A bomb of mass `9 kg` explodes into `2` pieces of mass `3 kg` and `6 kg`. The velocity of mass `3 kg is 1.6 m//s`. The `K.E. of mass 6 kg` is

Options

A. `3.84 J`
B. `9.6 J`
C. `1.92 J`
D. `2.92 J`

Answer

C

Explanation

As the bomb initial was at rest therefore
Initial momentum of bomb`= 0`
Final momentum of system `= m_(1) v_(1) + m_(2) v_(2)`
As there is no external force
`:. m_(1) v_(1) + m_(2) v_(2) = 0 implies 3 xx 1.6 + 6 xx v_(2) = 0`
velocity of `6 kg mass v_(2) = 0.8 m//s` (numerically)
Its kinetic energy `= (1)/(2) m_(2) v_(2)^(2) = (1)/(2) xx 6 xx (0.8) ^(2) - 1.92 J`