NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same kinetic energy E 0 . The energy of the explosion will be K times of E 0 , then the value of K is
Correct answer
6
Step-by-step solution
P 1 → = m v 0 i ^ P 2 → = m v 0 j ^ P 3 → = m v 0 j ^ P 4 → = m v → 0 = P 1 → + P 2 → + P 3 → + P 4 → V → = - v 0 ( i ^ + j ^ + k ^ ) v = v 0 3 Total energy = 3 × 1 2 m v 0 2 + 1 2 m v 2 = 3 E 0 + 3 E 0 = 6 E 0