---
title: "Direct central impact and restitution"
subject: "Engineering Mechanics"
chapter: 4
key: "semester_3/engineering_mechanics:direct-central-impact-derivation"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/direct-central-impact-derivation
site: BCT Engineering
---

# Direct central impact and restitution

Engineering Mechanics · Chapter 4: Kinetics of Particles: Impulse and Momentum

- Asked in: year not given, 8 marks
- Video: [Impact: Coefficient of Restitution (learn to solve any problem)](https://www.youtube.com/watch?v=uzfeib3MkF8) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/direct-central-impact-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-4

## Question

Explain direct central impact. Define coefficient of restitution. For two smooth spheres of masses m1 and m2 moving with velocities u1 and u2 (u1 > u2) along the line of centres, derive expressions for their velocities after impact and the loss of kinetic energy.

## Short answer

In direct central impact, velocities lie along the line of centres. Coefficient of restitution e = (v₂ – v₁)/(u₁ – u₂). Using conservation of momentum and the restitution equation, final velocities are obtained; the loss of kinetic energy equals m₁ m₂ (1 – e²)(u₁ – u₂)² / [2(m₁ + m₂)].

## Answer

#### Direct central impact
Two bodies collide in **direct central impact** when their centres of mass lie on the line of impact, which is also the common normal at the contact point, and the velocities before and after impact are along this line. If the surfaces are smooth, the impulses act only along the line of impact.

```text
  before:  (m1)--> u1      (m2)--> u2      u1 > u2
  after :  (m1)--> v1      (m2)--> v2      v2 >= v1
```

The impact has two phases: **deformation** (from first contact to maximum compression, when both bodies have a common velocity) and **restitution** (the bodies regain shape and separate).

#### Coefficient of restitution
$e$ is the ratio of the impulse during restitution to the impulse during deformation, which for a direct central impact equals

$$
e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}
$$

$e = 1$ for a perfectly elastic impact and $e = 0$ for a perfectly plastic impact; $0 < e < 1$ in practice.

#### Final velocities
Conservation of momentum (no external impulse):

$$
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (1)
$$

Definition of $e$: $v_2 - v_1 = e(u_1 - u_2)$ (2)

Solving (1) and (2):

$$
v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2}
\qquad
v_2 = \frac{m_1 u_1 + m_2 u_2 + m_1 e(u_1 - u_2)}{m_1 + m_2}
$$

#### Loss of kinetic energy

$$
\Delta T = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 - \tfrac12 m_1 v_1^2 - \tfrac12 m_2 v_2^2
$$

Substituting the velocities and simplifying:

$$
\Delta T = \frac{m_1 m_2}{2(m_1 + m_2)}\,(1 - e^2)\,(u_1 - u_2)^2
$$

For $e = 1$, $\Delta T = 0$ (kinetic energy conserved); for $e = 0$ the loss is maximum and both bodies move together with velocity $(m_1u_1 + m_2u_2)/(m_1+m_2)$. The lost energy becomes heat, sound and permanent deformation.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In direct central impact:
   A. Velocities are along the line of impact through the centres
   B. Spheres collide at any angle
   C. Only one body moves
   D. Friction acts along the surface
   Answer: A. Centres lie on the common normal and velocities are along it. (Key point: Direct central impact)

2. For a direct impact, e equals:
   A. Velocity of approach / velocity of separation
   B. Velocity of separation / velocity of approach
   C. Final KE / initial KE
   D. (v1 + v2)/(u1 + u2)
   Answer: B. e = (v2 - v1)/(u1 - u2). (Key point: Coefficient of restitution)

3. Which two equations are solved for v1 and v2?
   A. Conservation of momentum and conservation of energy
   B. Conservation of momentum and the definition of e
   C. Two energy equations
   D. Impulse and work equations
   Answer: B. KE is not conserved in general, so momentum and e are used. (Key point: Deriving velocities)

4. The loss of kinetic energy is:
   A. m1 m2/(2(m1+m2)) (1 - e^2)(u1 - u2)^2
   B. m1 m2/(2(m1+m2)) (1 + e)(u1 - u2)
   C. (1/2)(m1+m2)(1 - e)(u1 - u2)^2
   D. Zero for any e
   Answer: A. It vanishes for e = 1 and is maximum for e = 0. (Key point: Loss of kinetic energy)

## Related in Engineering Mechanics

- [Direct impact of two spheres](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Atwo-spheres-direct-impact-numerical) — Engineering Mechanics, chapter 4
- [Oblique impact of ball on a wall](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aball-oblique-impact-wall) — Engineering Mechanics, chapter 4
- [Kinetic energy and work of rigid body](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Akinetic-energy-rigid-body-work-energy) — Engineering Mechanics, chapter 7

## Related in other subjects

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- [Coefficient of restitution; car on track](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Adirect-central-impact-restitution-car-track) — Applied Mechanics (Dynamics), chapter 2
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- [Two balls impact e = 0.84, energy lost](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Atwo-balls-e084-ke-lost) — Applied Mechanics (Dynamics), chapter 2