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Edexcel A-level Further Maths revision notes

Elastic collisions in one dimension

Section 4
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
2 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 4

Checked against Edexcel 9FM0 section 4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Direct impact of elastic spheres. Newton's law of restitution. Loss of kinetic energy due to impact.

Notes
Evidence from your answers: none yet
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A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a direct impact, choose one positive direction and conserve signed momentum along the line of centres. Newton's law of restitution states that speed of separation equals ee times speed of approach, with 0≤e≤10\leq e\leq1.
  • In signed form for AA behind BB, vB−vA=e(uA−uB)v_B-v_A=e(u_A-u_B). Solve the momentum and restitution equations simultaneously, then check that the post-impact velocities describe separation.
  • Spheres may be modelled as particles.
  • Momentum is conserved for the isolated system, but kinetic energy is generally lost; calculate loss as total kinetic energy before minus total kinetic energy after.
  • Examiners expect the non-negative root for ee and consistent velocity signs.
Worked example

A 3 kg3\,\text{kg} sphere moving at 5 m s−15\,\text{m s}^{-1} strikes a stationary 2 kg2\,\text{kg} sphere directly. The coefficient of restitution is 12\tfrac12. Find both velocities after impact.

  1. 1.Momentum: 15=3vA+2vB15=3v_A+2v_B.
  2. 2.Restitution: vB−vA=12(5)=2.5v_B-v_A=\tfrac12(5)=2.5.
  3. 3.Substitute vB=vA+2.5v_B=v_A+2.5: 15=5vA+515=5v_A+5, so vA=2v_A=2 and vB=4.5v_B=4.5.

Answer: The velocities are 2 m s−12\,\text{m s}^{-1} and 4.5 m s−14.5\,\text{m s}^{-1} in the original direction.

Common mistakes

  • Don't write speed of approach minus speed of separation in the restitution equation.
  • Don't conserve kinetic energy as well as momentum when e<1e<1.
  • Don't accept a negative algebraic root for ee despite 0≤e≤10\leq e\leq1.

Exam tip

Write momentum and restitution as two labelled equations before eliminating either final velocity.

Tier 1 · Easy

ORIGINAL

1.

Sphere AA approaches stationary sphere BB at 7 m s−17\,\text{m s}^{-1}. After their direct impact, AA continues in the same direction at 2 m s−12\,\text{m s}^{-1}. The coefficient of restitution is 0.40.4. Find the velocity of BB.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A 2 kg2\,\text{kg} sphere travelling at 6 m s−16\,\text{m s}^{-1} strikes a stationary 3 kg3\,\text{kg} sphere directly. The coefficient of restitution is 1/21/2. Find both velocities after impact and the kinetic energy lost.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A 1 kg1\,\text{kg} sphere moving at 5 m s−15\,\text{m s}^{-1} collides directly with a stationary 2 kg2\,\text{kg} sphere. The collision loses 16/3 J16/3\,\text{J} of kinetic energy. Determine the coefficient of restitution and both velocities after impact.

(6)

(Total for Question 1 is 6 marks)

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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Successive direct impacts of spheres and/or a sphere with a smooth plane surface.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Successive impacts must be handled chronologically. Keep one positive direction throughout and use the velocities immediately after one impact as the approach velocities for the next.
  • At a fixed smooth plane, the sphere's normal velocity reverses and its speed is multiplied by ee; the plane remains stationary. After every calculation, use the ordering and signed velocities to decide which bodies can meet next.
  • A faster rear sphere may cause another sphere-sphere collision, while increasing velocities from rear to front prevent further impacts.
  • Spheres are modelled as particles in these direct-impact problems.
  • Examiners expect a justified impact order and a final no-further-collision argument, not merely a list of algebraic velocity pairs.
Two spheres in line with a fixed wall, a common setup for successive direct impacts.
Worked example

Identical spheres AA and BB lie in that order before a wall. Initially AA moves right at 8 m s−18\,\text{m s}^{-1} and BB is stationary. Their mutual coefficient of restitution is 12\tfrac12; BB rebounds from the wall with coefficient 34\tfrac34. Find the velocities after the first two impacts and decide whether AA and BB meet again.

  1. 1.For the first equal-mass impact, vA+vB=8v_A+v_B=8 and vB−vA=4v_B-v_A=4, giving vA=2v_A=2, vB=6v_B=6.
  2. 2.At the wall, BB reverses with velocity −(34)(6)=−4.5 m s−1-(\tfrac34)(6)=-4.5\,\text{m s}^{-1}.
  3. 3.AA moves right while BB moves left, so their separation decreases.

Answer: After AA-BB: (2,6) m s−1(2,6)\,\text{m s}^{-1}; after the wall: (2,−4.5) m s−1(2,-4.5)\,\text{m s}^{-1}, so they collide again.

Common mistakes

  • Don't reuse the original velocity of a sphere instead of its velocity after the preceding impact.
  • Don't multiply the wall-impact speed by ee and fail to reverse its direction.
  • Don't state that another impact occurs without comparing relative positions and velocities.

Exam tip

After each impact, draw the new signed velocity arrows before deciding which impact happens next.

Tier 1 · Easy

ORIGINAL

1.

A sphere moving normally towards a fixed smooth wall at 5 m s−15\,\text{m s}^{-1} has coefficient of restitution 0.70.7 with the wall. State its velocity immediately after impact, taking motion towards the wall as positive.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Identical spheres AA and BB lie on a line, with BB between AA and a wall. Sphere AA moves towards stationary BB at 6 m s−16\,\text{m s}^{-1}. Their coefficient of restitution is 1/21/2, and BB's coefficient with the wall is 2/32/3. Find the velocities just after the first sphere-sphere impact and explain why AA and BB collide again after BB rebounds from the wall.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Three identical spheres AA, BB and CC are arranged in that order on a straight line. Initially AA moves towards the other two at 8 m s−18\,\text{m s}^{-1} while BB and CC are at rest. Every impact has coefficient of restitution 1/21/2. Find the velocities after all impacts have occurred, and justify that no further collision follows.

(7)

(Total for Question 1 is 7 marks)

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