A block of mass $M$ is kept on a smooth horizontal surface. A jet of water emerging from a nozzle of cross-section, it strikes the block horizontally with a speed $u.$Assuming $\rho$ as the density of water.

 

Let velocity of the block at any time be $v.$

The thrust force on the block is $F_{thrust}=\rho A{v}_{rel}^2={(u-v)}^{2}A\rho \mathrm{.........}\left(i\right)$

The power due to impact force is $P=F_{thrust}\cdot v=A\rho v_{rel}^2\cdot v=A\rho {(u-v)}^{2}v$

For maximum power, $\frac{dP}{dv}=0$or $A\rho \{-v\times 2(u-v)+{(u-v)}^2\}=0\Rightarrow v=\frac{u}{3}$

The power applied on the block due to the thrust imparted by water jet on the block will be  maximum when the velocity of the block reaches one third of water jet velocity.

Putting $v=\frac{u}{3}$in Eqs. (i) and (ii) we get $F_{thrust}=\frac{4}{9}A\rho u^2$ and $P_{\max }=\frac{4}{27}A\rho u^3$

(ii) The thrust force on the block is $F_{thrust}=\rho A{v}_{rel}^2={(u-v)}^{2}A\rho$

${(u-v)}^{2}A\rho =\frac{Mdv}{dt}$

$\underset{0}{\overset{v}{\int }}\frac{dv}{{(u-v)}^2}=\frac{A\rho }{M}\underset{0}{\overset{t}{\int }}dt$

which gives, $v=\frac{u}{(1+\frac{M}{A\rho ut})}$