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InterviewSolution
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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. `m upsilon^(20`B. `(3)/(2)m upsilon^(2)`C. `2 m upsilon^(2)`D. `4 m upsilon^(2)` |
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Answer» Correct Answer - B Here,`m_(1)=m_(2)=m, m_(3)=4m-2m=2m,` `p_(1)=p_(2)=m upsilon` . A these pieces move perpendicular to each other, their resultant momentum `=p=sqrt(p_(1)^(2)+p_(2)^(2))=sqrt((m upsilon)^(2)+( m upsilon)^(2))=m upsilonsqrt(2)` Principle of conservation of linear momentum gives `p_(3)=m_(3) upsilon_(3) =p= m upsilonsqrt(2)` `upsilon_(3)=( m upsilon sqrt(2))/( m_(3))=( m upsilonsqrt(2))/( 2m)=(upsilon)/( sqrt(2))` Total `K.E.=(1)/(2) m _(1)upsilon_(1)^(2)+(1)/(2)m_(2)upsilon_(2)^(2)+(1)/(2) m_(3)upsilon_(3)^(2)` `=(1)/(2) m upsilon^(2)+ (1)/(2) m upsilon^(2) +(1)/(2) (2m) ((upsilon)/(sqrt(2)))^(2)` `E=(3)/(2) m upsilon^(2)` |
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