Edexcel A-level Further Maths revision notes

Elastic collisions in one dimension

Section FM1-4

Year 1

2 specification points

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

Checked against Edexcel 9FM0 section FM1-4

Checked against Edexcel 9FM0 section FM1-4. 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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FM1-4.1

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

Notes

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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 $e$ times speed of approach, with $0\leq e\leq1$ .
In signed form for $A$ behind $B$ , $v_B-v_A=e(u_A-u_B)$ . 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 $e$ and consistent velocity signs.

Worked example

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

1.Momentum: $15=3v_A+2v_B$ .
2.Restitution: $v_B-v_A=\tfrac12(5)=2.5$ .
3.Substitute $v_B=v_A+2.5$ : $15=5v_A+5$ , so $v_A=2$ and $v_B=4.5$ .

Answer: The velocities are $2\,\text{m s}^{-1}$ and $4.5\,\text{m s}^{-1}$ 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 $e<1$ .
Don't accept a negative algebraic root for $e$ despite $0\leq e\leq1$ .

Exam tip

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

Tier 1 · Easy

ORIGINAL

Sphere

A

approaches stationary sphere

B

7\,\text{m s}^{-1}

. After their direct impact,

A

continues in the same direction at

2\,\text{m s}^{-1}

. The coefficient of restitution is

0.4

. Find the velocity of

B

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

2\,\text{kg}

sphere travelling at

6\,\text{m s}^{-1}

strikes a stationary

3\,\text{kg}

sphere directly. The coefficient of restitution is

1/2

. Find both velocities after impact and the kinetic energy lost.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1\,\text{kg}

sphere moving at

5\,\text{m s}^{-1}

collides directly with a stationary

2\,\text{kg}

sphere. The collision loses

16/3\,\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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FM1-4.2

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

Notes

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Your confidence:

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

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

Answer: After $A$ - $B$ : $(2,6)\,\text{m s}^{-1}$ ; after the wall: $(2,-4.5)\,\text{m s}^{-1}$ , 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 $e$ 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

A sphere moving normally towards a fixed smooth wall at

5\,\text{m s}^{-1}

has coefficient of restitution

0.7

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

Identical spheres

A

and

B

lie on a line, with

B

between

A

and a wall. Sphere

A

moves towards stationary

B

6\,\text{m s}^{-1}

. Their coefficient of restitution is

1/2

, and

B

's coefficient with the wall is

2/3

. Find the velocities just after the first sphere-sphere impact and explain why

A

and

B

collide again after

B

rebounds from the wall.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Three identical spheres

A

B

and

C

are arranged in that order on a straight line. Initially

A

moves towards the other two at

8\,\text{m s}^{-1}

while

B

and

C

are at rest. Every impact has coefficient of restitution

1/2

. 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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