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A body of mass 1 kg at rest explodes int...

A body of mass 1 kg at rest explodes into three fragments of masses in the ratio 1 : 1 : 3 . The two pieces of equal mass fly in mutually perpendicular directions with a speed of `30 m//s` each . What is the velocity of the heavier fragment ?

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The correct Answer is:
14.12
The linear momentum of the body is zero before explosion, hence it should be zero just after explosion according to conservation of linear momentum. The resultant momentum of the fragments flying off perpendicular to each other should be equal and opposite to the third fragment. Resolving the momentum of the third fragment in the lines of momentum of the other two. `vecp_1, vecp_2 and vecp_3` are the linear momentum of first, second and third fragments respectively.
From conservation of linear momentum `vecp_3`should be equal and opposite to (resultant of `vecp_1 and vecp_2` ). So, let `v'` be the velocity of third fragment, then
`(3m)v' = sqrt(2)mv :. v' = (sqrt2)/(3)v`
Here, `v = 30 m//s ("given") " ":. v' = (sqrt2)/(3) xx 30 = 10sqrt(2) m//s`
This velocity is at `45^@` at shown in the figure.
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