A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be -

Options

A. `11.5 m//s`
B. `14.0 m//s`
C. `7.0 m//s`
D. `9.89 m//s`

Answer

D

Explanation

`P_(x)= m xx v_(x)= 1 xx 21= 21kg m//s`
`P_(y)= m xx v_(y)= 1 xx 21 = 21kg m//s`
`therefore` Resultant `= sqrt(P_(x)^(2) + P_(y)^(2))= 21 sqrt2kg m//s`
The momentum of heavier fragment should be numerically equal to resultant of `vec(P)_(x) and vec(P)_(y)`.
`3 xx v = sqrt(P_(x)^(2) + P_(y)^(2))= 21 sqrt2 therefore v= 7sqrt2= 9.89 m//s`