15.3.2 Successive Collisions with a Surface

Introduction

This note focuses on successive oblique impacts of a sphere with smooth plane surfaces. Each collision changes the velocity of the sphere, which continues moving until subsequent impacts occur. The velocity is decomposed into components parallel and perpendicular to the surface, where:

The perpendicular component changes according to the coefficient of restitution ( $e$ )
The parallel component remains unchanged because no impulse acts along the tangential direction. We will explore these concepts in problems involving collisions at arbitrary angles and with non-parallel surfaces.

Velocity Components in Oblique Collisions

When a sphere collides with a smooth plane:

Perpendicular Component ( $v_\perp$ ):

Changes due to the collision:

v_{\perp, \text{after}} = -e v_{\perp, \text{before}}

where $e$ (0 ≤ e ≤ 1) is the coefficient of restitution.

Parallel Component ( $v_\parallel$ ):

Remains unchanged:

v_{\parallel, \text{after}} = v_{\parallel, \text{before}}

Successive Collisions

After each impact:

The sphere rebounds with a modified velocity vector, based on the resolved perpendicular and parallel components.
The path of the sphere after the collision depends on the geometry of the surface and the rebound velocity.

Vector Representation of Velocity

If the velocity is expressed as a vector $\mathbf{v}$ , it can be decomposed as:

\mathbf{v} = \mathbf{v}_\parallel + \mathbf{v}_\perp

where:

$\mathbf{v}_\parallel$ is the projection of $\mathbf{v}$ onto the surface,
$\mathbf{v}_\perp$ is the projection of $\mathbf{v}$ onto the direction normal to the surface. For collisions at arbitrary angles, the direction of the surface normal determines the components.

Repeated Impacts and Height Reduction

For successive vertical impacts of a sphere with a horizontal surface:

After each impact, the perpendicular component reduces in magnitude:

v_{\perp, n} = (-e)^n v_{\perp, 0}

where $n$ is the number of collisions.

The rebound height after the $n$ th collision is:

h_n = e^{2n} h_0

where $h_0$ is the initial height of the sphere before the first collision.

Worked Examples

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Example 1: Successive Collisions with a Horizontal Surface

Problem

A sphere 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 = 0.7

Find:

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

Part 1: Speed Before and After the First Collision

Step 1: Speed before the first collision:

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

where g = 9.8 ms⁻², h = 2 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 the first collision:

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

Substitute e = 0.7, v_impact = 6.26

v_\text{rebound} = 0.7 \times 6.26 = 4.382 \, \text{ms}^{-1}

Part 2: Rebound Height After the Second Collision

Step 1: Rebound height after the first collision:

h_1 = \frac{v_\text{rebound}^2}{2g}

Substitute v_rebound = 4.382, g = 9.8

h_1 = \frac{(4.382)^2}{2 \times 9.8} = \frac{19.2}{19.6} \approx 0.98 \, \text{m}

Step 2: Rebound height after the second collision:

h_2 = e^2 h_1

Substitute e = 0.7, h₁ = 0.98

h_2 = (0.7)^2 \times 0.98 = 0.49 \times 0.98 = 0.4802 \, \text{m}

Final Answer:

Speed before the first collision: 6.26 ms⁻¹, speed after: 4.382 ms⁻¹
Rebound height after the second collision: 0.48 m

infoNote

Example 2: Successive Oblique Collisions with a Vertical Wall

Problem

A sphere of mass 1 kg is moving at $\mathbf{u} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} \, \text{ms}^{-1}$ and strikes a smooth vertical wall.

The coefficient of restitution is e = 0.6

Find the velocity of the sphere after the first collision and the total loss of kinetic energy.

