Oblique Collisions with a Surface (Edexcel A-Level Further Mathematics): Revision Notes

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15.3.1 Oblique Collisions with a Surface

Introduction

In an oblique collision, a smooth elastic sphere strikes a fixed smooth surface or another smooth elastic sphere at an angle. These collisions involve the decomposition of velocities into components perpendicular and parallel to the surface of contact. The behaviour in these two directions differs:

  • Perpendicular to the surface: Newton's law of restitution applies.
  • Parallel to the surface: No impulse acts, so velocity remains unchanged. This note covers:
  1. Oblique impacts with a fixed smooth surface.
  2. Loss of kinetic energy during such impacts.
  3. Vector forms for solving problems involving collisions at arbitrary angles.

Velocity Components in an Oblique Collision

The velocity of the sphere is resolved into:

  1. Perpendicular component (v⊥v_\perp): Acts normal to the surface.
  2. Parallel component (v∥v_\parallel): Acts tangentially along the surface.

Behaviour During Impact

Perpendicular Direction:

Newton's law of restitution applies:

v⊥,after=−ev⊥,beforev_{\perp, \text{after}} = -e v_{\perp, \text{before}}

where ee is the coefficient of restitution (0≤e≤10 \leq e \leq 1)

Parallel Direction:

The velocity remains unchanged as no impulse acts in this direction:

v∥,after=v∥,beforev_{\parallel, \text{after}} = v_{\parallel, \text{before}}

Loss of Kinetic Energy

The total kinetic energy before and after the collision is:

KEinitial=12m(v⊥,before2+v∥,before2)KE_\text{initial} = \frac{1}{2}m(v_{\perp,\text{before}}^2 + v_{\parallel,\text{before}}^2) KEfinal=12m(v⊥,after2+v∥,after2)KE_\text{final} = \frac{1}{2}m(v_{\perp,\text{after}}^2 + v_{\parallel,\text{after}}^2)

Using v⊥,after=−ev⊥,beforev_{\perp,\text{after}} = -e v_{\perp,\text{before}}, the loss of kinetic energy due to the collision is:

ΔKE=12m(1−e2)v⊥,before2\Delta KE = \frac{1}{2}m(1 - e^2)v_{\perp,\text{before}}^2

Vector Representation

For problems involving arbitrary angles, represent the velocity as a vector:

v=v∥+v⊥\mathbf{v} = \mathbf{v}_\parallel + \mathbf{v}_\perp
  • v⊥\mathbf{v}_\perp: The projection of v\mathbf{v} onto the normal direction.
  • v∥\mathbf{v}_\parallel: The projection of v\mathbf{v} onto the tangential direction.

Worked Examples

infoNote

Example 1: Oblique Impact with a Smooth Wall

Problem

A sphere of mass 2 kg is moving at 8 ms⁻¹ towards a smooth vertical wall at an angle of 60° to the wall.

The coefficient of restitution between the sphere and the wall is e = 0.5

Find:

  1. The velocity components after the collision.
  2. The loss of kinetic energy due to the collision.

Step 1: Velocity Components Before Collision

Resolve the initial velocity u\mathbf{u}

Perpendicular component (u⊥u_\perp):

u⊥=usin⁡60∘=8×32=:highlight[43] ms−1u_\perp = u \sin 60^\circ = 8 \times \frac{\sqrt{3}}{2} = :highlight[4\sqrt{3}] \, \text{ms}^{-1}

Parallel component (u∥u_\parallel):

u∥=ucos⁡60∘=8×12=:highlight[4] ms−1u_\parallel = u \cos 60^\circ = 8 \times \frac{1}{2} = :highlight[4] \, \text{ms}^{-1}

Step 2: Velocity Components After Collision

Perpendicular direction (v⊥v_\perp):

v⊥=−eu⊥=−0.5×43=:highlight[−23] ms−1v_\perp = -e u_\perp = -0.5 \times 4\sqrt{3} = :highlight[-2\sqrt{3}] \, \text{ms}^{-1}

Parallel direction (v∥v_\parallel): Remains unchanged:

v∥=u∥=:highlight[4] ms−1v_\parallel = u_\parallel = :highlight[4] \, \text{ms}^{-1}

