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 = L₀ the particle speed is v = v₀. The piston is moved inward at a very low speed V such that V << $\frac{dL}{L}$ V₀ , where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct ?

• A. The rate at which the particle strikes the piston is v/L
• B. After each collision with the piston, the particle speed increases by 2V
• C. The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L₀ to $\frac{1}{2}$ L₀
• D. If the piston moves inward by dL, the particle speed increases by 2v $\frac{dL}{L}$

Answer 1

Answer: (B), (C)

(1) average rate of collision = $\frac{2L}{v}$

(2) speed of particle after collision $=2V+v_0$
change in speed $=\left(2V+v_0\right)-v_0$
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

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

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

Now, $dv=-\frac{vdL}{L}$ {as L decrease}
$\underset{v_0}{\overset{v}{\int }}\frac{dv}{v}=-\underset{L_0}{\overset{L_0/2}{\int }}\frac{dL}{L}$

$\Rightarrow {\left[lnv\right]}_{v_0}^v=-{\left[lnL\right]}^{L_0/2}$
$\Rightarrow v=2v_0$

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

$K{E}_{L_0/2}=\frac{1}{2}m{\left(2v_0\right)}^2$
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 1/4
$vdx+xdv=0$