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A bomb at rest explodes into three fragm...

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 .

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The correct Answer is:
15
Let `vecp_(1).vecp_(2)` and `vecp_(3)` be the momentum `vecv_(1).vecv_(2)` and `vecv_(3)` be the velocities of the three fragments respectively.
Then `vecp_(1) = mvecv_(1), vecp_(2) = mvecv_(2),vecp_(3) = mvecv_(3)`
As `vecp_(1) bot vecp_(2)`, so the magnitude of their resultant is
`p= sqrt(p_(1)^(2) + p_(2)^(2)) = msqrt(v_(1)^(2) + v_(2)^(2))`
`= msqrt(9^(2) +12^(2)) =15m kg ms^(-1)`
`therefore v_(3) = p/m = (15m)/m = 15 ms^(-1)`
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