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)`

Let mass of rocket at any time t = M

∆M = mass of gas ejected in time interval ∆t.

\((KE)_{t+Δt}=\frac{1}{2}(m-Δm)(v+Δv)^2+\frac{1}{2}Δm(v-u)^2\)

For rocket For gas

\(\frac{1}{2}mv^2+MvΔv-Δmvu+\frac{1}{2}Δmu^2\)

Initial (KE)i = \(\frac{1}{2}mv^2\)

\((KE)_{t+Δt}-(KE)t=(MΔv-Δmu)v+\frac{1}{2}Δmu^2\)

By Newton’s third law, Reaction force on rocket (upward) = Action force by burnt gases (downward)

\(\frac{Mdv}{dt}=\frac{dm}{dt}|u|\)

(∵ F = mu)

or M∆v = ∆mu ⇒ M∆v − u∆m = 0

substitute this value in (i)

K = \(\frac{1}{2}\)∆mu²