Physics · Work, Energy And Power · NEET
Yes. In every type of collision (elastic, inelastic, or perfectly inelastic), the total linear momentum of the system is conserved, because the collision forces are internal action-reaction pairs and act for a very short time. So you can always write m1u1 + m2u2 = m1v1 + m2v2. This is the one rule that never breaks, so start almost every collision problem with momentum conservation.
Not always. Kinetic energy is conserved ONLY in an elastic collision. In an inelastic collision, some kinetic energy converts into heat, sound, or deformation, so final KE is less than initial KE. Momentum stays the same, but KE can drop. Never assume KE is conserved unless the question says the word 'elastic'.
Elastic: both momentum and kinetic energy are conserved, and the bodies separate (coefficient of restitution e = 1). Inelastic: momentum is conserved but KE is lost (0 < e < 1), and bodies separate with less relative speed. Perfectly inelastic: momentum is conserved, maximum KE is lost, and the bodies stick together and move as one (e = 0).
No. A collision only needs a strong mutual force acting for a short time. Two positive charges approaching and repelling, or a comet swinging past a planet, are collisions even without contact. NEET counts any brief strong interaction that changes velocities as a collision, so do not restrict yourself to bodies that touch.
Only momentum is conserved. Since the two bodies move together with one common velocity v, momentum gives (m1u1 + m2u2) = (m1 + m2)v. Kinetic energy is NOT conserved here; the loss of KE is the maximum possible for that momentum. This 'stick together' case is a favourite NEET setup, e.g. a bullet embedding in a block.
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
Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision will respectively be
A moving block of mass m collides with another stationary block of mass 4m. The lighter block comes to rest after the collision. When the initial velocity of the lighter block is v, the value of the coefficient of restitution (e) is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The forces during a collision are internal action-reaction pairs, so they cancel and cannot change the total momentum. Energy, however, can quietly convert into heat, sound, and permanent deformation, so kinetic energy is not guaranteed to stay the same.
It is the ratio of the relative speed of separation to the relative speed of approach: e = (separation speed)/(approach speed). For an elastic collision e = 1, for a perfectly inelastic collision e = 0, and for a general inelastic collision e is between 0 and 1.
Elastic (approximately): two smooth billiard balls or gas molecules colliding. Inelastic: a real ball bouncing on the ground and rising to a lower height. Perfectly inelastic: a bullet getting embedded in a wooden block, or two lumps of clay sticking together.
No. If the velocities before and after all lie along one straight line it is a one-dimensional (head-on) collision. If the bodies move off at angles it is a two-dimensional or oblique collision, and momentum must be conserved separately along two perpendicular directions.
The perfectly inelastic collision loses the maximum possible kinetic energy for the given momentum, because the bodies stick and move with one common velocity. It does not lose ALL the kinetic energy, since the combined mass still moves and carries kinetic energy.
