Elastic and Inelastic Collisions defined an elastic collision as one that keeps the total kinetic energy. Solving one means handling two equations at once, one of them quadratic, which is a job for Nonlinear Systems. A clever division turns it into two linear equations.
Why it works
For masses $m_1, m_2$ with velocities $v_1, v_2$ before and $v_1', v_2'$ after, move each object's terms to its own side: $$m_1(v_1 - v_1') = m_2(v_2' - v_2) \quad\text{(momentum)},$$ $$m_1(v_1^2 - v_1'^2) = m_2(v_2'^2 - v_2^2) \quad\text{(kinetic energy, times 2)}.$$ Factor the squares, $a^2 - b^2 = (a - b)(a + b)$, and divide the second equation by the first (the collision happened, so $v_1 \ne v_1'$): $$v_1 + v_1' = v_2 + v_2' \;\Rightarrow\; v_2' - v_1' = -(v_2 - v_1).$$
Key idea
In a one-dimensional elastic collision, the relative velocity reverses: the objects separate as fast as they approached. Together with momentum conservation, that gives $$v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}v_1 + \dfrac{2m_2}{m_1 + m_2}v_2, \qquad v_2' = \dfrac{2m_1}{m_1 + m_2}v_1 + \dfrac{m_2 - m_1}{m_1 + m_2}v_2.$$
Worked example: A light glider hits a heavy one
A $2\ \text{kg}$ glider at $6\ \text{m/s}$ collides elastically with a $4\ \text{kg}$ glider at rest. Find both velocities afterward.
- Momentum: $2v_1' + 4v_2' = 12$.
- Relative velocity reverses: $v_2' - v_1' = 6$.
- Substitute $v_2' = v_1' + 6$: $2v_1' + 4v_1' + 24 = 12$, so $v_1' = -2\ \text{m/s}$ and $v_2' = 4\ \text{m/s}$.
- Check the energy: $\tfrac{1}{2}(2)(36) = 36\ \text{J}$ before; $\tfrac{1}{2}(2)(4) + \tfrac{1}{2}(4)(16) = 4 + 32 = 36\ \text{J}$ after.
Answer: $v_1' = -2\ \text{m/s}$ (it bounces back), $v_2' = 4\ \text{m/s}$.
Worked example: Head-on, both moving
A $3\ \text{kg}$ cart at $4\ \text{m/s}$ meets a $1\ \text{kg}$ cart moving toward it at $2\ \text{m/s}$. The collision is elastic. Find the velocities afterward.
- Take the 3 kg cart's direction as positive: $v_1 = 4$, $v_2 = -2$. Momentum: $3v_1' + v_2' = 12 - 2 = 10$.
- Relative velocity: $v_2' - v_1' = 4 - (-2) = 6$.
- Then $3v_1' + v_1' + 6 = 10$, so $v_1' = 1\ \text{m/s}$ and $v_2' = 7\ \text{m/s}$.
Answer: $v_1' = 1\ \text{m/s}$, $v_2' = 7\ \text{m/s}$, both in the positive direction.
Common mistake
Writing only the momentum equation. With two unknown final velocities, one equation can't decide them; any pair on a line works, including ones that create energy. Check: count unknowns and equations. For an elastic collision use momentum plus the reversal of relative velocity (which is energy conservation in disguise); for a perfectly inelastic one, use momentum plus "same final velocity."
- Equal masses, one at rest: they swap velocities (the moving one stops).
- A very heavy object hitting a light one at rest: the light one leaves at nearly $2v_1$.
- A light object hitting a very heavy one at rest: it bounces back at nearly its original speed.