A shell of mass 15 kg, initially a rest, explodes into three fragments of masses in the ratio 1:1: 3. The fragments with equal masses fly off in mutually perpendicular directions with a speed of `6 ms^(-1)` The speed of the heaviest fragment will be :

Options

A. `12ms^(-1)`
B. `6ms^(-1)`
C. `sqrt6ms^(-1)`
D. `(2sqrt6)/(3)ms^(-1)`

Answer

d

Explanation

M= 15 kg since `m_(1) :m_(2) : m_(3) = 1:1:3`
`:. m_(1) = m_(2) = 3 kg` and `m_(3) = 9 kg`
Now according to law of conservation of momentum, momentum `(p_(3))` of heavier part must be equal and opposite to resultants of `P_(1)` and `P_(2)`

`|p_(3)|=sqrt(p_(1)^(2)+p_(2)^(2))`
`:.9xxupsilon_(3)=sqrt((3xx6^(2)+3xx6^(2)))`
`upsilon_(3)=(6sqrt6)/(9)=(2sqrt6)/(3)m//s`
Hence the correct choice is (d)