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  • A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u.

A rocket accelerates straight up by ejecting gas downwards. In a small time interval \Delta t, it ejects a gas of mass \Delta m at a relative speed u. Calculate KE of the entire system at t+\Delta t and t and show that the device that ejects gas does work = \left ( \frac{1}{2} \right ) \Delta m u^{2} in this time interval.

Answers (1)

We assume the mass of the rocket at any time to be m.

Let velocity of rocket be v

Mass of gas ejected in time t can be Δm

(KE) _{t+ \Delta t} = \frac{1}{2} ( M- \Delta m) (v + \Delta v)^{2} + \frac{1}{2} \Delta m (v-u)^2

= \frac{1}{2} [Mv^{2} + M \Delta v^{2} + 2Mv \Delta v - \Delta mv^{2} - m \Delta v^{2} - 2v \Delta m \Delta v+ \Delta mv^{2}+ \Delta mu^{2} - 2uv \Delta m ]

When we neglect small terms we get,

(KE) = \frac{1}{2} MV^{2}

\Delta K = \frac{1}{2} \Delta mu^{2} + v (M \Delta v - u \Delta m)

In accordance with newton’s third law,

Reaction upward force on the rocket is equal to action force by burning of gases in downward direction

M \frac{dv}{dt }= \frac{dm}{dt} |u|

M \Delta v = \Delta mu

Substituting this value, we get,

K = \frac{1}{2} u^{2} \Delta m

As we know from work-energy theorem, \Delta KE = work done

We have, work done = \frac{1}{2} \Delta m u^{2}

Posted by

infoexpert24

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