A-Level — physicsMedium5 min read

Linear Momentum and Impulse

p = mv; impulse = FΔt; conservation of momentum; elastic and inelastic collisions

Tutor AI Team
Tutor AI Team

# Linear Momentum and Impulse — A-Level Physics

Momentum is one of the most fundamental quantities in physics. The law of conservation of momentum, together with the impulse-momentum theorem, allows us to analyse collisions and explosions with great precision.

1. Momentum

p⃗=mv⃗\boxed{\vec{p} = m\vec{v}}

  • Vector quantity — direction matters
  • Units: kg m/s (or N s)
  • A stationary object has zero momentum

2. Newton's Second Law (Momentum Form)

F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

For constant mass: F=ΔpΔt=m(v−u)Δt=maF = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t} = ma

3. Impulse

Impulse (JJ) is the change in momentum:

J=FΔt=Δp=mv−mu\boxed{J = F\Delta t = \Delta p = mv - mu}

  • Units: N s (equivalent to kg m/s)
  • Impulse equals the area under a force-time graph
  • For a variable force: J=∫F dtJ = \int F \, dt

Impulse and Safety

For a given change in momentum, increasing the time reduces the force: F=ΔpΔtF = \frac{\Delta p}{\Delta t}

Smaller force over longer time → less damage. This explains:

  • Crumple zones, airbags, seatbelts (increase collision time)
  • Bending knees when landing (increases deceleration time)
  • Catching a cricket ball by moving hands back

4. Conservation of Momentum

In a closed system (no external forces), the total momentum before an interaction equals the total momentum after.

∑pbefore=∑pafter\sum p_{\text{before}} = \sum p_{\text{after}}

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Why?

From Newton's Third Law: during a collision, F12=−F21F_{12} = -F_{21} for the same time Δt\Delta t. So Δp1=−Δp2\Delta p_1 = -\Delta p_2, meaning total momentum is unchanged.

5. Types of Collisions

Perfectly Elastic

  • Momentum conserved ✓
  • Kinetic energy conserved ✓
  • Relative speed of approach = relative speed of separation
  • Objects bounce apart

Inelastic

  • Momentum conserved ✓
  • Kinetic energy NOT conserved ✗ (some → heat, sound, deformation)
  • Most real collisions

Perfectly Inelastic

  • Objects stick together (maximum KE loss)
  • Momentum still conserved
  • (m1+m2)v=m1u1+m2u2(m_1 + m_2)v = m_1u_1 + m_2u_2

Explosions (Reverse Collision)

  • Total initial momentum = 0 (if initially at rest)
  • m1v1=−m2v2m_1v_1 = -m_2v_2
  • KE increases (chemical/elastic energy → kinetic energy)

6. 2D Collisions

Conservation of momentum applies independently in each direction:

∑px (before)=∑px (after)\sum p_x \text{ (before)} = \sum p_x \text{ (after)} ∑py (before)=∑py (after)\sum p_y \text{ (before)} = \sum p_y \text{ (after)}

7. Coefficient of Restitution (Extension)

e=relative speed of separationrelative speed of approach=v2−v1u1−u2e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}

  • e=1e = 1: perfectly elastic
  • e=0e = 0: perfectly inelastic
  • 0<e<10 < e < 1: partially inelastic

Worked Example: Impulse

Problem

A 0.15 kg ball hits a wall at 20 m/s and bounces back at 15 m/s. Find the impulse.

Taking the initial direction as positive: J=mv−mu=0.15(−15)−0.15(20)=−2.25−3.0=−5.25 N sJ = mv - mu = 0.15(-15) - 0.15(20) = -2.25 - 3.0 = -5.25 \text{ N s}

The impulse has magnitude 5.25 N s, directed away from the wall.

Solution

Worked Example: Collision

Problem

A 2 kg trolley at 3 m/s collides with a stationary 4 kg trolley. They stick together. Find (a) the velocity, (b) the KE lost.

(a) 2(3)+4(0)=6v2(3) + 4(0) = 6v → v=1v = 1 m/s

(b) KE before: 12(2)(9)=9\frac{1}{2}(2)(9) = 9 J. KE after: 12(6)(1)=3\frac{1}{2}(6)(1) = 3 J. Lost: 6 J.

Solution

Worked Example: 2D Collision

Problem

A 3 kg ball moving at 4 m/s east collides with a 2 kg ball moving at 3 m/s north. They stick together. Find the velocity.

px=3(4)=12p_x = 3(4) = 12 kg m/s, py=2(3)=6p_y = 2(3) = 6 kg m/s

Total mass = 5 kg. vx=12/5=2.4v_x = 12/5 = 2.4 m/s, vy=6/5=1.2v_y = 6/5 = 1.2 m/s v=2.42+1.22=7.2=2.68v = \sqrt{2.4^2 + 1.2^2} = \sqrt{7.2} = 2.68 m/s θ=tan⁡−1(1.2/2.4)=26.6°\theta = \tan^{-1}(1.2/2.4) = 26.6° north of east

Solution

Worked Example: Force-Time Graph

Problem

A force acts on a 0.5 kg object initially at rest. The force-time graph is a triangle with peak 100 N at t = 0.02 s (total time 0.04 s). Find the final velocity.

Impulse = area = 12×0.04×100=2\frac{1}{2} \times 0.04 \times 100 = 2 N s Δp=mv−0=2\Delta p = mv - 0 = 2 → v=2/0.5=4v = 2/0.5 = 4 m/s

Solution

9. Practice Questions

    1. A 60 kg ice skater pushes a 40 kg skater. The 40 kg skater moves at 3 m/s. Find the velocity of the 60 kg skater. (3 marks)
    1. A 0.4 kg ball falls from 5 m and bounces to 3.2 m. Calculate the impulse during the bounce. (5 marks)
    1. Show that the collision between a 1 kg ball at 6 m/s and a 2 kg ball at rest where the 1 kg ball rebounds at 2 m/s is elastic. (4 marks)

    Answers

    1. 0=60v+40(3)0 = 60v + 40(3) → v=−2v = -2 m/s (opposite direction).
    1. Speed before bounce: v=2gh=2(9.81)(5)=9.90v = \sqrt{2gh} = \sqrt{2(9.81)(5)} = 9.90 m/s (down). Speed after bounce: v=2(9.81)(3.2)=7.92v = \sqrt{2(9.81)(3.2)} = 7.92 m/s (up). Impulse = m(vafter−vbefore)=0.4(7.92−(−9.90))=0.4(17.82)=7.13m(v_{\text{after}} - v_{\text{before}}) = 0.4(7.92 - (-9.90)) = 0.4(17.82) = 7.13 N s (upward).

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Summary

  • p⃗=mv⃗\vec{p} = m\vec{v}; F=dp/dtF = dp/dt
  • Impulse J=FΔt=ΔpJ = F\Delta t = \Delta p (area under F-t graph)
  • Conservation: total momentum before = after (closed system)
  • Elastic: KE conserved. Inelastic: KE not conserved
  • 2D: apply conservation independently in x and y directions
  • Safety: increase time → reduce force for same impulse
Tags:
#a-level#physics#mechanics#momentum#impulse#collisions#conservation

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