Part 1: Resolve Velocity Components

Perpendicular component ( $\mathbf{u}_\perp$ ): Normal to the wall:

\mathbf{u}_\perp = \begin{pmatrix} 6 \\ 0 \end{pmatrix}

Parallel component ( $\mathbf{u}_\parallel$ ): Tangential to the wall:

\mathbf{u}_\parallel = \begin{pmatrix} 0 \\ -4 \end{pmatrix}

Part 2: Velocity After Collision

Perpendicular direction ( $\mathbf{v}_\perp$ ):

\mathbf{v}_\perp = -e \mathbf{u}_\perp = -0.6 \begin{pmatrix} 6 \\ 0 \end{pmatrix} = \begin{pmatrix} -3.6 \\ 0 \end{pmatrix}

Parallel direction ( $\mathbf{v}_\parallel$ ):

\mathbf{v}_\parallel = \mathbf{u}_\parallel = \begin{pmatrix} 0 \\ -4 \end{pmatrix}

The total velocity after the collision:

\mathbf{v} = \mathbf{v}_\perp + \mathbf{v}_\parallel = \begin{pmatrix} -3.6 \\ -4 \end{pmatrix}

Part 3: Loss of Kinetic Energy

Initial kinetic energy:

KE_\text{initial} = \frac{1}{2}m|\mathbf{u}|^2

Substitute m = 1 kg, $\mathbf{u} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$

|\mathbf{u}|^2 = 6^2 + (-4)^2 = 36 + 16 = 52

KE_\text{initial} = \frac{1}{2}(1)(52) = 26 \, \text{J}

Final kinetic energy:

KE_\text{final} = \frac{1}{2}m|\mathbf{v}|^2

Substitute $\mathbf{v} = \begin{pmatrix} -3.6 \\ -4 \end{pmatrix}$

|\mathbf{v}|^2 = (-3.6)^2 + (-4)^2 = 12.96 + 16 = 28.96

KE_\text{final} = \frac{1}{2}(1)(28.96) = 14.48 \, \text{J}

Loss of kinetic energy:

\Delta KE = KE_\text{initial} - KE_\text{final} = 26 - 14.48 = 11.52 \, \text{J}

Final Answer:

Velocity after the collision: $\mathbf{v} = \begin{pmatrix} -3.6 \\ -4 \end{pmatrix} \, \text{ms}^{-1}$
Loss of kinetic energy: 11.52 J

Note Summary

infoNote

Common Mistakes

Incorrect resolution of components: Always project velocity onto perpendicular and parallel directions relative to the surface.
Forgetting that parallel velocity remains unchanged.
Ignoring rebound heights: Successive collisions reduce height based on $e^{2n}$

infoNote

Key Formulas

Perpendicular Velocity After Impact:

v_\perp = -e u_\perp

Parallel Velocity After Impact:

v_\parallel = u_\parallel

Rebound Height After $n$ th Collision:

h_n = e^{2n} h_0

Loss of Kinetic Energy:

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

Successive Collisions with a Surface Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

15.3.2 Successive Collisions with a Surface

Introduction

Velocity Components in Oblique Collisions

Perpendicular Component ( $v_\perp$ ):

Parallel Component ( $v_\parallel$ ):

Successive Collisions

Vector Representation of Velocity

Repeated Impacts and Height Reduction

Worked Examples

Example 1: Successive Collisions with a Horizontal Surface

Example 2: Successive Oblique Collisions with a Vertical Wall

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Successive Collisions with a Surface For their A-Level Exams.

Other Revision Notes related to Successive Collisions with a Surface you should explore

Successive Collisions with a Surface Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

15.3.2 Successive Collisions with a Surface

Introduction

Velocity Components in Oblique Collisions

Perpendicular Component (v⊥v_\perpv⊥​):

Parallel Component (v∥v_\parallelv∥​):

Successive Collisions

Vector Representation of Velocity

Repeated Impacts and Height Reduction

Worked Examples

Example 1: Successive Collisions with a Horizontal Surface

Example 2: Successive Oblique Collisions with a Vertical Wall

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Successive Collisions with a Surface For their A-Level Exams.

Other Revision Notes related to Successive Collisions with a Surface you should explore

Perpendicular Component ( $v_\perp$ ):

Parallel Component ( $v_\parallel$ ):