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].