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A particle of mass `4m` explodes into three pieces of masses m,m and `2m` The equal masses move along X-axis and Y-axis with velocities `4ms^(-1)` and `6ms^(-1)` respectively. The magnitude of the velocity of the heavier mass is

Answer» `M =4m, u =0, m_(1) =m, m_(2) =m, m_(3) =2m`
`v_(1) = 4ms^(-1), v_(2) =6ms^(-1) ,v_(3) =`
According to law of conservation of momentum
`vecP_(1) + vecP_(2)+vecP_(3) =0`
`vecP_(3) =- (vecP_(1) +vecP_(2)),|vecP_(3)|=|vecP_(1)+vecP_(2)|`
`P_(3) =sqrt(P_(1)^(2)+P_(2)^(2)+2P_(1)P_(2)Costheta)`
`P_(1)` and `P_(2)` are perpendicular to each other
`P_(3) = sqrt(P_(1)^(2) +P_(2)^(2)), m_(3) v_(3) =sqrt((m_(1)v_(1))^(2)+(m_(2)v_(2))^(2))`
`2mv_(3) = sqrt((mxx4)^(2)+(mxx6)^(6))`
`2v_(3)=sqrt(16 +36) rArrv_(3) =sqrt13 ms^(-1)` .


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