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A rocket accelerates straight up by ejec...

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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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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