Question

A bomb of mass `1 kg` is thrown vertically upwards with a speed of `100m//s`. After 5 seconds, it explodes into two fragments. One fragment of mass `400 gm` is found to go down with a speed of `25m//s`. What will happen to the second fragment just after the explosion ? `(g=10ms^(-1))`

Answer 1

After 5 sec, velocity of the bomb,
`v =u +at`
`vecv = u hatj - "gt" hatj = (100 -10 xx 5)hatj =50 hatj m//s`
`m =1kg, m_(1) =0.4kg,m_(2) =0.6kg ,v_(1) =25 ms^(-1)`
According to law of conservation of
momentum `mvecv = m_(1) barv_(1) = m_(2) barv_(2)`
`1 xx 50 hatj = - 0.4 xx 25 hatj + 0.6 vecv_(2)`
`rArr v_(2) 100hatj =v_(2) =100ms^(-1)` vertically upwards .