Question

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 $L=L_o$ 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?

• A. After each collision with the piston, the particle speed increases by $2V.$
• B. If the piston moves inward by dL, the particle speed increases by $2V\frac{dL}{L}$
• C. The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from $L_0$ to $\frac{L_0}{2}$
• D. The rate at which the particle strikes the piston is $v/L$

Answer 1

Answer: (A), (C)

The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from $L_0$ to $\frac{L_0}{2}$

Speed of particle after collision $=2V+v_0$ (As $m<<<m_{p}iston$)

change in speed $=(2V+v_0)-v_0$

after each collision $=2V$

no. of collision per unit time (frequency) $=\frac{v}{2L}$

change in speed in $dt$ time $=2V\times$ number of collision in $dt$ time

$\Rightarrow dv=2V\left(\frac{v}{2L}\right)\cdot \frac{dL}{V}$

$dv=\frac{vdL}{L}$

Now, $dv=-\frac{vdL}{L}\text{(as L decrease}\}$

${\int }_{v_0}^v\frac{dv}{v}=-{\int }_{L_0}^{L_0/2}\frac{dL}{L}$

$\Rightarrow [\ln v{]}_{v_0}^v=-[\ln L{]}^{L_0/2}$

$\Rightarrow v=2v_0$

$\Rightarrow K{E}_{L_0}=\frac{1}{2}m{v}_0^2\frac{K{E}_{l_{0/2}}}{K{E}_0}=4$

$K{E}_{L_0/2}=\frac{1}{2}m{\left(2v_0\right)}^2$
or

$\left(dt\right)\left(\frac{v}{2x}\right)\frac{2mv}{dt}=F$

$F=\frac{m{v}^2}{x}$

$-mv\frac{dv}{dx}=\frac{m{v}^2}{x}$

$-\frac{dv}{v}=\frac{dx}{x}$

$\ln \frac{v_2}{v_1}=\ln \frac{x_1}{x_2}$

$vx=constant\Rightarrow$ on decreasing length to half K.E. becomes $4$
$vdx+xdv=0$