A bomb of mass 10 kg at rest explodes into two parts of 3 kg and 7 kg.If smaller part moves with a speed `14 m s^(-1)` towards east ,find the velocity of the heavier part.

Explanation

Initial momentum = `mu = 6 xx 0=0,` Final momentum = `m_(1)vecv_(1) + m_(2)vecv_(2)=3vecv_(1) +3vecv_(2)` .
According to law of conservation of momentum,
`3vecv_(1) +3vecv_(2)=0` or `vecv_(1)= -vecv_(2)`
`vecv_(1)` and `vecv_(2)` are equal and opposite. Thus fragments must fly apart at an angle of `180^(@)` .