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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 ............

Text Solution

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The correct Answer is:
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`2mv'costheta =mv` …………i
`impliessintheta=costheta=1/(sqrt(2))`

Putting this value in eqn i we get
`(2mv')/(sqrt(2))=mvimpliesv'=v/(sqrt(2))`
total `KE=1/2mv^(2)+1/2mv^(2)+1/2(2m)(v/(sqrt(2)))^(2)`
`=1/2mv^(2)+1/2mv^(2)+(mv^(2))/2=3/2mv^(2)`
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Knowledge Check

  • A body of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed v in mutually perpendicular directions. Total energy released in the process is

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