Physics · Work, Energy And Power · NEET
| Kinetic energy | Conserved (KE before = KE after) | Not conserved (some KE lost) |
| Momentum | Conserved | Conserved |
| Coefficient of restitution e | e = 1 | 0 <= e < 1 (e = 0 if perfectly inelastic) |
| Bodies after impact | Separate, move independently | May move together (perfectly inelastic) or separately |
| Energy to heat/sound | None | Yes, part of KE is lost |
| Real example | Hard steel balls, atomic collisions | Car crash, clay ball hitting a wall, bullet in block |
Yes. Linear momentum is conserved in every collision, elastic or inelastic, as long as no external force acts. This comes from Newton's third law: the two bodies push on each other with equal and opposite forces. So writing m1u1 + m2u2 = m1v1 + m2v2 is always allowed. Only kinetic energy behaves differently between the two types.
No. In an inelastic collision, total kinetic energy after the collision is less than before. The lost energy becomes heat, sound, or permanent deformation of the bodies. It is not destroyed, it just changes form, so the law of conservation of energy still holds. In an elastic collision, total kinetic energy stays the same.
It is the special inelastic case where the two bodies stick together and move with one common velocity after the collision. Momentum is still conserved: m1u1 + m2u2 = (m1 + m2)v. This case loses the maximum possible kinetic energy that is allowed while keeping momentum conserved. A bullet embedding in a wooden block is the classic example.
Almost all everyday collisions are inelastic, because some energy always leaks away as sound or heat. A truly elastic collision (no KE loss) is an idealisation. The closest real examples are collisions between hard steel balls or between atoms and molecules, which lose very little energy.
Read the wording. If the problem says elastic, use both momentum conservation and kinetic energy conservation. If it says inelastic or the bodies stick together, use only momentum conservation. To check numerically, calculate total KE before and after: if they are equal the collision is elastic, if KE after is smaller it is inelastic.
Two bodies A and B of the same mass undergo a completely inelastic one-dimensional collision. Body A moves with velocity v1 while body B is at rest before the collision. The velocity of the system after collision is v2. The ratio v1 : v2 is
Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head-on and elastic. After the collision the fraction of energy lost by the colliding body A is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Linear momentum and total energy are conserved in both. The key gap is kinetic energy: it is conserved only in elastic collisions, not in inelastic ones.
The coefficient of restitution e is 1 for a perfectly elastic collision, 0 for a perfectly inelastic collision (bodies stick), and between 0 and 1 for a partially inelastic collision.
Not in ordinary collisions. It can increase only in an explosive or super-elastic event where stored energy (like a spring or chemical energy) is released. In normal elastic collisions KE stays the same; in inelastic collisions it decreases.
No. Even a perfectly inelastic collision keeps some kinetic energy, because the combined mass still moves to conserve momentum. It loses the maximum KE allowed, but not all of it, unless the total momentum was zero.
Collision questions appear almost every year in the Work, Energy and Power chapter. Knowing exactly what is conserved lets you pick the right equations quickly and avoid the trap of assuming KE conservation in inelastic cases.
