Ncert Exercises & Solutions

Q.46 A rocket accelerates straight up by ejecting gas downwards. In a small time interval $∆$ t, it ejects a gas of mass Am at a relative speed u. Calculate KE of the entire system at t + $∆$ t and t and show that the device that ejects gas does work = (1/2) $∆ {mu}^{2}$ in this time interval (negative gravity).

Hint: Change in KE = work done by the device.

Step 1: Find the change in KE in time t.

Let M be the mass of the rocket at any time t and $v_{1}$ the velocity of the rocket at the same time t.
Let $∆$ m is the mass of gas ejected in time interval $∆$ t.
The relative speed of gas = u.

Consider at time t + $∆$ t

$\begin{array}{l} (KE)_{t + Δt} = K E of rocket + KE of gas \\ = \frac{1}{2} (M - Δm) (v + Δv)^{2} + \frac{1}{2} Δm (v - u)^{2} \\ = \frac{1}{2} {Mv}^{2} + MvΔv - Δmvu + \frac{1}{2} {Δmu}^{2} \\ (KE)_{t} = KE of the rocket at time t = \frac{1}{2} {Mv}^{2} \\ ΔK = (KE)_{t + Δt} - (KE)_{t} \\ = (MΔv - Δmu) v + \frac{1}{2} {Δmu}^{2} \end{array}$

Since action-reaction forces are equal.

Hence,

$\begin{array}{l} M \frac{d v}{d t} = \frac{d m}{d t} u \\ \Rightarrow M Δ v = Δ m u \\ Δ K = \frac{1}{2} Δ m u^{2} \end{array}$

Step 2: Find work done by applying work-energy theorem

Now, by work-energy theorem,

$\begin{array}{l} Δ K = Δ W \\ \Rightarrow Δ W = \frac{1}{2} Δ m u^{2} \end{array}$