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UY1: Linear Momentum, Impulse & Collisions

Define momentum and impulse, apply conservation laws to systems, and classify collisions using energy and the coefficient of restitution (UY1 Mechanics).

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  • University Physics Year 1
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Enlarged diagram

Why this matters + quick links

This page gives the UY1 working model/result for Linear Momentum, Impulse & Collisions. You reuse it as part of the mechanics toolkit: choose coordinates, pick a method (Newton/energy/momentum), and check limits and signs.

1) At a glance

Δ vector P_system = ∫ ∑ vector Fₑₓₜ dt

If external impulse is negligible, vector P_system is conserved.

  • Common trap: conserving kinetic energy in every collision. Only momentum is always conserved for an isolated system.

2) Setup (model, assumptions, coordinates/signs)

Before you write equations, do the modelling step explicitly. Most “collision mistakes” are modelling mistakes.

A. Choose the system (what you include)

  • For two-body collisions, the system is usually “object 1 + object 2”.
  • Exclude the Earth/table unless the question forces you to include them. Excluding them means gravity/normal are external.

B. Choose the time window

  • Momentum conservation is applied over a time interval. In a collision, that interval is the short contact time Δ t.

C. Decide whether external impulse is negligible

  • Write the impulse form: Δ vector P = ∫ ∑ vector Fₑₓₜ dt.
  • If the collision is brief and horizontal, the external impulse (e.g. mg Δ t) is often small compared with the contact impulse, so vector P is approximately conserved in the horizontal direction.

D. Choose axes + sign convention

  • 1D: pick one positive direction and keep it; treat velocities as signed numbers (-2 m s⁻¹ means “the other way”).
  • 2D: choose x and y axes and conserve momentum component-by-component.
  • For oblique impacts, it is often cleaner to resolve along the line of impact (normal direction) and the tangential direction.
Method choice (how to pick the right tool)
  • Use momentum/impulse when the interaction force is large and short-lived (collisions, explosions, recoil), or when internal forces are messy/unknown.
  • Use energy when forces are conservative and you need speeds/turning points (but energy alone cannot give directions, and kinetic energy is not conserved in inelastic collisions).
  • Use Newton’s 2nd law when you need time-dependent motion or forces/accelerations during the interaction.
Sign conventions that break a lot of collision solutions
  • In 1D, keep the algebra signed. Don’t “put everything as a magnitude” and then guess directions later.
  • The coefficient of restitution formula uses relative speeds along the line of impact (the 1D axis in a head-on collision).
  • If your answer gives a negative speed, it’s not “wrong”: it means the final direction is opposite to your chosen positive axis.

3) Core method / derivation

For one particle:

vector p = m vector v

with units kg m s⁻¹.

From Newton’s 2nd law:

vector F = (d vector p)/dt ⇒ ∫_t₁^t₂ vector F dt = Δ vector p = vector J

where vector J is impulse.

For many particles:

vector P = ∑ᵢ vector pᵢ, (d vector P)/dt = ∑ᵢ vector Fᵢ

Internal forces cancel pairwise in the total sum, so:

(d vector P)/dt = ∑ vector Fₑₓₜ

Hence

Δ vector P = ∫ ∑ vector Fₑₓₜ dt

and with negligible external impulse, vector P is constant.

For 1D collisions:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Second relation:

  • perfectly inelastic (stick): v₁ = v₂ = v
  • with restitution:
e = (v₂-v₁)/(u₁-u₂), 0 ≤ e ≤ 1

Checks:

  • Units: impulse vector J has units N s = kg m s⁻¹.
  • Sign check: a negative solved velocity means motion opposite your chosen positive axis.
  • Limit check: e → 0 gives sticking behavior; e → 1 is elastic.
  • Energy check: in a perfectly inelastic collision, kinetic energy must decrease; in a perfectly elastic collision, kinetic energy is conserved (in addition to momentum).

