15.2.2 Energy in 1D and Successive Collisions

Introduction

This note explores successive direct impacts of elastic spheres and collisions between spheres and smooth surfaces in one dimension. Using the principles of momentum conservation, Newton's law of restitution, and energy considerations, we solve problems involving:

Velocities before and after collisions.
Conditions for further impacts or when collisions cease.

Successive Collisions

When multiple collisions occur between spheres (or between a sphere and a smooth surface), the velocities of each object after one collision become the initial velocities for the next collision.

Conservation of Momentum

The total momentum of the system before and after each collision is conserved:

m_A u_A + m_B u_B = m_A v_A + m_B v_B

where:

$m_A, m_B$ : Masses of the spheres,
$u_A, u_B$ : Initial velocities,
$v_A, v_B$ : Final velocities.

Newton's Law of Restitution

The coefficient of restitution ( $e$ ) relates the speeds of approach and separation:

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

or:

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

Successive Collisions with a Smooth Plane

For a sphere colliding with a smooth plane:

e = \frac{v_\text{rebound}}{v_\text{impact}}

where:

$v_\text{impact}$ : Velocity of the sphere just before hitting the plane,
$v_\text{rebound}$ : Velocity of the sphere just after rebounding.

Conditions for Further Impacts

Further impacts between spheres occur if:

v_B > v_A

where $v_B$ and $v_A$ are the final velocities of spheres $B$ and $A$ , respectively, after a collision.

If $v_A \geq v_B$ , the spheres will not collide again.

Worked Examples

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Example 1: Successive Collisions Between Two Spheres

Problem

Two spheres $A$ (mass 2 kg) and $B$ (mass 3 kg) move in a straight line.

Initially:

$A$ moves at 4 ms⁻¹
$B$ moves at 1 ms⁻¹ The coefficient of restitution between $A$ and $B$ is e = 0.8.

Determine:

The velocities of $A$ and $B$ after their first collision.
Whether the spheres collide again.

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 ( $m_A = 2, m_B = 3, u_A = 4, u_B = 1$ ):

2(4) + 3(1) = 2v_A + 3v_B

Simplify:

8 + 3 = 2v_A + 3v_B \quad \Rightarrow \quad 11 = 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 $e = 0.8, u_A = 4, u_B = 1$

8 = \frac{v_B - v_A}{4 - 1}

Simplify:

0.8 = \frac{v_B - v_A}{3} \quad \Rightarrow \quad v_B - v_A = 2.4. \quad \text{(2)}

Step 3: Solve Simultaneous Equations

From equations (1) and (2):

$11 = 2v_A + 3v_B$
$v_B = v_A + 2.4$ Substitute $v_B = v_A + 2.4$ into (1):

11 = 2v_A + 3(v_A + 2.4)

Simplify:

11 = 2v_A + 3v_A + 7.2 \quad \Rightarrow \quad 11 = 5v_A + 7.2

Solve for $v_A$ :

5v_A = 11 - 7.2 = 3.8 \quad \Rightarrow \quad v_A = \frac{3.8}{5} = 0.76 \, \text{ms}^{-1}

Substitute $v_A = 0.76$ into $v_B = v_A + 2.4$

v_B = 0.76 + 2.4 = 3.16 \, \text{ms}^{-1}

Step 4: Determine Further Impacts

Further collisions occur if $v_B > v_A$ , which is true:

v_B = 3.16 \, \text{ms}^{-1}, \quad v_A = 0.76 \, \text{ms}^{-1}

Final Answer:

Velocities after the first collision:

vₐ = 0.76 ms⁻¹ (sphere A)
vᵦ = 3.16 ms⁻¹ (sphere B)

The spheres will collide again.

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Example 2: Successive Impacts with a Smooth Plane

Problem

A sphere of mass 0.5 kg is dropped from a height of 2 m onto a smooth horizontal plane.

The coefficient of restitution between the sphere and the plane is e = 0.6

Find:

The speed of the sphere just before and just after the first impact.
The height of the sphere after the second rebound.

Part 1: Speed Before and After First Impact

Step 1: Speed before impact:

v_\text{impact} = \sqrt{2gh}

where $g = 9.8 \, \text{ms}^{-2}, h = 2 \, \text{m}$

Substitute:

v_\text{impact} = \sqrt{2 \times 9.8 \times 2} = \sqrt{39.2} \approx 6.26 \, \text{ms}^{-1}

Step 2: Speed after impact:

v_\text{rebound} = e v_\text{impact}

Substitute $e = 0.6, v_\text{impact} = 6.26$

v_\text{rebound} = 0.6 \times 6.26 = 3.76 \, \text{ms}^{-1}

Part 2: Height After Second Rebound

Step 1: Use the rebound velocity to find the height:

h = \frac{v^2}{2g}

Substitute $v = 3.76, g = 9.8$

h = \frac{3.76^2}{2 \times 9.8} = \frac{14.14}{19.6} \approx 0.72 \, \text{m}

Final Answer:

Speed:

Before impact: 6.26 ms⁻¹
After impact: 3.76 ms⁻¹

Height after the second rebound: 0.72 m

Note Summary

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Common Mistakes

Incorrect application of restitution: Always ensure the correct relative velocities are used.
Ignoring momentum conservation: Total momentum must always be conserved during collisions.
Misinterpreting conditions for further impacts: Check whether vᵦ > vₐ after collisions.
Forgetting energy loss: With e < 1, kinetic energy is not conserved, so successive rebounds decrease in height.

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Key Formulas

Newton's Law of Restitution:

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

Conservation of Momentum:

m_A u_A + m_B u_B = m_A v_A + m_B v_B

Free Fall Speed:

v = \sqrt{2gh}

Rebound Height:

h = \frac{v^2}{2g}

Energy in 1D and Successive Collisions Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

15.2.2 Energy in 1D and Successive Collisions

Introduction

Successive Collisions

Conservation of Momentum

Newton's Law of Restitution

Successive Collisions with a Smooth Plane

Conditions for Further Impacts

Worked Examples

Example 1: Successive Collisions Between Two Spheres

Example 2: Successive Impacts with a Smooth Plane

Note Summary

Common Mistakes

Key Formulas

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