Physics · Work, Energy And Power · NEET
Yes. Momentum conservation and kinetic energy conservation are two separate rules. Momentum is always conserved in every collision (elastic or inelastic) because no external force acts during the short collision time. Kinetic energy is conserved ONLY in a perfectly elastic collision. In an inelastic collision some kinetic energy turns into heat, sound or deformation, but the total momentum m1u1 + m2u2 is still equal to the final total momentum. NEET often traps students here, so always start an inelastic collision problem with momentum, never with energy.
An explosion is a collision run backwards. Before the blast the object is at rest, so total momentum = 0. Internal forces (the blast) push the fragments outward, but internal forces cannot change the total momentum. So after the explosion the vector sum of all the fragment momenta must still be zero. If two fragments fly off in known directions, the third fragment must carry exactly the momentum needed to make the total zero. This is the whole idea behind the 5m-breaks-into-fragments PYQ.
Use momentum conservation DURING the collision or explosion (the short instant of contact). Use energy conservation BEFORE or AFTER, when only gravity or a spring acts and no sudden impact happens. In the classic bullet-into-hanging-block problem you use momentum for the strike, then energy conservation (mgh = 1/2 mv^2) for the block rising afterwards. Mixing them up (using energy during the collision) is the most common NEET mistake.
Momentum has direction, so in two dimensions you conserve it separately along the x-axis and the y-axis. Total x-momentum before = total x-momentum after, and the same for y. You cannot just add speeds. In the exploding-particle problem, two fragments at 90 degrees give a resultant of sqrt(2) mv, so the third fragment must move opposite with that same magnitude. Forgetting the vector nature gives a wrong answer.
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
A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular directions with speed v each. The energy released during the process is
A bullet of mass 10 g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg suspended by a light inextensible string of length 5 m. As a result 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.
Yes. Total linear momentum is conserved in elastic, inelastic and perfectly inelastic collisions, because during the short collision time the external forces are negligible compared to the large internal impulsive forces. Only kinetic energy conservation depends on the type of collision.
For two bodies: m1 u1 + m2 u2 = m1 v1 + m2 v2, where u are velocities before and v are velocities after. In an explosion the left side is zero if the body starts at rest, so the fragment momenta must add (as vectors) to zero.
The forces that blow the object apart are internal forces. By Newton's third law they occur in equal and opposite pairs, so they cancel and cannot change the total momentum. If there is no external force, the total momentum before the blast equals the total after.
Yes, if it was zero before. A stationary bomb has zero momentum, so after it explodes the vector sum of all fragment momenta is still zero, even though each fragment moves fast. Individual momenta are non-zero but cancel out.
Many NEET questions on collisions, bullets, explosions and recoil are solved in seconds using momentum conservation. It is the one rule that works even when energy is lost, so it is the safe starting point for any impact problem.
