IB — physicsMedium2 min read

Momentum and Impulse

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

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Tutor AI Team

# Momentum and Impulse — IB Physics

Momentum is a fundamental quantity in physics, conserved in all interactions. Combined with impulse, it provides powerful tools for analysing collisions and explosions.

1. Momentum

p=mv\boxed{p = mv}

Units: kg·m/s or N·s. Momentum is a vector — direction matters.

2. Impulse

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

Impulse = area under force-time graph.

Applications: airbags, crumple zones, helmets all increase the time of impact, reducing the average force for the same change in momentum.

3. Conservation of Momentum

In a closed system (no external forces): ∑pbefore=∑pafter\sum p_{\text{before}} = \sum p_{\text{after}} m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

4. Types of Collisions

Elastic: KE is conserved; momentum conserved. (Atomic collisions)

Inelastic: KE is NOT conserved (some → heat/sound/deformation); momentum still conserved.

Perfectly inelastic: Objects stick together. Maximum KE loss.

Worked Example: Collision

Problem

3 kg at 4 m/s hits 2 kg at rest. They stick together. Find final velocity.

3(4)+2(0)=5v3(4) + 2(0) = 5v v=12/5=2.4v = 12/5 = 2.4 m/s

KE before: 12(3)(16)=24\frac{1}{2}(3)(16) = 24 J. KE after: 12(5)(5.76)=14.4\frac{1}{2}(5)(5.76) = 14.4 J. KE lost: 9.6 J → inelastic ✓

Solution

Worked Example: Explosion

Problem

A 5 kg object at rest splits into 2 kg and 3 kg pieces. The 2 kg piece moves at 6 m/s.

0=2(6)+3v0 = 2(6) + 3v → v=−4v = -4 m/s (opposite direction)

Solution

Worked Example: Impulse

Problem

A 0.15 kg cricket ball at 30 m/s is hit back at 40 m/s. Contact time = 0.02 s.

Δp=0.15(40)−0.15(−30)=6+4.5=10.5\Delta p = 0.15(40) - 0.15(-30) = 6 + 4.5 = 10.5 kg·m/s F=Δp/Δt=10.5/0.02=525F = \Delta p/\Delta t = 10.5/0.02 = 525 N

Solution

6. Practice Questions

    1. A 0.5 kg ball at 10 m/s hits a wall and bounces back at 8 m/s. Find the change in momentum. (2 marks)
    1. A 1500 kg car at 20 m/s collides with a 1000 kg car at 10 m/s (same direction). They stick together. Find the final velocity. (3 marks)
    1. Explain how crumple zones reduce injury in car crashes. (3 marks)

    Answers

    1. Δp=0.5(−8)−0.5(10)=−4−5=−9\Delta p = 0.5(-8) - 0.5(10) = -4 - 5 = -9 kg·m/s. Magnitude = 9 kg·m/s.
    1. 1500(20)+1000(10)=2500v1500(20) + 1000(10) = 2500v. v=40000/2500=16v = 40000/2500 = 16 m/s.

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Summary

  • p=mvp = mv (vector); J=FΔt=ΔpJ = F\Delta t = \Delta p
  • Momentum conserved in all collisions (closed system)
  • Elastic: KE conserved; Inelastic: KE not conserved
  • Safety: increase time → decrease force
Tags:
#ib#physics#mechanics#momentum#impulse#collisions#conservation

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