Physics • Year 11 • Module 2 • Lesson 13

Momentum Synthesis

Apply stage-identification, data interpretation and cause-and-effect reasoning to real multi-step scenarios.

Apply · Data & Reasoning

1. Classify each stage — four scenarios

The table below describes four real-world scenarios. For each row, identify the stage type, name the correct formula, and state the key input and output. 12 marks (1 per cell, 3 per row)

Scenario	Stage type	Correct formula	Key input → output
A rifle (5 kg) fires a 0.01 kg bullet. The rifle-bullet system was at rest.
The bullet strikes a 2 kg stationary block and embeds in it.
The block-plus-bullet slides 1.2 m on a rough floor before stopping.
A batsman hits a 0.16 kg cricket ball. Contact time is 0.004 s. Ball velocity changes from −40 m/s to +60 m/s.

Stuck? Use the Decision Flowchart from the lesson: explosion? → objects stick? → sliding/stopping?

2. Interpret a graph — post-collision slide distance vs initial speed

The graph below shows the slide distance of a combined block-plus-ball mass (total 2 kg, μ_k = 0.30) after a perfectly inelastic collision, for different initial ball speeds. The block was stationary before each collision. Ball mass = 0.5 kg. 7 marks

Figure 1. Slide distance of combined mass (2 kg) on a rough surface (μ_k = 0.30) versus initial ball speed. Ball mass = 0.5 kg; block initially stationary. Illustrative data calculated from momentum conservation + W_net = ΔKE.

2.1 Describe the shape of the curve and explain what it tells you about the relationship between initial speed and slide distance. 2 marks

2.2 Use the graph to estimate the slide distance when the initial ball speed is 7 m/s. Show how you arrive at your estimate. 2 marks

2.3 A student states: “Doubling the initial ball speed doubles the slide distance.” Use the graph to assess whether this statement is correct, and explain the underlying physics. 3 marks

Stuck? Think about how slide distance depends on KE, which depends on v².

3. Cause-and-effect chain — bullet-block-slide

The cause boxes below are filled. Write the matching effect in each empty box. Each effect should be one sentence. 5 marks (1 per effect)

Cause	→	Effect (write here)
The rifle-bullet system starts at rest so total momentum = 0.	→
The bullet (small mass) is fired with a very high speed.	→
The bullet embeds in the block — a perfectly inelastic collision.	→
After the collision, the combined block-plus-bullet carries only a tiny fraction of the original KE.	→
Friction acts on the sliding block-plus-bullet over a distance s.	→

Overall outcome (so…): ____________________________________________________________________

Stuck? Work through Cards 1 and 2 and the worked examples in the lesson.

4. Predict and justify — changing the bullet mass

A 5 kg rifle fires a bullet into a 2 kg stationary block. In Version A, the bullet mass is 0.005 kg. In Version B, the bullet mass is 0.05 kg. In both cases the rifle recoils at the same speed (−0.6 m/s). The block slides on a surface with μ_k = 0.12. 5 marks

4.1 Without fully solving, predict which version produces a greater slide distance for the block. Justify your prediction with reference to how momentum transfers through the chain. 3 marks

4.2 A student argues: “A heavier bullet must always produce a shorter slide distance because more mass is lost in the collision.” Identify the flaw in this argument. 2 marks

Stuck? Trace: rifle recoil → rifle momentum → bullet momentum → collision → post-collision v_f → KE_f → slide distance.

Answers — Do not peek before attempting

Q1 — Stage classification table

Row 1 (rifle fires bullet): Stage type: explosion from rest. Formula: 0 = m_bv_b′ + m_rv_r′. Input: masses + initial state at rest. Output: bullet velocity v_b′.

Row 2 (bullet embeds in block): Stage type: perfectly inelastic collision. Formula: m_bv_b′ + m_block×0 = (m_b+m_block)v_f. Input: bullet velocity from Stage 1. Output: post-collision velocity v_f.

Row 3 (block slides to rest): Stage type: Phase 2 bridge (friction stopping). Formula: −f_k×s = 0 − KE_f. Input: v_f from Stage 2. Output: slide distance s.

Row 4 (batsman hits ball): Stage type: impulse. Formula: J = FΔt = Δp = m(v_f−v_i). Input: mass, initial and final velocities, contact time. Output: average force on ball. J = 0.16×(60−(−40)) = 0.16×100 = 16 N s. F = 16/0.004 = 4000 N.

Q2.1 — Shape and relationship (2 marks)

The curve is parabolic (concave up), indicating that slide distance increases as the square of the initial ball speed. This is because KE ∝ v², and slide distance ∝ KE (from W_net = ΔKE). So doubling the speed quadruples the slide distance [1 for shape + trend; 1 for physics explanation].

Q2.2 — Estimate at 7 m/s (2 marks)

From the graph, at v = 6 m/s, s ≈ 0.45 m and at v = 8 m/s, s ≈ 0.80 m. Linear interpolation midway: s ≈ (0.45+0.80)/2 = 0.63 m. Accept answers in the range 0.60–0.65 m [1 for method; 1 for answer in acceptable range]. Exact calculation: v_f = 0.5×7/2 = 1.75 m/s; KE_f = ½×2×1.75² = 3.06 J; f_k = 0.30×2×9.8 = 5.88 N; s = 3.06/5.88 = 0.52 m.

Q2.3 — Doubling speed assessment (3 marks)

The statement is incorrect [1]. From the graph: at v = 4 m/s, s ≈ 0.20 m; at v = 8 m/s, s ≈ 0.80 m — four times larger, not double [1]. The physics: post-collision velocity v_f ∝ v_initial; KE_f = ½mv_f² ∝ v_initial²; and slide distance s = KE_f/f_k ∝ v_initial². Therefore doubling the initial speed quadruples the slide distance [1].

Q3 — Cause-and-effect chain (sample answers)

Cause 1 → Effect 1: The rifle recoils backward with a velocity equal and opposite (in momentum) to the bullet’s forward momentum.

Cause 2 → Effect 2: The bullet carries a very large momentum (p = mv) despite its small mass.

Cause 3 → Effect 3: Momentum is conserved but most kinetic energy is converted to heat and deformation; the combined mass moves forward at a much lower speed.

Cause 4 → Effect 4: The block slides only a short distance before stopping.

Cause 5 → Effect 5: Friction does negative work on the block, removing all remaining kinetic energy over the slide distance s.

Overall outcome: The block comes to rest after a calculable slide distance that confirms all three stages of the chain are physically consistent.

Q4.1 — Prediction (3 marks)

Version B (heavier bullet, 0.05 kg) produces a greater slide distance [1]. With the same rifle recoil speed, the rifle momentum is constant at 5×0.6 = 3 N s. By conservation, the bullet carries +3 N s in both versions. Version A: p_bullet = 3 N s → v_bullet = 3/0.005 = 600 m/s. Version B: v_bullet = 3/0.05 = 60 m/s. After the perfectly inelastic collision, v_f(A) = 0.005×600/(0.005+2) ≈ 1.496 m/s; v_f(B) = 0.05×60/(0.05+2) ≈ 1.463 m/s. These are nearly equal, so slide distances are nearly equal. Award marks for correct reasoning through the chain [2], correct conclusion relative to their working [1].

Q4.2 — Flaw in student’s argument (2 marks)

The flaw is that the student conflates “mass of the combined block+bullet system” with the transferred momentum [1]. The determining factor for slide distance is the post-collision kinetic energy, which depends on v_f². The post-collision velocity is set by momentum conservation, not simply by the bullet’s mass. A heavier bullet does not necessarily produce a shorter slide — it depends on how momentum is distributed through the chain [1].