In a one-dimensional elastic collision, both total momentum and total kinetic energy are conserved. Let $u_1,u_2$ be the initial velocities and $v_1,v_2$ the final velocities. Then

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2,$$

and

$$\frac12m_1u_1^2+\frac12m_2u_2^2 =\frac12m_1v_1^2+\frac12m_2v_2^2.$$

These two independent constraints determine the two final velocities when the masses and initial velocities are known.

Relative velocity reverses

Rewrite the energy equation as

$$m_1(u_1-v_1)(u_1+v_1) =-m_2(u_2-v_2)(u_2+v_2).$$

Momentum conservation gives

$$m_1(u_1-v_1)=-m_2(u_2-v_2).$$

For a nontrivial collision, dividing the two relations gives

$$u_1+v_1=u_2+v_2,$$

or equivalently

$$u_1-u_2=-(v_1-v_2).$$

The relative speed of approach therefore equals the relative speed of separation. This relation is a compact signature of a one-dimensional elastic collision.

Solving it together with momentum conservation gives

$$v_1=\frac{m_1-m_2}{m_1+m_2}u_1 +\frac{2m_2}{m_1+m_2}u_2,$$

$$v_2=\frac{2m_1}{m_1+m_2}u_1 +\frac{m_2-m_1}{m_1+m_2}u_2.$$

For equal masses, the objects exchange velocities.

Worked example

A $2,\mathrm{kg}$ cart moving at $3,\mathrm{m/s}$ strikes a stationary $1,\mathrm{kg}$ cart elastically. Then

$$v_1=\frac{2-1}{2+1}(3)=1,\mathrm{m/s},$$

and

$$v_2=\frac{2(2)}{2+1}(3)=4,\mathrm{m/s}.$$

The momentum before the collision is

$$P_i=(2)(3)=6,\mathrm{kg,m/s},$$

and afterward

$$P_f=(2)(1)+(1)(4)=6,\mathrm{kg,m/s}.$$

The kinetic energy is also unchanged:

$$K_i=\frac12(2)(3^2)=9,\mathrm J,$$

$$K_f=\frac12(2)(1^2)+\frac12(1)(4^2)=9,\mathrm J.$$

Momentum conservation by itself would allow many possible final velocities. The elastic condition supplies the additional constraint that selects this outcome.

Demonstration: testing the predictions

Walter Lewin's air-track demonstrations compare the predicted outcomes for equal and unequal masses with measured motion, making the limiting cases visible rather than leaving them as algebra alone.