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 the figure, when the distance of the piston from the closed end is $L = {L_0}$ the particle speed is $v = {v_0}$ . The piston is moved inward at a very low speed such that $V < < \dfrac{{dL}}{L}{v_0}$ . Where $dL$ is an infinitesimal displacement of the piston.
Which of the following statements is 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 $2V\dfrac{{dL}}{L}$
C. The particle’s kinetic energy increases by factor of $4$ when the piston is moved inward from ${L_0}$ to $\dfrac{1}{2}{L_0}$
D. The rate at which the particle strikes at the piston is $\dfrac{v}{L}$

Answer 1

Hint : In order to solve this question, we are going to see the conditions before and after the collision and the rate at which a particle hits the piston then, the speed of particle is found from the rate of change of velocity and afterwards the kinetic energies of a particle initially and finally.
The rate of change of velocity is
$\dfrac{{dv}}{{dt}} = f \times 2V$
Kinetic energy, $K = \dfrac{1}{2} \times m \times {v^2}$
m = mass \\\ v = velocity \\\

Complete Step By Step Answer:
The inferences that can be drawn from the question
Piston before the collision

After the collision,

The rate at which the particle strikes the piston is given by
$= f = \dfrac{v}{{2x}}$
If $x = L$ , then, $f = \dfrac{v}{{2L}}$
The rate of the change of speed of particle is
\dfrac{{dv}}{{dt}} = f \times 2V \\\ dv = \dfrac{v}{{2x}} \times 2Vdt \\\ dv = \dfrac{v}{{2x}} \times 2\left( { - dx} \right) \\\
Now integrating to find the velocity,
\Rightarrow \int\limits_{{v_0}}^v {\dfrac{{dv}}{v} = \int\limits_l^x {\dfrac{{ - dx}}{x}} } \\\ \Rightarrow \ln \dfrac{v}{{{v_0}}} = - \ln \dfrac{x}{l} \\\ \Rightarrow v = \dfrac{{{v_0}l}}{x} \\\
Now, taking the case, $x = \dfrac{{{L_0}}}{2}$
Then, if we find the velocity
$v = \dfrac{{{v_0}{L_0} \times 2}}{{{L_0}}}$
Now finding the kinetic energy at the value $x = \dfrac{{{L_0}}}{2}$
${K_f} = \dfrac{1}{2} \times m \times 4{v_0}^2$
Kinetic energy at $x = {L_0}$
${K_i} = \dfrac{1}{2} \times m \times {v_0}^2$
Therefore, the ratio of the kinetic energies is
$\dfrac{{{K_f}}}{{{K_i}}} = \dfrac{4}{1} = 4$
Therefore, the options that are correct are A and C.

Note :
After each collision with the piston, the particles hit the piston with more and more energy that is , the particle speed increases by $2V$ and if we find the corresponding particle’s kinetic energy , it increases by factor of $4$ when the piston is moved inward from ${L_0}$ to $\dfrac{1}{2}{L_0}$ .