horizontal with a spirit level. This ensures θ = 0°, so the vertical initial velocity component is zero and the model t = √(2h/g) (which assumes vy0 = 0) applies. A slight tilt of the ramp exit would introduce a non-zero vy0, causing the actual range to differ from the model prediction — a systematic error in every trial at a given height.

[Mark 4 — controls landing precision] Range measurement precision is improved by using carbon paper over white paper to mark the first-contact landing point (eliminating post-bounce ambiguity) and a plumb line to locate the point on the floor directly below the ramp exit (eliminating a constant horizontal offset). Three trials are performed at each height to average out random scatter in landing marks.

[Mark 5 — random error identified and addressed] Random error arises from trial-to-trial variation in the exact landing mark (due to minor differences in ball spin, surface roughness, or carbon-paper impression). This is addressed by averaging three trials at each height: the mean range R̄ reduces the influence of any single anomalous mark. The spread of the three values can be used to estimate a range uncertainty, quantifying how well the random error has been controlled.

[Mark 6 — systematic error identified and addressed] A key systematic error would arise if the ramp exit is not exactly horizontal — even a 2° downward tilt would introduce a small initial downward vy, reducing range by the same proportion at every height and causing the experimental gradient to be consistently below the theoretical value. This is addressed (but not eliminated) by the spirit-level check before trials begin. A second systematic error is air resistance: the model assumes it is negligible, but in reality a light ball would experience drag that reduces R below the theoretical prediction at higher speeds (i.e., larger h). Using a dense steel ball bearing minimises this effect.

[Mark 7 — evaluates the <10% criterion] The <10% percentage difference criterion is a reasonable practical threshold for a Year 12 school experiment, but it is not a rigorous statistical test. A gradient difference of 9.9% could still be physically significant if the experimental uncertainties are only ±2–3%. Strictly, the model is validated at the precision of the experiment if the experimental gradient is consistent with the theoretical value within the combined uncertainty of both, not merely within an arbitrary 10% band. Nevertheless, for this investigation, a percentage difference well under 10% provides strong qualitative support for the model.

[Mark 8 — unverifiable assumption] One assumption that this experiment cannot verify is that horizontal and vertical motions are truly independent (i.e., that no horizontal force acts during the flight). The experiment confirms that R vs √h is linear (consistent with t = √(2h/g) controlling flight time) and that the implied vx is approximately constant, but it cannot separately measure the horizontal velocity during flight to confirm vx is unchanged from launch to landing. A photogate at both the exit and the landing position would be needed to confirm this directly.

[Mark 9 — overall evaluative judgement] Overall, the design is well-suited to its purpose: it tests the specific prediction R ∝ √h, systematically controls the three key variables, addresses both types of error through its procedure, and provides a quantitative comparison via the gradient. Its main limitation is that it cannot independently verify that vx is constant during flight (only at launch) and cannot eliminate rolling friction as a small systematic confound. With the addition of a photogate at the ramp exit, the design would provide a direct, independent measurement of vx to compare with the gradient-derived value, substantially strengthening the validation.

Marking criteria summary (9 marks):

  • 1 = states specific testable prediction (R ∝ √h) and justifies linearisation with correct equation.
  • 1 = correctly assesses control of launch speed (same release position, energy conservation argument).
  • 1 = correctly assesses control of launch angle (spirit level, θ = 0°).
  • 1 = correctly assesses control of landing precision (carbon paper, plumb line, 3 trials).
  • 1 = identifies a specific random error and explains how the design addresses it.
  • 1 = identifies a specific systematic error (ramp tilt or air resistance) and explains how the design addresses it.
  • 1 = evaluates the <10% criterion with reasoning (e.g., compares to experimental uncertainty; notes it is a practical threshold, not a rigorous statistical test).
  • 1 = identifies one assumption that cannot be verified by this experiment (e.g., independence of horizontal and vertical motion, constant vx during flight).
  • 1 = reaches an explicit integrative evaluative conclusion about the overall quality of the design, including a targeted improvement.

Band descriptors: Band 3 (1–3 marks): names a variable and a formula; describes the experiment narratively. Band 4 (4–5 marks): uses the linearisation argument; correctly identifies random vs systematic error. Band 5 (6–7 marks): evaluates the 10% criterion and connects design features to specific errors. Band 6 (8–9 marks): identifies an unverifiable assumption, proposes a targeted improvement, and reaches a coherent, quantitatively grounded evaluative judgement.