4) Worked examples

Worked example 1

A perfectly inelastic collision

Problem

A 2 kg cart moving at 5 m s⁻¹ hits a 3 kg cart at rest, and they stick. Find their common velocity.

Show full solution

  1. Conserve momentum

    Method

    Equate total momentum before and after.

    Reason

    No external horizontal impulse acts during the collision; after it the carts share one velocity.

    Working

    Answer

    m₁u₁ + m₂u₂ = (m₁ + m₂)v.

  2. State the answer

    Working

    Answer

    v = (2(5) + 3(0))/5 = 2 m s⁻¹, in the original direction.

Worked example 2

A collision with a coefficient of restitution

Problem

A 1 kg ball at 4 m s⁻¹ hits a 2 kg ball at rest head-on. The coefficient of restitution is e = 0.5. Find both final velocities.

Show full solution

  1. Conserve momentum

    Method

    Write the momentum equation.

    Reason

    Momentum is conserved whatever the value of e.

    Working

    Answer

    4 = v₁ + 2v₂.

  2. Apply restitution

    Method

    Use e = (v₂-v₁)/(u₁-u₂).

    Reason

    e compares the separation speed with the approach speed, giving the second equation.

    Working

    Answer

    0.5 = (v₂-v₁)/4 ⇒ v₂ = v₁ + 2.

  3. Solve

    Method

    Substitute into the momentum equation.

    Reason

    Two equations fix two unknowns.

    Working

    Answer

    4 = v₁ + 2(v₁ + 2) ⇒ v₁ = 0.

  4. State the answer

    Working

    Answer

    v₁ = 0 and v₂ = 2 m s⁻¹.

Worked example 3

A sticking collision in two dimensions

Problem

On frictionless ice, a 0.20 kg puck moving east at 3.0 m s⁻¹ hits a 0.30 kg puck moving north at 2.0 m s⁻¹, and they stick. Find their final velocity.

Show full solution

  1. Add the momenta as vectors

    Method

    Write each momentum in components.

    Reason

    Momentum is a vector, so each component is conserved separately.

    Working

    Answer

    vector Pᵢ = (0.20)(3.0,0) + (0.30)(0,2.0) = (0.60,0.60) kg m s⁻¹.

  2. Divide by the total mass

    Method

    Use vector Pf = M vector v with M = 0.50 kg.

    Reason

    After sticking, the pucks move as one body.

    Working

    Answer

    vector v = (0.60,0.60)/0.50.

  3. State the answer

    Working

    Answer

    vector v = (1.2,1.2) m s⁻¹: 1.7 m s⁻¹ at 45° north of east.

5) Practice set

Check your understanding 1

A 0.20 kg ball’s velocity changes from + 8 to -6 m s⁻¹ in a collision. Find the impulse on the ball.

Show hint

Impulse equals the change in momentum, m(v-u).

Show answer

J = m(v-u) = 0.20(-6-8) = -2.8 N s: 2.8 N s in the negative direction. The rebound makes the change larger than either speed alone.

Check your understanding 2

Two carts stick together: 0.5 kg at 3 m s⁻¹ and 1.0 kg at -1 m s⁻¹. Find their common velocity.

Show hint

Divide the total initial momentum by the total mass.

Show answer

v = (0.5(3) + 1.0(-1))/1.5 = 0.5/1.5 ≈ 0.33 m s⁻¹, in the positive direction.

Check your understanding 3

In a head-on collision u₁ = 6 and u₂ = 2 m s⁻¹ (same direction), and afterwards v₁ = 3 and v₂ = 4.5 m s⁻¹. Find e.

Show hint

Compare the separation speed with the approach speed.

Show answer

e = (v₂-v₁)/(u₁-u₂) = (4.5-3)/(6-2) = 0.375: the balls separate at well under half their approach speed.

6) Summary + next steps

  • Momentum tracks system motion reliably when collision forces are brief and large.
  • Impulse links force-over-time to momentum change directly.
  • Collision problems need momentum plus one more relation (sticking, restitution, or energy when valid).

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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