InterviewSolution
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InterviewSolution
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A bomb at rest explodes into two fragments of masses `3.0kg` and `1.0kg`. The total kinetic energy of the fragments is `6xx10^(4)J`. Calculate (i) , kinetic energy of the bigger fragment (ii) momentum of the smaller fragment. |
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Answer» If `upsilon_(1), upsilon_(2)` are velocities of fragments of masses `m_(1)=3kg` and `m_(2)=1kg`, then according to the principle of conservation of linear momentum, `m_(1)upsilon_(1)+m_(2)upsilon_(2)` `3upsilon_(2)+1upsilon_(2)=0` or ` (upsilon_(1))/(upsilon_(2))=-(1)/(3)` `(E_(1))/(E_(2))=((1)/(2)m_(1)upsilon_(1)^(2))/((1)/(2)m_(2)upsilon_(2)^(2))=(3)/(1)(-(1)/(3))^(2)=(1)/(3)` As total `K.E.=E=E_(1)+E_(2)=6xx10^(4)J` `:. E_(1)=(6xx10^(4)xx1)/((1+3))=1.5xx10^(4)J` `E_(2)=6xx10^(4)-1.5xx10^(4)=4.5xx10^(4)J` Linear momentum of smaller fragment `P_(2)=sqrt(2mE_(2))=sqrt(2xx1xx4.5xx10^(4))` `3xx10^(2)kg ms^(-1)` As the direction is opposite `:. P_(2)=-300kg ms^(-1)` |
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