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A particle of mass 4 m which is at rest ...

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

`3/2mv^(2)`

B

`2/3mv^(2)`

C

`(mv^(2))/2`

D

`4mv^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
After explosion, the masses of the three fragments will be 3m, n and m.
Their momentum before explosion= 0
The resultant momentum of the two fragments each of the mass 'm' after collision.
`=sqrt((m xx 60)^(2) + ( m xx 60)^(2))`
as they are mutually perpendicular. The momentum of the third fragment `= 3m xx v `
`:. ( m xx 60)^(2) + ( m xx 60)^(2) = ( 3m xx v ) ^(2)`
`:. 3600m^(2) + 3600m^(2) = 9m^(2) v^(2)`
`:. 9 v^(2) = 7200 ` `:. v^(2) = 800`
`:. v = sqrt( 800) = 20 sqrt(2)m//s`
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