A car of mass M is free to move on a frictionless  horizontal surface. A gun fires bullets on the car.  The bullets leave the stationary gun with speed u and mass rate $b {\mathrm{kgs}}^{-1}$ The bullets hit the vertical  rear surface of the car while travelling horizontally  and collisions are elastic. If the car starts at rest 
find its speed  as a function of time.  Mass of the car M >> mass of each bullet. 

Let the speed of the car at time t be v. Let dm mass of bullets hit the car in a small time interval dt. Velocity of approach = velocity of separation 
$u-v=v_b+u\Rightarrow v_b=2u-v$
v_b is velocity of bullet after the hit in direction opposite to its original direction. We are assuming that change in speed of car due to impact of one bullet is negligible. Momentum transferred to the car in interval dt is 
$\mathrm{dp}=2(\mathrm{u}-\mathrm{v})\mathrm{dm}$
$\therefore \text{ Force on car }\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}=2(\mathrm{u}-\mathrm{v})\frac{\mathrm{dm}}{\mathrm{dt}}$
The bullets do not hit the car at the rate at which they leave the gun. 
$$\begin{gathered} \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{b}(\mathrm{u}-\mathrm{v})}{\mathrm{u}} \\ \therefore \mathrm{M}\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{2(\mathrm{u}-\mathrm{v}{)}^2}{\mathrm{u}}\mathrm{b}\Rightarrow {\int }_0^{\mathrm{v}} \frac{\mathrm{dv}}{(\mathrm{u}-\mathrm{v}{)}^2}=\frac{2\mathrm{b}}{\mathrm{Mu}}{\int }_0^{\mathrm{t}} \mathrm{dt} \\ \Rightarrow \frac{1}{\mathrm{u}-\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{2\mathrm{bt}}{\mathrm{Mu}} \\ \Rightarrow \mathrm{v}=\frac{\left(\frac{2\mathrm{bt}}{\mathrm{M}}\right)\mathrm{u}}{1+\frac{2\mathrm{bt}}{\mathrm{M}}}=\frac{\mathrm{u}}{\frac{\mathrm{M}}{2\mathrm{bt}}+1} \\ \mathrm{As}\mathrm{t}\to \mathrm{\infty };\mathrm{v}\to \mathrm{u} \end{gathered}$$