Step 3: Final Velocity

The final velocity (v\mathbf{v}) is the vector sum of v⊥v_\perp and v∥v_\parallel:

v=v∥2+v⊥2v = \sqrt{v_\parallel^2 + v_\perp^2}

Substitute:

v=42+(−23)2=16+12=28≈:highlight[5.29] ms−1v = \sqrt{4^2 + (-2\sqrt{3})^2} = \sqrt{16 + 12} = \sqrt{28} \approx :highlight[5.29] \, \text{ms}^{-1}

Step 4: Loss of Kinetic Energy

Initial kinetic energy:

KEinitial=12m(u⊥2+u∥2)KE_\text{initial} = \frac{1}{2}m(u_\perp^2 + u_\parallel^2)

Substitute m=2,u⊥=43,u∥=4m = 2, u_\perp = 4\sqrt{3}, u_\parallel = 4

KEinitial=12(2)((43)2+42)=(48+16)=:highlight[64J]KE_\text{initial} = \frac{1}{2}(2)((4\sqrt{3})^2 + 4^2) = (48 + 16) = :highlight[64 J]

Final kinetic energy:

KEfinal=12m(v⊥2+v∥2)KE_\text{final} = \frac{1}{2}m(v_\perp^2 + v_\parallel^2)

Substitute v⊥=−23,v∥=4v_\perp = -2\sqrt{3}, v_\parallel = 4

KEfinal=12(2)((−23)2+42)=(12+16)=:highlight[28J]KE_\text{final} = \frac{1}{2}(2)((-2\sqrt{3})^2 + 4^2) = (12 + 16) = :highlight[28 J]

Loss of kinetic energy:

ΔKE=KEinitial−KEfinal=64−28=:highlight[36J]\Delta KE = KE_\text{initial} - KE_\text{final} = 64 - 28 = :highlight[36 J]

Final Answer:

Final velocity components:

  • Perpendicular: -2√3 ms⁻¹
  • Parallel: 4 ms⁻¹
  • Magnitude: 5.29 ms⁻¹ Loss of kinetic energy: 36 J
infoNote

Example 2: Oblique Collision in Vector Form

Problem

A sphere has an initial velocity u=(3−4) ms−1\mathbf{u} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} \, \text{ms}^{-1} and strikes a smooth horizontal plane.

The coefficient of restitution is e = 0.8

Find the velocity of the sphere after the collision.

Step 1: Resolve Velocity Components

Perpendicular component (u⊥\mathbf{u}_\perp): Normal to the plane (vertical):

u⊥=(0−4)\mathbf{u}_\perp = \begin{pmatrix} 0 \\ -4 \end{pmatrix}

Parallel component (u∥\mathbf{u}_\parallel): Tangential to the plane (horizontal):

u∥=(30)\mathbf{u}_\parallel = \begin{pmatrix} 3 \\ 0 \end{pmatrix}

Step 2: Apply Newton's Law of Restitution

Perpendicular component after collision:

v⊥=−eu⊥=−0.8(0−4)=(03.2)\mathbf{v}_\perp = -e \mathbf{u}_\perp = -0.8 \begin{pmatrix} 0 \\ -4 \end{pmatrix} = \begin{pmatrix} 0 \\ 3.2 \end{pmatrix}

Parallel component after collision:

v∥=u∥=(30)\mathbf{v}_\parallel = \mathbf{u}_\parallel = \begin{pmatrix} 3 \\ 0 \end{pmatrix}

Step 3: Combine Components

The final velocity is:

v=v⊥+v∥=(33.2)\mathbf{v} = \mathbf{v}_\perp + \mathbf{v}_\parallel = \begin{pmatrix} 3 \\ 3.2 \end{pmatrix}

Final Answer:

The velocity after the collision is v=(33.2)\mathbf{v} = \begin{pmatrix} 3 \\ 3.2 \end{pmatrix} ms⁻¹

Note Summary

infoNote

Common Mistakes

  1. Confusing components: Always resolve velocities into perpendicular and parallel components.
  2. Forgetting restitution applies only perpendicular to the surface.
  3. Ignoring signs: Ensure the direction of the perpendicular component reverses after the collision.
  4. Mixing up kinetic energy: Only the perpendicular component contributes to energy loss.
infoNote

Key Formulas

  1. Perpendicular Velocity After Collision:
v⊥=−eu⊥v_\perp = -e u_\perp
  1. Parallel Velocity After Collision:
v∥=u∥v_\parallel = u_\parallel
  1. Loss of Kinetic Energy:
ΔKE=12m(1−e2)u⊥2\Delta KE = \frac{1}{2}m(1 - e^2)u_\perp^2

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