InterviewSolution
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InterviewSolution
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A rocket accelerates straight up by ejecting gas downwards. In a small time interval `Deltat`, it ejects a gas of mass `Delta m` at a relative speed `u` . Calculate KE of the entire system at `t+Deltat` and `t` and show that the device that ejects gas does work `=((1)/(2))Delta m . u^(2)` in this time interval (neglect gavity). |
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Answer» LET M be the mass of rocket at any time t and `v_(1)` the velocity of rocket at the same time t. Let `Deltam`=mass of gas ejected in time interval `Deltat.` Relative speed of gas ejected=u. Consider at time `t+Deltat` `(KE)_(t)+Deltat="KE of rocket+KE of gas"` `=(1)/(2)(M-Deltam)(v+Deltav)^(2)+(1)/(2)Deltam(v-u)^(2)` `=(1)/(2)Mv^(2)+MvDeltav-Deltamvu+(1)/(2)Delta"mu"^(2)` `(KE)_(t)="KE of the rocket at time"t=(1)/(2)Mv^(2)` `DeltaK=(KE)_(t)+Deltat-(KE)_(t)` `=(MDeltav=Delta"mu")v+(1)/(2)"mu"^(2)` Hence, `M(dt)/(dt)=(dm)/(dt)|u|` `DeltaK=(1)/(2)"mu"^(2)` Now by work energy theoram=`DeltaK=DeltaW` `Rightarrow DeltaW=(1)/(2)Delta"mu"^(2)` |
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