Momentum & Impulse in 1D Revision Notes for Edexcel A-Level Further Mathematics

13.1.1 Momentum & Impulse in 1D

Introduction

Momentum and impulse are fundamental concepts in mechanics that describe how forces act on objects over time. The impulse-momentum principle connects these ideas, while the principle of conservation of momentum governs interactions like collisions between particles.

This note focuses on:

Momentum and impulse in one-dimensional motion.
Using the impulse-momentum principle.
Applying the principle of conservation of momentum to 1D collisions between two spheres.

Momentum

Momentum ( $p$ ) is the product of an object's mass and velocity:

p = mv

where:

$m$ is the mass ( $\text{kg}$ )
$v$ is the velocity ( $\text{ms}^{-1}$ ) Momentum is a vector quantity, but in 1D problems, we treat it as positive or negative depending on direction.

Impulse

Impulse ( $J$ ) is the product of a force ( $F$ ) and the time ( $\Delta t$ ) over which the force acts:

J = F\Delta t

Impulse is also equal to the change in momentum:

J = \Delta p = mv - mu

where:

$u$ is the initial velocity,
$v$ is the final velocity.

Impulse-Momentum Principle

The impulse acting on a particle is equal to the change in its momentum:

F\Delta t = mv - mu

Principle of Conservation of Momentum

In a closed system (no external forces), the total momentum before an interaction is equal to the total momentum after:

\text{Total Momentum Before} = \text{Total Momentum After}

For two spheres of masses $m_1$ and $m_2$ , with initial velocities $u_1$ and $u_2$ , and final velocities $v_1$ and $v_2$ :

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Worked Examples

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Example 1: Impulse and Momentum

Problem

A particle of mass $2 \, \text{kg}$ is initially moving at $3 \, \text{ms}^{-1}$ .

A force acts on it for $4 \, \text{s}$ , causing its velocity to increase to 7 $\, \text{ms}^{-1}$ .

Find:

The impulse on the particle.
The magnitude of the force.

Step 1: Recall the formula for impulse in terms of momentum:

J = mv - mu

where:

$m$ is the mass of the particle ( $2 \, \text{kg}$ )
$u$ is the initial velocity ( $3 \, \text{ms}^{-1}$ )
$v$ is the final velocity ( $7 \, \text{ms}^{-1}$ )

Step 2: Substitute the values:

J = 2(7) - 2(3) = 14 - 6 = 8 \, \text{Ns}

Step 3: Recall the relationship between impulse and force:

J = F\Delta t

where:

$J$ is the impulse ( $8 \, \text{Ns}$ )
$\Delta t$ is the time the force acts ( $4 \, \text{s}$ )
$F$ is the magnitude of the force.

Step 4: Rearrange the formula to solve for $F$ :

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

Step 5: Substitute the values:

F = \frac{8}{4} = 2 \, \text{N}

Final Answer:

Impulse: 8 Ns
Force: 2 N

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Example 2: Conservation of Momentum in a Collision

Problem

Two spheres $A$ and $B$ of masses $3 \, \text{kg}$ and $5 \, \text{kg}$ , respectively, are moving in a straight line.

Before the collision:

Sphere $A$ moves at $4 \, \text{ms}^{-1}$
Sphere $B$ moves at $2 \, \text{ms}^{-1}$ After the collision, sphere $A$ moves at $1 \, \text{ms}^{-1}$

Find the velocity of sphere $B$ after the collision.

Step 1: Write the conservation of momentum formula

The total momentum before the collision equals the total momentum after:

m_Au_A + m_Bu_B = m_Av_A + m_Bv_B

where:

$m_A$ and $m_B$ are the masses of spheres $A$ and $B$
$u_A$ and $u_B$ are their initial velocities
$v_A$ and $v_B$ are their final velocities.

Step 2: Substitute the known values

3(4) + 5(2) = 3(1) + 5v_B

Step 3: Simplify

12 + 10 = 3 + 5v_B

22 = 3 + 5v_B

Step 4: Solve for $v_B$

5v_B = 19 \quad \Rightarrow \quad v_B = \frac{19}{5} = 3.8 \, \text{ms}^{-1}

Final Answer:

The velocity of sphere $B$ after the collision is 3.8 ms⁻¹

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Example 3: Impulse During a Collision

Problem

In Example 2, find the impulse exerted by sphere $B$ on sphere $A$ .

Step 1: Recall the impulse-momentum principle

Impulse is the change in momentum of a particle:

J = mv - mu

where:

$m = 3 \, \text{kg}$
$u = 4 \, \text{ms}^{-1}$ (initial velocity of $A$ )
$v = 1 \, \text{ms}^{-1}$ (final velocity of $A$ )

Step 2: Substitute the values

J = 3(1) - 3(4)

J = 3 - 12 = -9 \, \text{Ns}

The negative sign indicates the impulse is opposite to sphere $A$ 's initial direction.

Step 3: Use Newton's Third Law

By Newton's third law, the impulse exerted by $A$ on $B$ is equal in magnitude but opposite in direction.

Hence, the impulse on $B$ is:

J_B = +9 \, \text{Ns}

Final Answer:

The impulse exerted by sphere $B$ on $A$ is -9 Ns, and the impulse on $B$ is +9 Ns

Note Summary

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Common Mistakes

Sign errors: Forgetting that velocity (and momentum) can be negative depending on direction.
Impulse misunderstanding: Not recognising that impulse equals the total change in momentum.
Momentum conservation errors: Failing to account for both objects in the system.
Force-impulse confusion: Misinterpreting $F$ as impulse instead of $F\Delta t$
Omitting mass: Forgetting that momentum and impulse depend on mass.

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Key Formulas

Momentum:

p = mv

Impulse:

J = F\Delta t = mv - mu

Impulse-Momentum Principle:

F\Delta t = \Delta p

Conservation of Momentum:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Momentum & Impulse in 1D (Edexcel A-Level Further Mathematics): Revision Notes

13.1.1 Momentum & Impulse in 1D

Introduction

Momentum

Impulse

Impulse-Momentum Principle

Principle of Conservation of Momentum

Worked Examples

Example 1: Impulse and Momentum

Example 2: Conservation of Momentum in a Collision

Example 3: Impulse During a Collision

Note Summary

Common Mistakes

Key Formulas

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