Physics · Laws Of Motion · NEET
When two bodies A and B interact, A pushes B with force F and B pushes A with force -F (third law), for the same time interval dt. So the impulse on B is +F.dt and on A is -F.dt. These change the two momenta by equal and opposite amounts, so the total change is zero. That is exactly why the total momentum of the pair does not change.
Momentum is conserved only when the NET EXTERNAL force is zero. Friction and gravity are external forces, so in general total momentum is NOT conserved when they act over the interaction. But in a very short collision or explosion the internal impulsive forces are huge compared to gravity or friction, so we treat momentum as conserved during that short instant. NEET expects you to spot 'isolated system' or 'no external force'.
No. Momentum is always conserved with zero external force, but kinetic energy is conserved only in ELASTIC collisions. In inelastic collisions and explosions, momentum stays the same while kinetic energy changes (it drops in inelastic collisions, and it INCREASES in an explosion because stored energy is released).
Total momentum stays zero, not each piece. Before the explosion total momentum is zero. After it, the pieces fly out in different directions but their momentum vectors add up to zero. That is why fragments move in opposite or balancing directions — the vector sum must still cancel.
Conserve momentum separately along each axis. Write total x-momentum before = total x-momentum after, and the same for y. For perpendicular pieces, combine the two momenta with the Pythagoras rule: resultant momentum = sqrt(px^2 + py^2). This is the key trick in explosion PYQs.
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 shell of mass m is initially at rest. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The two equal fragments fly off along mutually perpendicular directions, each with speed v. The speed of the third (lighter) fragment 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.
If no net external force acts on a system, the total linear momentum of the system remains constant in both magnitude and direction. This is called an isolated system.
The net external force on the system must be zero. Internal forces between the parts do not affect total momentum because they occur in equal and opposite pairs (Newton's third law).
Yes. Momentum is conserved in every collision with no external force, whether elastic or inelastic. Only kinetic energy is lost in an inelastic collision, not momentum.
Yes. If the body was at rest, total momentum stays zero, so the fragments fly out such that their momentum vectors add up to zero. Kinetic energy, however, increases because stored energy is released.
Write total momentum before = total momentum after, treating momentum as a vector. In 2D, conserve each axis separately or use resultant momentum = sqrt(px^2 + py^2). This solves recoil, explosion, and collision problems quickly.
