A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is L = $L_0$ the particle speed is v = $v_0$. The piston is moved inward at a very low speed V such that V << $\frac{dL}{L}$ $v_0$, where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct ?

$\text{ Let say after collision velocity of particle becomes }{v}^{\mathrm{\prime }}\text{ and piston keeps moving with same speed }V$

$\text{ Velocity of separation }=\text{ Velocity of approach }$

$\Rightarrow v^{\mathrm{\prime }}-V=v+V$

$\Rightarrow v^{\mathrm{\prime }}=v+2V$

$\text{ Thus after every collision with piston, speed of particle increases by }2\mathrm{V}\text{ . }$

$\text{ Time between two successive collision }=T=\frac{\text{ Total dist to and fro }}{\text{ velocity }}=\frac{2L}{v}$

$\text{ Rate of collision }=f=\frac{1}{T}=\frac{v}{2L}$

$\text{ Let say piston move by distance }\mathrm{dL}\text{ in time }\mathrm{dt}$

$\begin{array}{l}\Rightarrow dL=V\times dt\\ \Rightarrow dt=\frac{dL}{V}\end{array}$

$\text{ In this dt time, No of collision occured with particle }=n=dt\times f=\frac{dL}{V}\times \frac{v}{2L}$

$\text{ After these }n\text{ collisions, the increament in speed of particle }=dv=n\left(2\mathrm{V}\right)=\frac{\mathrm{dL}}{V}\times \frac{\mathrm{v}}{2L}\times 2\mathrm{V}$

$\Rightarrow dv=\frac{dL}{L}\times v$

$\text{ Since, }L\text{ is decresing, we will put a negative sign, }$

$\begin{array}{l}\Rightarrow {\int }_{V_0}^v \frac{dv}{v}=-{\int }_{L_0}^{L_0/2} \frac{dL}{L}\\ \Rightarrow v=2v_0\\ \text{ Since, }KE\propto v^2\text{ , }\\ \text{ KE becomes }4\text{ times. }\end{array}$