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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 ............
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Answer: B::CSolution:
rArr sin theta =cos theta =1/sqrt2(2mv')/sqrt2=mv, v'=v/sqrt2Total K.E. =1/2mv^2+1/2mv^2+1/2(2m)(v/sqrt2)^2=1/2mv^2+1/2mv^2+(mv^2)/2=3/2mv^2\displaystyle
rArr sin theta =cos theta =1/sqrt2(2mv')/sqrt2=mv, v'=v/sqrt2Total K.E. =1/2mv^2+1/2mv^2+1/2(2m)(v/sqrt2)^2=1/2mv^2+1/2mv^2+(mv^2)/2=3/2mv^2\displaystyle
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Question Text | 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 ............ |
Answer Type | Text solution:1 |
Upvotes | 150 |