For simplicity we take m1=1 kg, m2=2 kg and m3= 3kg.

We assume that the bomb is at rest initially, and P1, P2 and P3 are the momentum (as vectors) or the 3 masses after explosion. The law of the conservation of total linear momentum before and after explosion can be written as

0 = P1+P2+P3so that P3= -(P1 +P2)

and in absolute value

`P3 = sqrt((m1*V1)^2 +(m2*V2)^2) = sqrt((1*40)^2 +(2*40)^2) = 89.44 kg*m/s`

But `P3 = m3*V3`  and therefore

`V3 = P3/m3 =89.44/3 = 29.81 m/s`

The angle that V3 is making with V1 is simply

`alpha = 180-arctan((P2)/(P1)) = 180-arctan((2*40)/(1*40)) = 180-arctan(2) =116.57 degree`

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