AIIMS2018PhysicsCenter of Mass, Momentum and Collision
A bomb moving with velocity (40 i +50 j -25 k ) m s ⁻¹ explodes into two pieces of mass ratio 1: 4 . After explosion the smaller piece moves away with velocity (200 i +70 j +15 k ) m s ⁻¹ . The velocity of larger piece after explosion is
Options
- A45 j -35 k
- B45 i -35 j
- C45 k -35 j
- D-35 i +45 k
Correct answer
A. 45 j -35 k
Step-by-step solution
( Let the mass of the unexploded bomb be 5 ~m . It explodes into the two pieces of masses m and 4 m respectively. Initial momentum of the unexploded bomb =5 m(40 i +50 j -25 k ) After explosion, momentum of the smaller piece =m v ₁=m(200 i +70 j +15 k ) and momentum of the larger piece =4 m v ₂ where v ₁ and v ₂ are the velocities of the two pieces respectively. According to the law of conservation of momentum, we get aligned & 5 m(40 i +50 j -25 k )=m(200 i +70 j +15 k )+4 m v ₂ & 4 m v ₂=5 m(40 i +50 j -25 k )-m(