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प्रश्न
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. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = `(1/2)∆m u^2` in this time interval (neglect gravity).
उत्तर
Let M be the mass of the rocket at any time t and v1 the velocity of the rocket at the same time t.
Let ∆m = mass of gas ejected in time interval ∆t.
The relative speed of gas ejected = u.
Consider at time t + ∆t
(KE)t + ∆t = KE of rocket + KE of gas
= `1/2 (M - ∆m) (v + ∆v)^2 + 1/2 ∆m (v - u)^2`
= `1/2 Mv^2 + Mv ∆v - ∆mvu + 1/2 ∆m u^2`
(KE)t = KE of the rocket at time t = `1/2 Mv^2`
∆K = (KE)t + ∆t – (KE)t
= (M∆v ∆mu)v + `1/2` ∆mu2
Since action-reaction forces are equal.
Hence, `M (dv)/(dt) = (dm)/(dt)|u|`
⇒ M∆v = ∆mu
∆K = `1/2` ∆mu2
Now, by work-energy theorem,
∆K = ∆W
⇒ ∆W = `1/2` ∆mu2
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