InterviewSolution
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InterviewSolution
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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. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas work = (1/2) ∆ mu2 in this time interval (neglect gravity). |
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Answer» 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 |
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