Question

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).

Answer 1

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