A ball hits a wall directly at 6\,m s^-1 with e = 0.5. Find the rebound speed. [2 marks]

A 2\,kg sphere at 3\,m s^-1 collides directly with a stationary 2\,kg sphere, e = 1. Find the final speeds. [3 marks]

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How does the coefficient of restitution determine the outcome of collisions?

Newton's experimental law of restitution, direct and oblique impact of smooth spheres, impact with a fixed surface, and kinetic energy lost in a collision.

A focused answer to the Edexcel A-Level Further Mathematics Further Mechanics content on elastic collisions, covering Newton's experimental law of restitution, direct and oblique impact of smooth spheres, impact with a fixed surface, and the kinetic energy lost in a collision.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The law of restitution
Direct impact
Oblique impact and energy loss
Examples in context
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What this dot point is asking

Edexcel Further Mechanics wants you to apply Newton's experimental law of restitution alongside conservation of momentum, analyse direct and oblique impacts of smooth spheres, handle impact with a fixed smooth surface, and compute the kinetic energy lost in a collision. The combination of momentum (a vector conservation law) and restitution (a relative-speed condition) is the core technique, and oblique impacts add a resolve-into-components layer.

The law of restitution

When two bodies collide, momentum is always conserved (the collision forces are internal and equal-and-opposite), but kinetic energy generally is not. Newton's experimental law of restitution supplies the second equation you need, relating the relative speeds before and after impact along the line of centres.

The two extremes are $e = 1$ (perfectly elastic, no kinetic energy lost) and $e = 0$ (perfectly inelastic, the bodies coalesce and move together).

Direct impact

In a direct (head-on) impact the velocities lie along the line joining the centres, so you have a one-dimensional problem. The two equations, momentum and restitution, give the two unknown final speeds. Choose a positive direction and stick to it, letting signs handle reversals.

Direct impact of two spheres

A $1\,\text{kg}$ sphere moving at $4\,\text{m s}^{-1}$ strikes a stationary $1\,\text{kg}$ sphere directly, with $e = 0.5$ . Find the speeds of the two spheres after the collision.

Step 1: Write the conservation of momentum equation

Momentum is always conserved in a collision because the collision forces are internal and equal-and-opposite. Taking the initial direction of travel as positive and letting $v_1$ and $v_2$ be the post-collision speeds:

1(4) + 1(0) = 1 v_1 + 1 v_2 \implies v_1 + v_2 = 4.

Step 2: Apply Newton's law of restitution

The law of restitution gives the second equation: the speed of separation equals $e$ times the speed of approach along the line of impact. The approach speed is $4 - 0 = 4\,\text{m s}^{-1}$ , so:

v_2 - v_1 = e \times 4 = 0.5 \times 4 = 2.

Step 3: Solve the simultaneous equations

Adding the two equations eliminates $v_1$ : $2v_2 = 6$ , so $v_2 = 3\,\text{m s}^{-1}$ , and substituting back gives $v_1 = 1\,\text{m s}^{-1}$ . Both spheres move in the original direction, and the struck sphere moves faster, confirming they separate cleanly.

Final answer: $v_1 = 1\,\text{m s}^{-1}$ , $v_2 = 3\,\text{m s}^{-1}$ .

Oblique impact and energy loss

For a smooth sphere striking a fixed surface or another smooth sphere obliquely, the key idea is to resolve into components along and perpendicular to the line of impact (the common normal at the point of contact). Because the surfaces are smooth there is no tangential impulse, so the perpendicular-to-the-line component of velocity is unchanged, while the along-the-line component reverses and is scaled by $e$ .

Kinetic energy lost in a direct impact

For the impact above ( $1\,\text{kg}$ at $4\,\text{m s}^{-1}$ into a stationary $1\,\text{kg}$ , $e = 0.5$ , giving $v_1 = 1\,\text{m s}^{-1}$ and $v_2 = 3\,\text{m s}^{-1}$ ), find the kinetic energy lost in the collision.

Step 1: Calculate the kinetic energy before the collision

The kinetic energy before the impact comes entirely from the moving sphere, since the target is stationary. Using $\text{KE} = \frac{1}{2}mv^2$ :

\text{KE}_{\text{before}} = \tfrac{1}{2}(1)(4^2) + 0 = 8\,\text{J}.

