15.2.1 Newton's Law of Restitution

Introduction

Newton's Law of Restitution describes the behaviour of two elastic spheres during a direct impact. It introduces the coefficient of restitution ( $e$ ), which measures the elasticity of the collision and determines the velocities of the spheres after the collision.

This note covers:

Newton's law of restitution and the range $0 \leq e \leq 1$
The significance of special cases: $e = 0$ and $e = 1$
Loss of kinetic energy due to impact.

Newton's Law of Restitution

Newton's law of restitution states:

e = \frac{\text{speed of separation}}{\text{speed of approach}}

For two spheres $A$ and $B$ colliding directly:

e = \frac{v_B - v_A}{u_A - u_B}

where:

$u_A, u_B$ : Initial velocities of spheres $A$ and $B$
$v_A, v_B$ : Final velocities of spheres $A$ and $B$
$e$ : Coefficient of restitution ( $0 \leq e \leq 1$ )

Coefficient of Restitution ( $e$ )

$e = 1$ : Perfectly elastic collision. Kinetic energy is conserved, and the spheres separate with the same relative speed as they approached.
$e = 0$ : Perfectly inelastic collision. The spheres stick together after the collision, and there is maximum loss of kinetic energy. For all real-world collisions:

0 < e < 1

Conservation of Momentum

In a direct collision between two spheres, the total momentum before and after the collision is conserved:

m_A u_A + m_B u_B = m_A v_A + m_B v_B

where $m_A$ and $m_B$ are the masses of spheres $A$ and $B$ .

Loss of Kinetic Energy

Kinetic energy is not generally conserved during a collision. The loss of kinetic energy is given by:

\Delta KE = KE_\text{initial} - KE_\text{final}

The initial kinetic energy is:

KE_\text{initial} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2

and the final kinetic energy is:

KE_\text{final} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2

The loss of kinetic energy is maximum when $e = 0$

Worked Examples

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Example 1: Finding Velocities After a Collision

Problem

Two spheres, $A$ (2 kg) and $B$ (3 kg), collide directly. Initially:

$A$ is moving at 5 ms⁻¹
$B$ is stationary ( $u_B = 0$ ) The coefficient of restitution is $e = 0.8$

Find the velocities of $A$ and $B$ after the collision.

Step 1: Apply Conservation of Momentum

m_A u_A + m_B u_B = m_A v_A + m_B v_B

Substitute the values:

2(5) + 3(0) = 2v_A + 3v_B

Simplify:

10 = 2v_A + 3v_B \quad \text{(1)}

Step 2: Apply Newton's Law of Restitution

e = \frac{v_B - v_A}{u_A - u_B}

Substitute $e = 0.8, u_A = 5$ , and $u_B = 0$

0.8 = \frac{v_B - v_A}{5}

Simplify:

v_B - v_A = 4 \quad \text{(2)}

Step 3: Solve Simultaneous Equations

From equations (1) and (2):

$10 = 2v_A + 3v_B$
$v_B - v_A = 4 \quad \Rightarrow \quad v_B = v_A + 4$ Substitute $v_B = v_A + 4$ into (1):

10 = 2v_A + 3(v_A + 4)

Simplify:

10 = 2v_A + 3v_A + 12 \quad \Rightarrow \quad 10 = 5v_A + 12

Solve for $v_A$ :

5v_A = -2 \quad \Rightarrow \quad v_A = -0.4 \, \text{ms}^{-1}

Substitute $v_A = -0.4$ into $v_B = v_A + 4$ :

v_B = -0.4 + 4 = 3.6 \, \text{ms}^{-1}

Final Answer:

Velocity of $A$ : -0.4 ms⁻¹ (moves left),
Velocity of $B$ : 3.6 ms⁻¹

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Example 2: Loss of Kinetic Energy

Problem

Using the data from Example 1, calculate the loss of kinetic energy during the collision.

Step 1: Initial Kinetic Energy

KE_\text{initial} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2

Substitute $m_A = 2 \, \text{kg}, u_A = 5 \, \text{ms}^{-1}, m_B = 3 \, \text{kg}, u_B = 0$

KE_\text{initial} = \frac{1}{2}(2)(5^2) + \frac{1}{2}(3)(0^2)

Simplify:

KE_\text{initial} = \frac{1}{2}(2)(25) = :highlight[25 J]

Step 2: Final Kinetic Energy

KE_\text{final} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2

Substitute $v_A = -0.4 \, \text{ms}^{-1}, v_B = 3.6 \, \text{ms}^{-1}$

KE_\text{final} = \frac{1}{2}(2)(-0.4^2) + \frac{1}{2}(3)(3.6^2)

Simplify:

KE_\text{final} = \frac{1}{2}(2)(0.16) + \frac{1}{2}(3)(12.96)

KE_\text{final} = 0.16 + 19.44 = :highlight[19.6 J]

Step 3: Loss of Kinetic Energy

\Delta KE = KE_\text{initial} - KE_\text{final}

Substitute:

\Delta KE = 25 - 19.6 = :highlight[5.4 J]

Final Answer:

The loss of kinetic energy is 5.4 J

Note Summary

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Common Mistakes

Forgetting to conserve momentum: Always use $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
Misinterpreting ee: Ensure $e$ is applied correctly as $\frac{\text{speed of separation}}{\text{speed of approach}}$
Mixing up initial and final velocities: Clearly distinguish $u$ (initial) from $v$ (final).
Ignoring special cases:

$e = 1$ : Perfectly elastic collision (no kinetic energy lost).
$e = 0$ : Perfectly inelastic collision (maximum energy lost).

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Key Formulas

Newton's Law of Restitution:

e = \frac{v_B - v_A}{u_A - u_B}

Conservation of Momentum:

m_A u_A + m_B u_B = m_A v_A + m_B v_B

Loss of Kinetic Energy:

\Delta KE = KE_\text{initial} - KE_\text{final}

Newton's Law of Restitution Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

15.2.1 Newton's Law of Restitution

Introduction

Newton's Law of Restitution

Coefficient of Restitution ( $e$ )

Conservation of Momentum

Loss of Kinetic Energy

Worked Examples

Example 1: Finding Velocities After a Collision

Example 2: Loss of Kinetic Energy

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Newton's Law of Restitution For their A-Level Exams.

Other Revision Notes related to Newton's Law of Restitution you should explore

Newton's Law of Restitution Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

15.2.1 Newton's Law of Restitution

Introduction

Newton's Law of Restitution

Coefficient of Restitution (eee)

Conservation of Momentum

Loss of Kinetic Energy

Worked Examples

Example 1: Finding Velocities After a Collision

Example 2: Loss of Kinetic Energy

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Newton's Law of Restitution For their A-Level Exams.

Other Revision Notes related to Newton's Law of Restitution you should explore

Coefficient of Restitution ( $e$ )