Physics · Laws Of Motion · NEET
Yes. An explosion is caused by an internal force (stored chemical energy), and internal forces come in action-reaction pairs that cancel. Since no net external force acts during the short burst, total momentum is conserved. Kinetic energy is NOT conserved — it increases because stored chemical energy is converted into motion. So the two rules are separate: momentum stays constant, energy is released.
Step 1: write the momentum of each known fragment as a vector (mass times velocity, keeping direction). Step 2: add those known momenta as vectors to get their resultant. Step 3: the unknown fragment's momentum must be equal and opposite so the total equals the original momentum (zero if the body was at rest). Step 4: divide that momentum by the third fragment's mass to get its speed. Example: two perpendicular momenta of mv each give a resultant of mv√2, so the third piece must carry mv√2 in the opposite direction.
Momentum has direction. Two fragments flying at 90 degrees do not cancel or simply add — you must combine them like force vectors, using the parallelogram or Pythagoras rule. Adding only the speeds ignores direction and gives a wrong answer. This is the single most common mistake in explosion problems: treat every fragment's momentum as an arrow, then add the arrows.
Before the explosion the body may be at rest with zero kinetic energy, yet afterwards the fragments are moving. The extra energy comes from stored internal energy — chemical energy in gunpowder, elastic energy in a compressed spring, or nuclear energy. This is called the energy released. You compute it as: (total KE of all fragments after) minus (KE before).
Before the explosion the body was at rest, so its momentum was zero. Because momentum is conserved and no external force acts, the total momentum after must also be zero. This means all the fragment momentum vectors add up to zero — they form a closed vector polygon. That is why fragments fly off in different directions that exactly balance.
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:
An object flying in air with velocity (20î + 25ĵ + 12k̂) suddenly breaks into two pieces whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity (100î + 35ĵ + 8k̂). The velocity of the larger piece will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Conservation of linear momentum. The explosive force is internal, so no net external force acts during the short burst and total momentum stays constant. Kinetic energy is not conserved — it increases because stored energy is released.
No. Kinetic energy increases in an explosion. Stored chemical, elastic or nuclear energy is converted into the motion of the fragments. This gain is called the energy released, equal to the total final kinetic energy minus the initial kinetic energy.
First use momentum conservation to find every fragment's velocity. Then add up the kinetic energy of all fragments after the explosion and subtract the kinetic energy before. For a body starting at rest, energy released equals the total kinetic energy of the fragments.
In a collision two or more bodies come together and momentum is conserved while kinetic energy may or may not be conserved. In an explosion one body breaks apart, momentum is still conserved, but kinetic energy always increases because internal stored energy is released. Both use the same momentum equation.
Since no external force acts, the centre of mass keeps moving exactly as it did before the explosion. If the body was at rest, the centre of mass stays at rest; if it was moving, the centre of mass continues on the same path even though the fragments scatter.
