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 ............
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The correct Answer is:
NA
`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
A
`mv^2`
B
`3/2 mv^2`
C
`2mv^2`
D
`3mv^2`
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A
`mv^(2)`
B
`2mv^(2)`
C
`1//2mv^(2)`
D
`4mv^(2)`
A body of mass 5m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass 'm' are found to move with a speed 60m//s in mutually perpendicular direction. The velocity of third fragment is
