Further Mathematics · Pearson Edexcel 9FM0 · Option FM1 · Topic 4

Elastic collisions in one dimension

Collisions in a straight line, using Newton's law of restitution and conservation of momentum, including loss of kinetic energy.

  • 2 sub-topics
  • Taught in Year 12
  • 32 practice questions
  • 2 animated lessons
Practice questions (32) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 4.1Direct impact and Newton's law of restitutionY12
  2. 4.2Successive direct impactsY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 32 questions in this topic, free, no account needed.

Elastic collisions in one dimension

Easy

Newton's law of restitution · Year 12 · 1 mark

  1. A sphere hits a fixed smooth wall at right angles at 6 m s⁻¹. The coefficient of restitution is 0.5. What is its speed after the impact?
    (1)

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  1. Speed of separation = e × speed of approach.A: 3.00 m s−13.00\ \mathrm{m\,s^{-1}}

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Elastic collisions in one dimension

Medium

Direct impact · Year 12 · 8 marks

Two smooth spheres AA and BB, of equal radius and masses 3 kg and 1 kg, move in the same straight line on a smooth horizontal surface. A moves with speed 6 m s⁻¹ towards BB, which moves towards AA with speed 3 m s⁻¹. The coefficient of restitution between them is 14\frac{1}{4}.
  1. (a)
    Find the velocities of AA and BB immediately after the collision.
    (5)
  2. (b)
    State whether the direction of motion of AA is changed.
    (1)
  3. (c)
    Find the kinetic energy lost in the collision.
    (2)
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Part (a)5 marks

  1. Conservation of linear momentum.3(6)+1(−3)=3vA+vB3(6) + 1(-3) = 3v_A + v_B
  2. Newton's law of restitution: separation = e × approach.vB−vA=14(6−(−3))v_B - v_A = \frac{1}{4}(6 - (-3))
  3. vAv_A=5116, vB{}= \frac{51}{16}, \ v_B=8716{}= \frac{87}{16}

Part (b)1 mark

  1. A's direction is unchanged.

Part (c)2 marks

  1. KE before−KE after\text{KE before} - \text{KE after}
  2. =3645128 J= \frac{3645}{128}\text{ J}

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Elastic collisions in one dimension

Hard

Successive direct impacts · Year 12 · 7 marks

A sphere AA of mass 2m2m moves with speed 6 m s⁻¹ on a smooth horizontal floor and collides directly with a sphere BB of mass 6m6m, which is at rest. The coefficient of restitution between AA and BB is 23\frac{2}{3}. BB then hits a vertical wall at right angles; the coefficient of restitution between BB and the wall is 34\frac{3}{4}.
  1. (a)
    Find the speeds of AA and BB after their collision.
    (4)
  2. (b)
    Find the speed of BB after it rebounds from the wall.
    (1)
  3. (c)
    Determine whether there is a second collision between AA and BB.
    (2)
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Part (a)4 marks

  1. Momentum (the mass m cancels).6⋅2=2vA+6vB6 \cdot 2 = 2v_A + 6v_B
  2. Restitution.vB−vA=4v_B - v_A = 4
  3. vAv_A=−32, vB{}= - \frac{3}{2}, \ v_B=52{}= \frac{5}{2}

Part (b)1 mark

  1. 34×52\frac{3}{4} \times \frac{5}{2}=158{}= \frac{15}{8}

Part (c)2 marks

  1. Compare the velocities after B's rebound.A moves away from the wall at 32, B moves back at 158\text{A moves away from the wall at } \frac{3}{2}, \ \text{B moves back at } \frac{15}{8}
  2. BB is faster than AA, so it catches AA: a second collision.

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