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A rocket accelerates straight up by ejec...

A rocket accelerates straight up by ejecting gas downwards . In small time interval `Deltat` , it ejects a gas of mass `Deltam ` at a relative speed u . Calculate kE of the entire system at `t+Deltat and t ` and show that the device that ejects gas does work = `(1/2) Delta"mu"^(2)` in this time interval (negative gravity ).

Text Solution

Verified by Experts

M =mass of rocket at t
v = velocity of rocket at t
`Deltam ` = mass of ejected gas in `Deltat`
u = relative speed of ejected gas
Consider at time t + `Deltat`
`(KE)_(t) +(KE)_(Deltat) = `KE of rocket + Ke of `= 1/2 (M-Deltam) (v+Deltav)^(2) +1/2Deltam (v-u)^(2)`
`1/2 Mv^(2) +Mv Deltav -Deltamvu +1/2 Delta"mu"^(2)`
`(KE)_(t) =` KE of the rocket at time t = `1/2 Mv^(2)`
`DeltaK - (KE)_(t) +(KE)_(Deltat) -(KE)_(t)`
`(MDeltav - Delta"mu") v +1/2 Delta" mu"^(2)`
Hene , `M (dv)/(dt) =(dc)/(dt) |u|`
` :. M Deltav = Delta"mu"`
`:. DeltaK = 1/2 Delta"mu"^(2)`
Now , by work - energy theorem ,
`DeltaK = Delta W `
` :. Delta W = 1/2 Delta"mu"^(2)`
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