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A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ............

A

0

B

`3/2mv^2`

C

`1/2mv^2`

D

`2mv^2`

Text Solution

Verified by Experts

The correct Answer is:
B
By momentum conservation
`2mv. = sqrt2 mv`
`implies v. =v/sqrt2`
KE = `2(1/2 mv^2)+1/2(2m)(v^2/2)`
`=mv^2 +(mv^2)/2 =3/2 mv^2`
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