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. `sqrt17 ms^(-1)`
B. `2sqrt13 ms^(-1)`
C. `sqrt13 ms^(-1)`
D. `(sqrt13)/(2) ms^(-1)`

Answer

C

Explanation

Initial mass of particle = 4M
Initial velocity of particle = 0
After explosion into three pieces of masses M, M and 2M particle of equal masses move along X and Y axes with velocities 4m/s and 6m/s, respectively. So, according to law of conservation of momentum mass 2M particle will move in the direction shown in the figure.

Law of conservation of momentum, Momentum before collision = Momentum after collision.
`4M xx 0 = 2Mv + Mv_(x) + Mv_(y)`
`rArr -2Mv = M(v_(x) + v_(y))`
`rArr -2v = 4hat(i) + 6hat(j) rArr v = -2hat(i) + 3hat(j)`
So, `|v| = sqrt(2^(2) + 3^(2)) = sqrt(4 + 9) = sqrt13 ms^(-1)`