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

Options

A. `sqrt(17 ms^(-1)`
B. `2sqrt(13 ms^(-1)`
C. `sqrt(13 ms^(-1)`
D. `sqrt(13)/(2)ms^(-1)`

Answer

c

Explanation

According to law of conservation of linear momentum,
`barp1+barp2+barp3=0`
therefor`barp3=-(barp1+barp2)`
[negative sign only indicate opposite direction, becouse velocity can not be a negative value]
`[(2m)v3]^(2)=[(4m)^(2)+(6m)^(2)]`
` (2m)v_(3)=sqrt((4m)^(2)+(6m)^(2))`
`v_(3) =sqrt(52m^(2))/(4m^(2))` [Mass of highest value i.e,2m]