Step 2: Calculate the kinetic energy after the collision

Both spheres are now moving, so we sum their individual kinetic energies using the speeds found previously:

\text{KE}_{\text{after}} = \tfrac{1}{2}(1)(1^2) + \tfrac{1}{2}(1)(3^2) = 0.5 + 4.5 = 5\,\text{J}.

Step 3: Find the energy lost

The kinetic energy lost is the difference. For any $e < 1$ this must be positive, since some energy is always dissipated as heat, sound, and deformation:

\text{Energy lost} = 8 - 5 = 3\,\text{J}.

Final answer: $3\,\text{J}$ of kinetic energy is lost. Setting $e = 1$ would instead give equal masses exchanging velocities with no energy loss.

Examples in context

Elastic collisions build directly on momentum and impulse, where conservation of momentum is established, and on work-energy methods, since the kinetic energy lost in an impact is dissipated (as heat, sound and deformation). The resolve-into-components technique for oblique impacts is the same vector decomposition used in further vectors and in resolving forces for circular motion. Newton's cradle, billiard-ball collisions, and a ball bouncing repeatedly off a floor (where the rebound height scales by $e^2$ each bounce) are standard modelling contexts, and the repeated-bounce problem connects to geometric series.

Try this

Q1. A ball hits a wall directly at $6\,\text{m s}^{-1}$ with $e = 0.5$ . Find the rebound speed. [2 marks]

Cue. Speed after $= e \times 6 = 3\,\text{m s}^{-1}$ .

Q2. State the range of values the coefficient of restitution can take. [1 mark]

Cue. $0 \le e \le 1$ .

Q3. A $2\,\text{kg}$ sphere at $3\,\text{m s}^{-1}$ collides directly with a stationary $2\,\text{kg}$ sphere, $e = 1$ . Find the final speeds. [3 marks]

Cue. Equal masses, perfectly elastic: they exchange velocities, so $v_1 = 0$ and $v_2 = 3\,\text{m s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20187 marksA sphere

A

of mass

2\,\text{kg}

moving at

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

collides directly with a sphere

B

of mass

3\,\text{kg}

moving in the same direction at

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

. The coefficient of restitution is

e = 0.4

. Find the speeds of

A

and

B

after impact.

Show worked answer →

Apply conservation of momentum and Newton's law of restitution as simultaneous equations.

Momentum: $2(5) + 3(1) = 2v_A + 3v_B$ , so $13 = 2v_A + 3v_B$ (M1 A1).

Restitution: separation $= e \times$ approach, so $v_B - v_A = 0.4(5 - 1) = 1.6$ (M1 A1).

From the second equation $v_B = v_A + 1.6$ . Substitute: $13 = 2v_A + 3(v_A + 1.6) = 5v_A + 4.8$ (M1), so $5v_A = 8.2$ , $v_A = 1.64\,\text{m s}^{-1}$ (A1) and $v_B = 3.24\,\text{m s}^{-1}$ (A1). Both move in the original direction and $v_B > v_A$ , so they separate, confirming the answer is consistent.

Edexcel 20226 marksA smooth sphere of mass

0.5\,\text{kg}

moving at

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

strikes a smooth vertical wall, its direction of motion making

60^\circ

with the wall. The coefficient of restitution between sphere and wall is

0.5

. Find the speed of the sphere immediately after impact and the kinetic energy lost.

Show worked answer →

Split the velocity into components parallel and perpendicular to the wall; only the perpendicular component is affected.

Component parallel to the wall: $8\cos 60^\circ = 4\,\text{m s}^{-1}$ (unchanged) (M1 A1). Component perpendicular to the wall: $8\sin 60^\circ = 6.93\,\text{m s}^{-1}$ , which reverses and scales by $e$ : $0.5 \times 6.93 = 3.46\,\text{m s}^{-1}$ (M1 A1).

Speed after: $\sqrt{4^2 + 3.46^2} = \sqrt{16 + 12} = \sqrt{28} \approx 5.29\,\text{m s}^{-1}$ (A1).

KE before $= \frac{1}{2}(0.5)(8^2) = 16\,\text{J}$ ; KE after $= \frac{1}{2}(0.5)(28) = 7\,\text{J}$ ; energy lost $= 9\,\text{J}$ (A1).

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Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)