Elastic vs inelastic collisions
Momentum survives every collision; kinetic energy only survives the elastic ones — and in one dimension elastic collisions have a clean shortcut.
Perfectly inelastic
The extreme case: the objects stick together and share one final velocity. Momentum is still conserved even though kinetic energy is not — the missing energy went into denting, heating, and sound. Add the momenta before, divide by the combined mass after.
v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}Elastic collisions
An elastic collision conserves kinetic energy as well as momentum. In one dimension this gives a shortcut: the relative velocity reverses, so the closing speed before equals the separating speed after. Equal masses are a special case — they simply swap velocities, which is how billiards pass speed down the line.
v_1 - v_2 = -(v_1' - v_2')Worked example
Inelastic: a 2.0 kg block sliding at 6.0 m/s hits a 4.0 kg block at rest and they stick. Final speed is 12/6 = 2.0 m/s; kinetic energy drops from 36 J to 12 J, so 24 J went into the crunch. Elastic: a 2.0 kg ball at 3.0 m/s elastically hits a 1.0 kg ball at rest; they separate at 1.0 m/s and 4.0 m/s, which conserves both momentum (6 = 2 + 4, in kg m/s) and kinetic energy (9 J = 1 J + 8 J).
v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 = \frac{1}{3}(3.0) = 1.0\ \text{m/s}Flashcard
Perfectly inelastic final velocity
Check your understanding
A 3.0 kg block sliding at 8.0 m/s collides with a 5.0 kg block at rest and they stick together. What is their common speed after the collision?