Physics · Laws Of Motion · NEET
Yes. Momentum is conserved in every collision (elastic or inelastic) as long as no external force acts, because internal action-reaction forces cancel. So m1u1 + m2u2 = (m1 + m2)v. This is the equation you always start with.
No. Kinetic energy is NOT conserved in a perfectly inelastic collision. Some KE is lost as heat, sound and permanent deformation when the bodies stick. In fact this type of collision has the maximum possible KE loss for a given momentum. So use momentum conservation, never energy conservation, to find the common velocity.
In a general inelastic collision some KE is lost but the bodies still separate and move with different velocities. In a perfectly (completely) inelastic collision the bodies stick together and move with ONE common velocity. Perfectly inelastic is the extreme case with the largest KE loss.
Add the two momenta before impact and divide by the total mass: v = (m1u1 + m2u2)/(m1 + m2). Keep the sign of each velocity (right = +, left = -). If both bodies move the same way, the answer is between the two speeds; if they move toward each other, it can even be zero.
That is the definition of 'perfectly inelastic'. The bodies deform and grip each other (mud on a ball, a bullet lodging in a block, two clay lumps) so there is no relative velocity left between them. One mass, one velocity.
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:
A bullet of mass 10 g moving horizontally at 400 m/s strikes a wooden block of mass 2 kg suspended by a light inextensible string of length 5 m. The centre of gravity of the block rises a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block 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 they hit and then move as a single body with one common velocity. Momentum is conserved but kinetic energy is not.
v = (m1u1 + m2u2)/(m1 + m2), where m1, m2 are the masses and u1, u2 are their velocities before the collision (with signs for direction).
Yes. When the bullet stays inside the block, they move together as one mass, so it is a perfectly inelastic collision. If the bullet passes through and comes out, it is only partially inelastic.
For a given total momentum, a perfectly inelastic collision loses the maximum possible kinetic energy. The lost KE turns into heat, sound and deformation. You compute it as KE_before minus KE_after.
Yes. If the two momenta are equal and opposite (for example, equal masses moving toward each other at the same speed), the numerator m1u1 + m2u2 becomes zero, so the combined body stops.
