$\Delta\text{m}$ is the mass of gas ejected in time interval $\Delta\text{t}$
$(\text{KE})_{\text{t}+\Delta\text{t}}=\frac{1}{2}(\text{M}-\Delta\text{m})(\text{v}+\Delta\text{v})^2+\frac{1}{2}\Delta\text{m}(\text{v}-\text{u})^2$
$=\frac{1}{2}\Big[(\text{M}-\Delta\text{M})(\text{v}+\Delta\text{v})^2+\frac{1}{2}\Delta\text{m}(\text{v}-\text{u})^2\Big]$
$(\text{KE})_{\text{t}+\Delta\text{t}}=\frac{1}{2}\begin{bmatrix}\text{Mv}^2+\text{M}\Delta\text{v}^2+2\text{Mv}\Delta\text{v}-\Delta\text{Mv}^2\\-\Delta\text{m}\Delta\text{v}^2-2\text{v}\Delta\text{m}\Delta\text{v}+\Delta\text{mv}^2+\\\Delta\text{mu}^2-2\text{uv}\Delta\text{m}\end{bmatrix}$
$(\text{KE})_{\text{t}+\Delta\text{t}}=\frac{1}{2}\text{Mv}^2+\text{Mv}\Delta\text{v}+\frac{1}{2}\Delta\text{mu}^2-\text{uu}\Delta\text{m}$
$\big[$neglecting the very small terms $\text{M}\Delta\text{v}^2,\Delta\text{m}\Delta\text{v}^2,2\text{v}\Delta\text{m}\Delta\text{v}$ contains $\Delta\text{v}^2$ and $\Delta\text{m}\Delta\text{v}\big]$
$(\text{KE})_\text{t}=\frac{1}{2}\text{Mv}^2$
$(\text{KE)}_{\text{t}+\Delta\text{t}}-(\text{KE})_\text{t}=\frac{1}{2}\text{Mv}^2+\text{Mv}\Delta\text{v}+\frac{1}{2}\Delta\text{mu}^2-\text{uu}\Delta\text{m}-\frac{1}{2}\text{Mv}^2$
$\Delta\text{K}=\frac{1}{2}\Delta\text{mv}^2+\text{v}(\text{M}\Delta\text{v}-\text{v}\Delta\text{m})$
By Newton’s third law, Reaction force on Rocket (upward) = Action force by burnt gases (downward)$\text{M}\frac{\text{dv}}{\text{dt}}=\frac{\text{dm}}{\text{dt}}|\text{u}|\ (\because\text{ F}=\text{ma})$
Or $\text{M}\Delta\text{v}=\Delta\text{mu}$$\text{M}\Delta\text{v}-\text{u}\Delta\text{m}=0$
Substitute this value in (i)$\text{K}=\frac{1}{2}\text{u}^2\Delta\text{m}$
By work energy theorem $\Delta(\text{KE})=\text{WD}$ Or $\text{W}=\Delta\text{K}=\frac{1}{2}\Delta\text{mu}^2$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
