elastic and inelastic collisions

In any collision, the total momentum of an isolated system stays constant, because the interaction forces are internal and act in equal and opposite pairs (Newton's third law). Kinetic energy, however, need not stay constant: it can be converted into internal energy, sound, and deformation. A collision is called elastic if the total kinetic energy after the impact equals the total kinetic energy before it; otherwise the collision is inelastic. In a perfectly inelastic collision the bodies stick together and move with a common velocity, and the kinetic-energy loss is the largest allowed by momentum conservation. Momentum conservation is a vector statement, so it holds separately along each axis. To determine the type of a collision, compare the value of $\sum \frac{1}{2}m v^2$ before and after the impact, not the speeds alone.

Useful relations for the one-dimensional case:

$$p_1 + p_2 = p_1' + p_2' \quad \text{(always)},$$

$$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2 \quad \text{(elastic only)}.$$

For a perfectly inelastic collision the second equation is replaced by the condition $v_1' = v_2' = V$, which gives $V = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.

A cart of mass $m_1 = 2$ kg moves at $v_1 = 4$ m/s along a straight frictionless horizontal track and hits a stationary cart of mass $m_2 = 3$ kg. The carts lock together and move as one body. Find their common velocity after the collision.

A ball of mass $m_1 = 2$ kg moving at $v_1 = 3$ m/s collides elastically head-on with a stationary ball of mass $m_2 = 1$ kg on a smooth horizontal surface. Find the velocities of both balls after the collision and verify both conservation laws.

A bullet of mass $m = 0.010$ kg flying at $v = 400$ m/s strikes a stationary wooden block of mass $M = 2.00$ kg hanging on a long string and embeds in it (ballistic pendulum). Find: (a) the speed of the block immediately after the bullet embeds; (b) the maximum height the block rises to; (c) the fraction of the kinetic energy converted into heat and deformation.

Self-study simulation plan (safe, no real collisions at hazardous speeds).

1. Open a collisions simulation (for example, the PhET "Collision Lab" in a browser) and choose two balls on a straight frictionless line.
2. Set $m_1 = 2$ kg, $m_2 = 3$ kg, $v_1 = 4$ m/s, $v_2 = 0$ m/s and turn on the momentum vectors and the kinetic-energy readouts.
3. Move the elasticity slider to 0 % (perfectly inelastic collision). Record in a table the total momentum and the total energy before and after the impact.
4. Repeat the run at 100 % (elastic collision) and at 50 % (partially inelastic collision).
5. For each case compare the measured values with calculations: $V = \frac{m_1 v_1}{m_1 + m_2}$ at 0 %, and $v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1$, $v_2' = \frac{2m_1}{m_1 + m_2}v_1$ at 100 %.
6. Draw a conclusion: which of the two conservation laws held in every run, and which held only in the elastic case. The simulation is used because real collisions at speeds of hundreds of metres per second (like a bullet) are hazardous and unavailable for home measurement.

Show that total momentum is conserved in every collision, whereas total kinetic energy is conserved only in an elastic one, and that in a perfectly inelastic collision the bodies move together.

A two-part sketch: on the left "before the impact" and on the right "after the impact" — two carts on a track with velocity arrows $v_1$, $v_2$ and $v_1'$, $v_2'$ (in the inelastic case a single common arrow $V$). Below the sketch, two bar charts: the upper one for total momentum, with equal bar heights before and after; the lower one for total kinetic energy, where the bars are equal for an elastic collision, while for an inelastic collision the "after" bar is lower and the shaded part marks the energy converted into heat and deformation.

A ball of mass $0.17\ \text{kg}$ rolls at $2.0\ \text{m/s}$ and hits head-on a stationary ball of the same mass. In a nearly elastic collision the first ball stops and the second rolls away at about $2.0\ \text{m/s}$.

This is an elastic collision example: the system momentum is conserved and kinetic energy is conserved too. For equal masses $m_1=m_2$ the conservation equations give $v_1'=0$ and $v_2'=v_1$; check: $\frac{1}{2} m v_1^2 = \frac{1}{2} m v_2'^2$.

A railcar of mass $20\,000\ \text{kg}$ moves at $1.5\ \text{m/s}$ and couples with a stationary railcar of the same mass. After a perfectly inelastic collision the cars move together at $0.75\ \text{m/s}$.

This is a perfectly inelastic collision: momentum is conserved, $m v = (m+m)v'$, so $v' = v/2 = 0.75\ \text{m/s}$. Kinetic energy is not conserved: initial $E_k = \frac{1}{2}m v^2 = 22\,500\ \text{J}$, final $E_k' = \frac{1}{2}(2m)v'^2 = 11\,250\ \text{J}$; the difference $11\,250\ \text{J}$ goes into deformation, heat and sound.

A ball of mass $0.60\ \text{kg}$ falls from a height of $1.8\ \text{m}$ and hits the floor at $5.9\ \text{m/s}$; after bouncing, its upward speed is $4.5\ \text{m/s}$.

This is a partially inelastic collision: kinetic energy before impact is $E_k = \frac{1}{2}mv^2 \approx 10.4\ \text{J}$, after impact $E_k' = \frac{1}{2}mv'^2 \approx 6.1\ \text{J}$, so about $4.3\ \text{J}$ goes into deformation of the ball, heat and sound. The ball's momentum changes, but for the ball–Earth system it is conserved; the coefficient of restitution $e = v'/v \approx 0.76$ shows how far the collision is from elastic.