Question

A bomb moving with velocity `(40i+50j-25k)m//s` explodes into two pieces of mass ratio 1:4. After explosion the smaller piece moves away with velocity `(200i +70j +15k)m//s`. The velocity of larger piece after explosion is .

Answer 1

From Law of conservation of linear momentum
`Mu = m_(1) v_(1) +m_(2) v_(2), M =5x, m_(1) =x, m_(2) =4x`
`u =40hati + 50hatj -25hatkms^(-1)` ,
`v_(1) = (200 hati + 70hatj +15hatk)ms^(-1)`
here `v_(2)` is the velocity of the larger piece
`5x (40hati +50hatj -25hatk)=`
`x(200hati +70 hatj +15hatk) +4x (v_(2))`
One simplification we get `v_(2) = 45hatj -35hatk`.