Physics • Year 11 • Module 2 • Lesson 1
Forces and Interactions
Build HSC Band 5–6 extended-response technique: apply Newton’s Laws to real data and design investigations around force interactions.
1. Data + scenario: rocket sled deceleration test at Woomera (Band 5–6)
8 marks Band 5–6
Scenario. In the 1960s, engineers at Woomera, South Australia, conducted rocket sled deceleration tests to evaluate the forces experienced by aircraft seats during emergency ejections. A 120 kg test sled (plus a 75 kg dummy) reached 300 m/s before a set of water brake troughs brought it to rest. Force sensors recorded the net braking force every 0.1 s. The table below summarises key data from one trial.
| Time (s) | Sled velocity (m/s) | Net braking force (N) | Direction of net force |
|---|---|---|---|
| 0.0 | 300 | 0 | — |
| 0.1 | 280 | 38 950 | backward (opposing motion) |
| 0.2 | 240 | 58 425 | backward |
| 0.3 | 180 | 58 425 | backward |
| 0.4 | 110 | 48 125 | backward |
| 0.5 | 40 | 38 950 | backward |
| 0.6 | 0 | 0 | — |
Illustrative data inspired by rocket sled test programs, Woomera Range Complex, South Australia. Total mass = 195 kg (sled + dummy).
Q1. Analyse and evaluate the data to explain the force interactions during the braking phase. In your response you must:
- State Newton’s First Law and explain what it predicts about the sled’s motion before and at the end of braking (t = 0.0 s and t = 0.6 s).
- Identify the Newton’s Third Law pair for the braking force at t = 0.2 s, stating the force type, magnitude, and direction of each force in the pair.
- Explain, using the data, why the net braking force is not constant throughout the test and what this implies about the sled’s acceleration at each time step.
- Assess whether the sled is in equilibrium at any point during the trial, with specific reference to the data.
- Identify one limitation of using net force values alone (without the individual force components) to fully describe this experiment.
2. Experimental design — testing Newton’s Third Law with bathroom scales (Band 5–6)
7 marks Band 5–6
Research question. A Year 11 student claims: “Newton’s Third Law only applies when both objects are moving. If one object is stationary, the forces are not equal.” Design an investigation using two bathroom scales placed face-to-face, pushed together by two students of different masses, to test this claim.
Constraints: You have two identical bathroom scales (reading to 0.1 kg), a smooth floor, two students of different body masses, and a 30 cm ruler. The investigation must be completable in one 60-minute lesson.
Q2. Design the investigation and present it in the format below.
- State a testable hypothesis that includes the independent and dependent variables.
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will measure and compare the forces on each scale.
- Predict the result if Newton’s Third Law holds, and state what result would falsify your hypothesis.
- State two limitations of the bathroom scale method and suggest one improvement.
Q1 — Sample Band 6 response (8 marks), annotated
Newton’s First Law and motion prediction: Newton’s First Law states that an object will remain at rest or continue moving at constant velocity in a straight line unless acted upon by a net external force [1]. At t = 0.0 s the sled is moving at 300 m/s and net force = 0; without the water brakes the sled would continue at 300 m/s indefinitely (dynamic equilibrium). At t = 0.6 s the sled has velocity = 0 and net force = 0; the sled is now in static equilibrium and will remain at rest [1].
Third Law pair at t = 0.2 s: At t = 0.2 s, the net braking force is 58 425 N directed backward (opposing motion) [1]. The Newton’s Third Law pair: (1) the water brake troughs exert a contact force of 58 425 N backward on the sled; (2) the sled exerts a contact force of 58 425 N forward on the water. Both are contact forces, equal in magnitude, opposite in direction, acting on different objects (sled and water) [1].
Non-constant net force and acceleration: The net braking force is not constant: it rises from 0 at t = 0 to a peak of 58 425 N at t = 0.2–0.3 s, then decreases as the sled slows [1]. Because a net force causes acceleration (F = ma), a varying net force implies a varying acceleration. At t = 0.2 s, the magnitude of deceleration = 58 425 / 195 ≈ 300 m/s²; at t = 0.1 s it is 38 950 / 195 ≈ 200 m/s². The non-uniform deceleration is consistent with a water brake that generates more force at higher relative speed [1].
Equilibrium assessment: The sled is in dynamic equilibrium only at t = 0.0 s (constant velocity, net force = 0) and in static equilibrium at t = 0.6 s (at rest, net force = 0) [1]. During the interval t = 0.1 s to t = 0.5 s the sled is not in equilibrium: net force is non-zero and the velocity is changing, confirming a net force causes the change in motion.
Limitation: The net force data does not distinguish between individual force components (e.g. the braking force from the water brakes versus any residual friction from the rails or air resistance). Knowing only the net force is insufficient to fully characterise the interaction between the sled and each separate element of the braking system [1].
Marking criteria summary (8 marks): 1 = correct statement of Newton’s First Law; 1 = correct prediction for t = 0.0 and t = 0.6 s with equilibrium terminology; 1 = Third Law pair identified with force type; 1 = correct magnitude and direction of both forces in the pair; 1 = explanation of varying force and linked varying acceleration; 1 = at least one calculated acceleration value using F = ma; 1 = correct assessment of equilibrium at t = 0.0 and t = 0.6 with explicit data reference; 1 = valid, specific limitation of net-force-only data.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If Newton’s Third Law applies regardless of the motion of the objects, then the reading on Scale A (force felt by Person B) will equal the reading on Scale B (force felt by Person A) for every push force applied, even when both students are stationary. Independent variable: push force applied (varied by asking students to push gently, moderately, firmly). Dependent variable: the readings on each scale (kg, converted to N by × 9.8). Controlled variables: floor surface (smooth level floor to minimise friction), scale type (identical models), body posture of each student (arms straight, feet planted) [1].
Procedure: (1) Place Scale A face against Scale B so their display faces both point outward. Student X stands behind Scale A; Student Y (different mass) stands behind Scale B. Both students press firmly against their scale so their palms touch the scale face. (2) At a given signal, both students push against the scales simultaneously. Read Scale A and Scale B simultaneously — have a third student photograph the two displays at the same instant. Record the readings in a table. (3) Repeat for five different push intensities (gentle, moderate, firm, very firm, maximum). Record all pairs of readings. (4) Convert readings from kg to N (multiply by 9.8). Plot Scale A reading (y-axis) versus Scale B reading (x-axis); if Newton’s Third Law holds, the points will lie on the line y = x [1].
Predicted result and falsification: Newton’s Third Law predicts both scales will always read the same value, producing a straight line through the origin with gradient = 1 [1]. The hypothesis would be falsified if one scale consistently reads more than the other — for example, if the heavier student’s scale consistently exceeds the lighter student’s scale by a measurable amount beyond experimental error [1].
Limitations: (1) Bathroom scales sample force over a period of time (averaging); they cannot record rapid force spikes, so any dynamic effects during the push are lost [1]. (2) It is difficult to ensure both readings are taken at the exact same instant; a time lag between readings introduces error because force is changing throughout the push [1].
Improvement: Use two digital force plates connected to a data logger sampling at 100 Hz so both force readings are recorded simultaneously and continuously, eliminating the timing error [1].
Marking criteria summary (7 marks): 1 = testable hypothesis naming IV and DV; 1 = four clear steps including simultaneous reading method; 1 = predicted result (equal readings, y = x line); 1 = specific falsification condition; 1 = one valid limitation (timing or averaging); 1 = second valid limitation (different issue from first); 1 = specific improvement using appropriate equipment.