13.1.2 Momentum & Impulse with Vectors

Introduction

Momentum and impulse are vector quantities in mechanics, meaning they have both magnitude and direction. In two or three dimensions, the impulse-momentum principle applies component-wise, allowing us to analyze changes in momentum due to forces acting at an angle.

This note covers:

Momentum and impulse in vector form.
The impulse-momentum principle in vector equations.
Calculating the angle of deflection caused by an impulse.

Momentum as a Vector

Momentum ( $\mathbf{p}$ ) in vector form is:

\mathbf{p} = m\mathbf{v}

where:

$m$ is the mass ( $\text{kg}$ )
$\mathbf{v}$ is the velocity vector ( $\text{ms}^{-1}$ )

Impulse as a Vector

Impulse ( $\mathbf{J}$ ) is the product of a force ( $\mathbf{F}$ ) and the time interval ( $\Delta t$ ):

\mathbf{J} = \mathbf{F}\Delta t

Impulse also represents the change in momentum:

\mathbf{J} = \Delta \mathbf{p} = m\mathbf{v}_\text{final} - m\mathbf{v}_\text{initial}

Angle of Deflection

The angle of deflection caused by an impulse is the angle between the initial and final velocity vectors.

If $\mathbf{v}_\text{initial}$ and $\mathbf{v}_\text{final}$ are the initial and final velocity vectors:

\cos\theta = \frac{\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final}}{|\mathbf{v}_\text{initial}| |\mathbf{v}_\text{final}|}

where:

$\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final}$ is the dot product of the velocity vectors,
$|\mathbf{v}_\text{initial}|$ and $|\mathbf{v}_\text{final}|$ are the magnitudes of the initial and final velocities.

Worked Examples

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Example 1: Impulse in Vector Form

Problem

A particle of mass $2 \, \text{kg}$ moves with an initial velocity $\mathbf{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \, \text{ms}^{-1}$

An impulse $\mathbf{J} = \begin{pmatrix} 4 \\ -6 \end{pmatrix} \, \text{Ns}$ acts on it.

Find:

The final velocity of the particle.
The change in speed caused by the impulse.

Part 1: Calculate Final Velocity

Step 1**:** Use the impulse-momentum principle in vector form:

\mathbf{J} = m\mathbf{v}_\text{final} - m\mathbf{v}_\text{initial}

Step 2**:** Rearrange to find $\mathbf{v}_\text{final}$

m\mathbf{v}_\text{final} = m\mathbf{v}_\text{initial} + \mathbf{J}

Step 3**:** Substitute the values ( $m = 2 \, \text{kg}, \mathbf{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \mathbf{J} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}$ ):

2\mathbf{v}_\text{final} = 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ -6 \end{pmatrix}

Step 4: Simplify:

2\mathbf{v}_\text{final} = \begin{pmatrix} 6 \\ 8 \end{pmatrix} + \begin{pmatrix} 4 \\ -6 \end{pmatrix} = \begin{pmatrix} 10 \\ 2 \end{pmatrix}

Step 5**:** Divide by 2 to find $\mathbf{v}_\text{final}$

\mathbf{v}_\text{final} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \, \text{ms}^{-1}

Part 2: Calculate Change in Speed

Step 1: Find the magnitudes of $\mathbf{u}$ and $\mathbf{v}_\text{final}$

|\mathbf{u}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \, \text{ms}^{-1}

|\mathbf{v}_\text{final}| = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} \, \text{ms}^{-1}

Step 2: Calculate the change in speed:

\Delta \text{Speed} = |\mathbf{v}_\text{final}| - |\mathbf{u}|

\Delta \text{Speed} = \sqrt{26} - 5 \, \text{ms}^{-1}

Final Answer:

Final velocity: $\mathbf{v}_\text{final} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \, \text{ms}^{-1}$

Change in speed: $\sqrt{26} - 5 \, \text{ms}^{-1}$

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Example 2: Angle of Deflection

Problem

A particle of mass $3 \, \text{kg}$ is initially moving with velocity $\mathbf{u} = \begin{pmatrix} 6 \\ 0 \end{pmatrix} \, \text{ms}^{-1}$

After a force acts on it, its velocity becomes $\mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \, \text{ms}^{-1}$

Find:

The impulse acting on the particle.
The angle of deflection caused by the impulse.

Part 1: Calculate Impulse

Step 1: Use the impulse-momentum principle:

\mathbf{J} = m\mathbf{v} - m\mathbf{u}

Step 2: Substitute $m = 3 \, \text{kg}, \mathbf{u} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}, \mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$

\mathbf{J} = 3\begin{pmatrix} 4 \\ 3 \end{pmatrix} - 3\begin{pmatrix} 6 \\ 0 \end{pmatrix}

Step 3: Simplify:

\mathbf{J} = \begin{pmatrix} 12 \\ 9 \end{pmatrix} - \begin{pmatrix} 18 \\ 0 \end{pmatrix} = \begin{pmatrix} -6 \\ 9 \end{pmatrix} \, \text{Ns}

Part 2: Calculate Angle of Deflection

Step 1: Use the cosine formula for the angle between two vectors:

\cos\theta = \frac{\mathbf{u} \times \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}

Step 2: Calculate $\mathbf{u} \times \mathbf{v}$

\mathbf{u} \times \mathbf{v} = (6)(4) + (0)(3) = 24

Step 3: Find the magnitudes of $\mathbf{u}$ and $\mathbf{v}$

|\mathbf{u}| = \sqrt{6^2 + 0^2} = 6

|\mathbf{v}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5

Step 4: Substitute into the cosine formula:

\cos\theta = \frac{24}{6 \times 5} = \frac{24}{30} = 0.8

Step 5: Find $\theta$ using the inverse cosine:

\theta = \cos^{-1}(0.8) \approx 36.87^\circ

Final Answer:

Impulse: $\mathbf{J} = \begin{pmatrix} -6 \\ 9 \end{pmatrix} \, \text{Ns}$

Angle of deflection: 36.87°

Note Summary

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Common Mistakes

Sign errors: Forgetting that momentum and impulse are vectors, leading to incorrect direction signs.
Misusing the dot product: Not applying the correct formula for the angle of deflection.
Confusing magnitudes and vectors: Using vector components directly instead of calculating magnitudes.
Forgetting vector notation: Ignoring the vector form in impulse-momentum calculations.
Neglecting non-negativity: Forgetting that speed (magnitude of velocity) is always positive.

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Key Formulas

Momentum: $\mathbf{p} = m\mathbf{v}.$
Impulse: $\mathbf{J} = \mathbf{F}\Delta t = m\mathbf{v}_\text{final} - m\mathbf{v}_\text{initial}$
Angle of deflection:

\cos\theta = \frac{\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final}}{|\mathbf{v}_\text{initial}| |\mathbf{v}_\text{final}|}

Momentum & Impulse with Vectors Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

13.1.2 Momentum & Impulse with Vectors

Introduction

Momentum as a Vector

Impulse as a Vector

Angle of Deflection

Worked Examples

Example 1: Impulse in Vector Form

Example 2: Angle of Deflection

Note Summary

Common Mistakes

Key Formulas

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