Momentum & Impulse with Vectors (Edexcel A-Level Further Mathematics): Revision Notes

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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 analyse changes in momentum due to forces acting at an angle.

This note covers:

  1. Momentum and impulse in vector form.
  2. The impulse-momentum principle in vector equations.
  3. Calculating the angle of deflexion caused by an impulse.

Momentum as a Vector

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

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

where:

  • mm is the mass (kg\text{kg})
  • v\mathbf{v} is the velocity vector (ms−1\text{ms}^{-1})

Impulse as a Vector

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

J=FΔt\mathbf{J} = \mathbf{F}\Delta t

Impulse also represents the change in momentum:

J=Δp=mvfinal−mvinitial\mathbf{J} = \Delta \mathbf{p} = m\mathbf{v}_\text{final} - m\mathbf{v}_\text{initial}

Angle of Deflexion

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

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

cos⁡θ=vinitial×vfinal∣vinitial∣∣vfinal∣\cos\theta = \frac{\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final}}{|\mathbf{v}_\text{initial}| |\mathbf{v}_\text{final}|}

where:

  • vinitial×vfinal\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final} is the dot product of the velocity vectors,
  • ∣vinitial∣|\mathbf{v}_\text{initial}| and ∣vfinal∣|\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 kg2 \, \text{kg} moves with an initial velocity u=(34) ms−1\mathbf{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \, \text{ms}^{-1}

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

Find:

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

Part 1: Calculate Final Velocity

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

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

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

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

Step 3**:** Substitute the values (m=2 kg,u=(34),J=(4−6)m = 2 \, \text{kg}, \mathbf{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \mathbf{J} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}):

2vfinal=2(34)+(4−6)2\mathbf{v}_\text{final} = 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ -6 \end{pmatrix}

Step 4: Simplify:

2vfinal=(68)+(4−6)=(102)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 vfinal\mathbf{v}_\text{final}

vfinal=(51) ms−1\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 u\mathbf{u} and vfinal\mathbf{v}_\text{final}

∣u∣=32+42=9+16=5 ms−1|\mathbf{u}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \, \text{ms}^{-1}∣vfinal∣=52+12=25+1=26 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:

ΔSpeed=∣vfinal∣−∣u∣\Delta \text{Speed} = |\mathbf{v}_\text{final}| - |\mathbf{u}|ΔSpeed=26−5 ms−1\Delta \text{Speed} = \sqrt{26} - 5 \, \text{ms}^{-1}

Final Answer:

Final velocity: vfinal=(51) ms−1\mathbf{v}_\text{final} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \, \text{ms}^{-1}

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

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Example 2: Angle of Deflexion

Problem

A particle of mass 3 kg3 \, \text{kg} is initially moving with velocity u=(60) ms−1\mathbf{u} = \begin{pmatrix} 6 \\ 0 \end{pmatrix} \, \text{ms}^{-1}

After a force acts on it, its velocity becomes v=(43) ms−1\mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \, \text{ms}^{-1}

Find:

  1. The impulse acting on the particle.
  2. The angle of deflexion caused by the impulse.

Part 1: Calculate Impulse

Step 1: Use the impulse-momentum principle:

J=mv−mu\mathbf{J} = m\mathbf{v} - m\mathbf{u}

Step 2: Substitute m=3 kg,u=(60),v=(43)m = 3 \, \text{kg}, \mathbf{u} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}, \mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}

J=3(43)−3(60)\mathbf{J} = 3\begin{pmatrix} 4 \\ 3 \end{pmatrix} - 3\begin{pmatrix} 6 \\ 0 \end{pmatrix}

Step 3: Simplify:

J=(129)−(180)=(−69) Ns\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⁡θ=u×v∣u∣∣v∣\cos\theta = \frac{\mathbf{u} \times \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}

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

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

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

∣u∣=62+02=6|\mathbf{u}| = \sqrt{6^2 + 0^2} = 6∣v∣=42+32=16+9=5|\mathbf{v}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5

Step 4: Substitute into the cosine formula:

cos⁡θ=246×5=2430=0.8\cos\theta = \frac{24}{6 \times 5} = \frac{24}{30} = 0.8

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

θ=cos⁡−1(0.8)≈36.87∘\theta = \cos^{-1}(0.8) \approx 36.87^\circ

Final Answer:

Impulse: J=(−69) Ns\mathbf{J} = \begin{pmatrix} -6 \\ 9 \end{pmatrix} \, \text{Ns}

Angle of deflexion: 36.87°

Note Summary

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Common Mistakes

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

  1. Momentum: p=mv.\mathbf{p} = m\mathbf{v}.

  2. Impulse: J=FΔt=mvfinal−mvinitial\mathbf{J} = \mathbf{F}\Delta t = m\mathbf{v}_\text{final} - m\mathbf{v}_\text{initial}

  3. Angle of deflection:

cos⁡θ=vinitial×vfinal∣vinitial∣∣vfinal∣\cos\theta = \frac{\mathbf{v}_\text{initial} \times \mathbf{v}_\text{final}}{|\mathbf{v}_\text{initial}| |\mathbf{v}_\text{final}|}

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