Momentum and impulse
Objective. Model collisions and impulse in one dimension.
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- 2. Worked example
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Learn: the key idea
Momentum and impulse analysis uses conservation of momentum, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, for collisions in an isolated system, alongside impulse J = FΔt = Δ(momentum) for forces acting over short times. In a perfectly elastic collision kinetic energy is also conserved, while in a perfectly inelastic collision the objects stick together afterwards with a common velocity. For masses 2kg and 3kg colliding and sticking, with initial speeds 4 m/s and 0, momentum conservation gives (2)(4)+(3)(0) = 5v, so v = 1.6 m/s. Impulse equals the area under a force-time graph.
Worked example
A 1 kg ball moving at 6 m/s collides and sticks to a stationary 2 kg ball; use momentum and impulse ideas to find their common velocity.
- 1Conservation of momentum: m₁u₁ + m₂u₂ = (m₁+m₂)v
- 2Substitute: (1)(6) + (2)(0) = 3v
- 3Simplify: 6 = 3v
- 4Solve: v = 2
Answer: 2 m/s
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Start practisingCommon mistake. Assuming kinetic energy is conserved in every collision; it is only conserved in perfectly elastic collisions. In inelastic collisions, momentum is conserved but kinetic energy is lost to heat and deformation.
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