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 $V \ll \left(\frac{dL}{L}\right) v_0$, where $dL$ is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct ?

Step 1: Find the change in speed of the particle per collision.

Analyze the collision in the frame of reference of the heavy piston moving inward with speed $V$.

$$v_{\text{rel, before}} = v + V$$ $$v_{\text{rel, after}} = v + V$$

Transforming back to the laboratory frame, the speed after collision is:

$$v' = (v + V) + V$$ $$v' = v + 2V$$

The increase in speed per collision is:

$$\Delta v = 2V$$

Step 2: Determine the rate of collisions with the piston.

The distance for one round trip between the closed end and the piston is $2L$.

$$\Delta t = \frac{2L}{v}$$

The rate of collision with the piston is:

$$\text{rate} = \frac{1}{\Delta t}$$ $$\text{rate} = \frac{v}{2L}$$

Step 3: Find the change in speed for an inward displacement $dL$.

In an infinitesimal time interval $dt$, the piston moves inward by a distance $dL = -V dt$.

$$dN = \text{rate} \cdot dt$$ $$dN = \frac{v}{2L} dt$$

The change in speed $dv$ is:

$$dv = 2V dN$$ $$dv = 2V \left(\frac{v}{2L} dt\right)$$ $$dv = -v \frac{dL}{L}$$

Step 4: Establish the relation between $v$ and $L$.

Integrate the differential relation.

$$\int \frac{1}{v} \, dv = -\int \frac{1}{L} \, dL$$ $$\ln v = -\ln L + \text{constant}$$ $$vL = \text{constant}$$

Step 5: Calculate the final kinetic energy when $L \to L_0/2$.

Use the conservation of $vL$.

$$v_0 L_0 = v \left(\frac{L_0}{2}\right)$$ $$v = 2v_0$$

The final kinetic energy $K$ is:

$$K = \frac{1}{2}mv^2$$ $$K = \frac{1}{2}m(2v_0)^2$$ $$K = 4 \left(\frac{1}{2}mv_0^2\right)$$ $$K = 4K_0$$

Step 6: Evaluate the given statements.

Compare the derived results with the statements:

• Statement (1): $K \to 4K$ when $L \to L_0/2$ is correct.
• Statement (2): $\Delta v = 2v \frac{dL}{L}$ is incorrect (the correct relation is $dv = -v \frac{dL}{L}$).
• Statement (3): rate $= v/L$ is incorrect (the correct rate is $v/2L$).
• Statement (4): $\Delta v = 2V$ per collision is correct.

Answer: The correct statements are:

1. The kinetic energy of the particle becomes $4$ times its initial value when the distance of the piston from the closed end is $L_0/2$.
2. After each collision with the piston, the speed of the particle increases by $2V$.