Work, Energy, Impulse and Momentum

Many dynamics problems are faster with energy or momentum methods than with F = ma, because they skip the accelerations and go straight from one state to another.

Work and energy

• Work by a constant force: W = F s cos θ. Work by a spring stretched from x₁ to x₂: ½k(x₁² − x₂²).
• Kinetic energy: translation ½mv²; rotation ½Iω²; rolling ½mv² + ½I_Gω².
• Work–energy theorem: total work done on a body = change in its kinetic energy.
• Conservation of energy (no friction): KE + PE = constant, with PE = mgh (gravity) + ½kx² (spring).
• Power = rate of doing work = F·v = Tω.

Impulse and momentum

• Linear impulse = ∫F dt = change in momentum m(v₂ − v₁).
• If no external force acts on a system in some direction, its total momentum in that direction is conserved.
• Angular impulse = change of angular momentum Iω. With no external moment, Iω is conserved (a skater who pulls in their arms spins faster).

Collisions

For a direct central impact between masses m₁ and m₂:

• Momentum is conserved: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
• Coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) = relative speed of separation / relative speed of approach.
• e = 1: perfectly elastic (kinetic energy conserved). e = 0: perfectly plastic (bodies move together).
• Loss of kinetic energy = [m₁m₂ / (2(m₁ + m₂))] (1 − e²)(u₁ − u₂)².

A ball dropped from height h onto a floor rebounds to e²h.

Worked example

A 2 kg ball at 6 m/s hits a stationary 4 kg ball head-on with e = 0.5. Find the speeds after impact.

• Momentum: 2 × 6 = 2v₁ + 4v₂, so v₁ + 2v₂ = 6.
• Restitution: v₂ − v₁ = 0.5 × 6 = 3.
• Adding: 3v₂ = 9, v₂ = 3 m/s, v₁ = 0 m/s.
• KE loss = ½·2·36 − ½·4·9 = 36 − 18 = 18 J.

Practice