Physics · Work, Energy And Power · NEET
Yes. In every collision (elastic, inelastic or perfectly inelastic) the total momentum of the isolated system is conserved, because the two bodies push on each other with equal and opposite forces (Newton's third law). So m1 u1 + m2 u2 = (m1 + m2) v always holds. Only kinetic energy behaves differently between the collision types.
During contact the bodies deform, get warm and may make a sound. This work of deformation converts some kinetic energy into heat, sound and internal energy. That energy does not come back, so the final KE is less than the initial KE. Total energy (all forms) is still conserved — only mechanical KE drops.
Use momentum conservation only. v = (m1 u1 + m2 u2)/(m1 + m2). If the second body starts at rest (u2 = 0), this simplifies to v = m1 u1/(m1 + m2). Do NOT use energy conservation here — energy is lost, so an energy equation would give a wrong answer.
Energy lost = initial KE minus final KE. For a body of mass m1 (speed u) hitting a body m2 at rest, the fraction of KE lost is m2/(m1 + m2). When both masses are equal, exactly half the kinetic energy is lost. A perfectly inelastic collision loses the maximum KE that momentum conservation allows.
In any inelastic collision some KE is lost but the bodies separate afterwards (coefficient of restitution 0 < e < 1). In a perfectly inelastic collision the bodies do NOT separate — they move together as one (e = 0). Perfectly inelastic is the extreme case with the greatest possible energy loss for that momentum.
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
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a collision where the two objects stick together after hitting and then move as a single combined object with one common velocity. A bullet embedding in a wooden block or two railway wagons that couple together are classic examples.
Common velocity: v = (m1 u1 + m2 u2)/(m1 + m2), from momentum conservation. If the target is at rest, v = m1 u1/(m1 + m2). Kinetic energy lost = ½·(m1 m2)/(m1 + m2)·(u1 − u2)², using the reduced mass.
Yes. 'Plastic collision' is another name for a perfectly inelastic collision. In both, the coefficient of restitution e = 0 and the bodies move together after impact.
Yes. e = (velocity of separation)/(velocity of approach). Since the bodies move together, their separation velocity is zero, so e = 0 for a perfectly inelastic collision.
NEET regularly asks direct one-line numericals (like NEET 2024) where you must pick momentum conservation, not energy conservation, and compute the common velocity or the fraction of KE lost. It is a quick, high-scoring sub-topic if you avoid the energy-conservation trap.
