Question

A bomb at rest explodes into three fragments of equal massses Two fragments fly off at right angles to each other with velocities of `9m//s` and `12//s` Calculate the speed of the third fragment .

Answer 1

Correct Answer - `-15m//s` .
Let m be the mass of each fragment
`:. P_(1) = mxx 9,p_(2) = m xx 12`
As`vec(p_(1))` and` vec(p_(2))` are perpendicular to each other therefore resultant momentum of the two fragments
`p = sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt((9m)^(2) +(12m)^(2)) = 15m`
According to the principle of conservation of linear momentum
`vec(p) + vec(p_(3)) = 0`
`vec(p_(3)) = - vec(p) = - 15 m `
`m xx upsilon_(3) = - 15m`
`upsilon_(3) = -15m//s`
Negative sign implies that third fragment will fly in a direction opposite to the direction of resultant momentum of the two fragments.