Collisions in one dimension

A collision is a brief, strong interaction between two bodies — brief enough that external forces (like friction or gravity acting during the contact) can be ignored. Two facts anchor everything that follows.

Momentum is conserved in every collision. As long as the two-body system is isolated from external forces, the total linear momentum before and after the collision is the same, regardless of what type of collision it is:

where: $m_1, m_2$ are the masses, $u_1, u_2$ are the initial velocities, and $v_1, v_2$ are the final velocities, all measured along the same line (with sign for direction).

Kinetic energy is conserved only in an elastic collision. An elastic collision conserves both momentum and total kinetic energy:

An inelastic collision conserves momentum but not kinetic energy — some of it converts to heat, sound, and permanent deformation. The extreme case, a perfectly inelastic collision, is where the two bodies stick together and move off with one common final velocity $v$. The figure below shows the equal-mass elastic case: the moving ball transfers its entire velocity to the stationary one and stops, which is exactly the "billiard ball" result from the overview.

The general one-dimensional elastic collision formula (derived below) lets you find $v_1$ and $v_2$ for any masses and initial speeds:

Setting $u_2 = 0$ (target at rest) gives three named special cases: $m_1 = m_2$ gives $v_1 = 0$, $v_2 = u_1$ (complete velocity exchange). $m_1 \gg m_2$ gives $v_1 \approx u_1$, $v_2 \approx 2u_1$ (the heavy body barely slows; the light body is flung forward at nearly double the incoming speed). $m_1 \ll m_2$ gives $v_1 \approx -u_1$, $v_2 \approx 0$ (the light body bounces straight back, like a ball off a wall; the heavy body barely moves). For a perfectly inelastic collision, momentum conservation alone (with $v_1 = v_2 = v$) gives the common final velocity:

and the kinetic energy lost is $\Delta K = K_i - K_f$, converted to heat, sound, and deformation.

A ball of mass $2\ \text{kg}$ moving at $6\ \text{m/s}$ collides elastically, head-on, with an identical $2\ \text{kg}$ ball at rest. Find both final velocities, and then check what happens if the second ball's mass is tripled to $6\ \text{kg}$ instead.

Step 1. Identify the case. Here $m_1=m_2=2\ \text{kg}$, $u_1=6\ \text{m/s}$, $u_2=0$ — the equal-mass special case, so we expect complete velocity exchange.

Step 2. Apply the general formulas. $v_1 = \dfrac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} = \dfrac{0+0}{4} = 0\ \text{m/s}$, and $v_2 = \dfrac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2} = \dfrac{0+2(2)(6)}{4} = \dfrac{24}{4} = 6\ \text{m/s}$.

Step 3. Interpret. The first ball stops completely and the second moves off at $6\ \text{m/s}$ — the incoming ball's entire velocity is transferred, exactly as the equal-mass rule predicts.

Answer: $v_1 = 0\ \text{m/s}$, $v_2 = 6\ \text{m/s}$.

Competitive stretch

Now keep $m_1=2\ \text{kg}$, $u_1=6\ \text{m/s}$, $u_2=0$, but make the target heavier: $m_2=6\ \text{kg}$. Using the same formulas, $v_1 = \dfrac{(2-6)(6)+0}{8} = \dfrac{-24}{8} = -3\ \text{m/s}$ and $v_2 = \dfrac{0+2(2)(6)}{8} = \dfrac{24}{8} = 3\ \text{m/s}$. The incoming ball rebounds backward at $3\ \text{m/s}$ (negative sign, reversed direction) while the heavier target moves forward at $3\ \text{m/s}$ — a preview of the $m_1 \ll m_2$ trend, though here the mass ratio is only $1:3$, not extreme enough for the full $v_1 \approx -u_1$, $v_2 \approx 0$ approximation to kick in yet.

• Assuming kinetic energy is conserved in every collision — only elastic collisions conserve it; momentum is conserved in all collisions (isolated system).
• Mixing up the three special cases — $m_1=m_2$, $m_1\gg m_2$, and $m_1\ll m_2$ give distinctly different outcomes and must be matched to the actual mass ratio, not applied by memory alone.
• Dropping the sign on a rebound velocity — when the light body bounces back off a heavy one, $v_1$ is negative, and that sign must carry through any energy or further-motion calculation.
• Thinking "perfectly inelastic" means the final kinetic energy is zero — the combined body still moves (unless total momentum is exactly zero) and only the conservation of kinetic energy fails, not motion itself.

Ask the tutor: Want to work through a few more one-dimensional collisions with different mass ratios, or see how the coefficient of restitution generalises the elastic and inelastic cases together?