Elastic and Inelastic Collisions

What conserves in any collision

Consider two masses $m_1$ and $m_2$, with $m_1$ moving at speed $v_{1i}$ and $m_2$ initially at rest. When they collide, each exerts an impulsive force on the other over the contact time $\Delta t$. The change in momentum of each is $\Delta p_1 = F_{12}\,\Delta t$ and $\Delta p_2 = F_{21}\,\Delta t$, where $F_{12}$ is the force on the first particle due to the second and vice versa.

By Newton's third law $F_{12} = -F_{21}$ at every instant, so $\Delta p_1 + \Delta p_2 = 0$. This conclusion holds even though the forces vary in a complex fashion during $\Delta t$, because the third law is true at every instant. The total impulse on one object is equal and opposite to that on the other. Hence the total linear momentum is conserved in any collision — elastic, inelastic, or perfectly inelastic.

Kinetic energy enjoys no such guarantee. The impact deforms the bodies, and the deformation may generate heat and sound. Part of the initial kinetic energy is then transformed into other forms. NCERT offers a vivid picture: imagine a "compressed spring" connecting the two masses during contact. If that spring regains its original shape with no loss, the final kinetic energy equals the initial — an elastic collision. If the deformation is not relieved, energy is lost.

Classifying collisions

NCERT names three categories by how much of the deformation is relieved. The dividing question is always "what happens to the kinetic energy?" — never "what happens to the momentum?", which is fixed.

The perfectly (completely) inelastic collision is the limiting member of the inelastic family: the two particles move together after the collision with a single common velocity. It loses the largest fraction of kinetic energy that momentum conservation will permit. The intermediate case, where deformation is only partly relieved and some — but not all — of the kinetic energy is lost, is what NCERT simply calls an inelastic collision and is the most common in everyday life.

Coefficient of restitution e

The single number that grades a collision on the elastic-to-inelastic scale is the coefficient of restitution, $e$. It compares how fast the bodies separate after impact to how fast they approached before it, measured along the line of impact:

$$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_{2f} - v_{1f}}{u_1 - u_2}$$

This definition is exactly what NCERT's elastic algebra produces. For a head-on elastic collision the chapter derives $v_{1i} + v_{1f} = v_{2f}$, i.e. $v_{2f} - v_{1f} = v_{1i}$ — the relative velocity simply reverses with the same magnitude, so $e = 1$. The three reference values anchor every restitution problem.

1-D elastic collision formulas

When the initial and final velocities of both bodies lie along the same straight line, the collision is one-dimensional, or head-on. With $m_2$ at rest, NCERT writes the two conservation laws side by side:

$$m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f} \quad\text{(momentum)}$$

$$\tfrac{1}{2} m_1 v_{1i}^{2} = \tfrac{1}{2} m_1 v_{1f}^{2} + \tfrac{1}{2} m_2 v_{2f}^{2} \quad\text{(kinetic energy)}$$

Dividing the energy equation by the momentum equation eliminates the squares and yields the restitution relation $v_{1i} + v_{1f} = v_{2f}$. Substituting back gives the two final velocities purely in terms of the knowns $\{m_1, m_2, v_{1i}\}$:

$$v_{1f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1i}, \qquad v_{2f} = \left(\frac{2 m_1}{m_1 + m_2}\right) v_{1i}$$

These are NCERT Eqs. (5.26) and (5.27). Commit them to memory in this "target at rest" form; nearly every 1-D NEET collision reduces to plugging masses into them, and the three special cases below are just limits of these two expressions.

Three special mass cases

The general formulas collapse into three results worth knowing by sight. NCERT spells out the first two explicitly; the third follows from the same expressions in the opposite mass limit.

Case II is the engine behind a famous application. NCERT's Example 5.11 shows that a fast neutron loses most of its kinetic energy in an elastic collision with a light nucleus such as deuterium or carbon — the moderator in a nuclear reactor. The fractional energy retained is $\left(\frac{m_1-m_2}{m_1+m_2}\right)^2$, so for deuterium ($m_2 = 2m_1$) the neutron keeps just $1/9$ of its energy and gives up $8/9$ to the moderator; for carbon about $71.6\%$ is retained and $28.4\%$ transferred.

Perfectly inelastic collisions

When the two bodies stick together, there is a single unknown — the common final velocity $v_f$ — and momentum conservation alone fixes it. With $m_2$ at rest:

$$m_1 v_{1i} = (m_1 + m_2) v_f \quad\Rightarrow\quad v_f = \frac{m_1}{m_1 + m_2}\, v_{1i}$$

This is NCERT Eq. (5.22). The kinetic energy is not conserved; the loss works out to

$$\Delta K = \tfrac{1}{2} m_1 v_{1i}^{2} - \tfrac{1}{2}(m_1 + m_2) v_f^{2} = \tfrac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right) v_{1i}^{2}$$

which is a positive quantity, as it must be. Dividing by the initial kinetic energy $\tfrac{1}{2} m_1 v_{1i}^2$ gives the fraction lost:

$$\frac{\Delta K}{K_i} = \frac{m_2}{m_1 + m_2}$$

The reading is intuitive. If the stationary target is far heavier ($m_2 \gg m_1$) almost all the kinetic energy is lost — think of a dart embedding in a massive block. If the target is very light ($m_2 \ll m_1$) almost none is lost, because the combined body barely slows down.

Collisions in two dimensions

If the velocities do not stay on one line, the collision is two-dimensional and the bodies scatter into a plane. Conservation of the out-of-plane (z) momentum forces the whole event into a single plane, leaving two scalar momentum equations. Taking the incoming direction as $x$:

$$m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2$$

$$0 = m_1 v_{1f}\sin\theta_1 - m_2 v_{2f}\sin\theta_2$$

Here $\theta_1$ and $\theta_2$ are the scattering angles of the two bodies on opposite sides of the original line. There are four unknowns $\{v_{1f}, v_{2f}, \theta_1, \theta_2\}$ but only two equations. Even adding kinetic-energy conservation for an elastic collision gives just one more equation — still one short. At least one quantity, usually a scattering angle, must be supplied, for instance by the position of a detector swung from the $x$ to the $y$ axis.

One beautiful 2-D result is worth memorising. NCERT's Example 5.12 (the billiard "cue and target") proves that when two equal masses undergo a glancing elastic collision with one initially at rest, they move off at right angles to each other. If the target is sunk into a pocket at $\theta_2 = 37^\circ$, the cue ball must travel at $\theta_1 = 53^\circ$, since the two angles sum to $90^\circ$. When the interaction acts at a distance rather than on contact — a comet past the Sun, an alpha particle near a nucleus — the same momentum bookkeeping applies, and the event is called scattering.

Worked example 1 — equal-mass velocity exchange

Worked example 2 — ballistic pendulum (perfectly inelastic KE loss)

Worked example 3 — NEET-style numerical (coefficient of restitution)