Physics · Laws Of Motion · NEET
Yes. Linear momentum is conserved in ALL collisions, elastic and inelastic, as long as no external force acts. This is the most common NEET trap. The two bodies push on each other with equal and opposite internal forces (Newton's third law), so the total momentum m1u1 + m2u2 = m1v1 + m2v2 stays the same. What changes between elastic and inelastic is only the kinetic energy, not the momentum.
Yes, only in an elastic collision. An elastic collision has TWO conditions: total momentum is conserved AND total kinetic energy is conserved. In an inelastic collision, only momentum is conserved; kinetic energy decreases because part of it becomes heat, sound, or permanent deformation. So KE(final) is less than KE(initial) for inelastic, but equal for elastic.
When two bodies of EQUAL mass collide elastically in one dimension, they simply exchange (swap) their velocities. So if A has velocity u1 and B has velocity u2, after the collision A moves with u2 and B moves with u1. This shortcut solves many NEET questions in seconds. It comes from solving momentum and KE conservation together with m1 = m2.
e = (velocity of separation) / (velocity of approach) = (v2 - v1) / (u1 - u2). It measures how bouncy a collision is. For a perfectly elastic collision e = 1. For a perfectly inelastic collision (bodies stick together) e = 0. For a real partially inelastic collision 0 < e < 1. NEET often asks you to find e after using momentum conservation to get the final velocities.
It is the extreme inelastic case where the two bodies stick together and move with one common final velocity after impact. Here e = 0 and the kinetic energy loss is the maximum possible (while still conserving momentum). Common velocity v = (m1u1 + m2u2)/(m1 + m2). This is covered separately in the perfectly inelastic collision topic.
Two identical balls A and B having velocities 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision, respectively, will be:
A moving block of mass m collides with a stationary block of mass 4m. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is v, the coefficient of restitution (e) is:
Two bodies A and B of the same mass undergo a completely inelastic one-dimensional collision. A moves with velocity v1 while B is at rest before the collision. The velocity of the combined 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.
Total linear momentum is always conserved in every collision, elastic or inelastic, when no external force acts. Kinetic energy is conserved only in elastic collisions.
Not in ordinary elastic or inelastic collisions. In these, KE either stays the same (elastic) or decreases (inelastic). KE can only increase in an explosion-type event where stored energy is released, which is a different case.
If the question says 'elastic', you may use both momentum and KE conservation. If it says 'inelastic', 'stick together', 'coalesce', or 'combined system', use momentum only and expect KE loss. Perfectly inelastic means e = 0.
For a perfectly elastic collision e = 1. For a perfectly inelastic collision e = 0. For real partly inelastic collisions e lies between 0 and 1.
Collisions are formally in the Work, Energy and Power chapter of NCERT, but they use momentum conservation from Laws of Motion. NEET tests both, so learn the momentum rules here and the KE-loss formulas alongside.
