A rocket ejects the fuel (hot gases) of density `rho` from the chamber which is maintained at a pressure `P_(1)`, say. The atmospheric pressure is `P_(0)`. If the base area of the cylindrical chamber is `A_(1)` and the area of the hole through which hot gases escape from the rocket is `A_(2)` find (a) velocity of efflux of the fuel, (b) thrust exerted on the rocket given by the escaping fuel.

Answer

`(P-P_(0))`

Explanation

Let us consider two points `1` and `2`. Applying Bernoulli's equatioin. We have
`P_(1)+1/2rhov_(1)^(2)+rhogh_(1)=P_(2)+1/2rhov_(2)^(2)+rhogh_(20`
Since `rhog(h_(1)-h_(2))=rhogh` is negligible compared to `P_(1)=P` (pressure inside the chamber) and `P_(2)=P` (atmospheric pressure) we have
`P+1/2rhov_(1)^(2)=P_(0)+1/2rhov_(2)^(2)`..........i
Equation of continuity gives
`A_(1)v_(1)=A_(2)v_(2)`......ii

Since `v_(1)=((A_(2))/(A_(1)))v_(2)` and `A_(2)lt ltA_(1)` we can ignore `v_(1)` when compared to `v`.
Then eqn i gives `v_(2)=sqrt((2(P-P_(0)))/rho)`
b. The impact force of the ejected gas is given as
`F=rhoA_(2)v_(2)^(2)`
where `v_(rel)=v_(2)=sqrt((2(P-P_(0)))/rho)`
Then we have `F=2A_(2)(P-P_(0))`