A bomb in the steady state explodes into three fragments. Two fragments of equal masses move with velocity 15 ms^(-1) in mutually perpendicular directions. The mass of the third fragment is equal to three times the mass of each of these two fragments. The magnitude of velocity of third fragment is ms^(-1) .

A bomb in the steady state explodes into three fragments. Two fragment

1.	A bomb in the steady state explodes into three fragments. Two fragments of equal masses move with velocity 15 ms^(-1) in mutually perpendicular directions. The mass of the third fragment is equal to three times the mass of each of these two fragments. The magnitude of velocity of third fragment is ms^(-1) .
Answer» 20 `10sqrt(2)` `5sqrt(2)` none of these Solution :Accordingto lawof conservationof momentuminitialmomentum ofbomb= finalmomentumof bomb `0=mv_(1) HAT( I ) =mv^(2) hat(J ) + 2mv_(3)` `:. ,vec( V )_(3)= (-5)hat(I ) + (-5)hat(j )` `\|vec(v)_(3)\| = sqrt(-5)^(2) + (-5)^(2))` `=sqrt(25 xx 2) = 5 squr(2) m//s`