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 < < (dL)/(L) v_0`, where dL is the infinitesimal displacement of the piston. Which of the following statements(s) is(are) correct?

Options

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(dL)/(L)`
C. The rate at which the particle strikes the piston is v/L
D. The particle’s kinetic energy increases by a factor of 4 when the piston is moved inward from `L_0 "to" 1/2 L_0`

Explanation

To solve the problem, we need to analyze the motion of the particle and the effects of the moving piston on the particle's speed and kinetic energy. Here’s the step-by-step solution:

### Step 1: Understand the System
We have a small particle of mass \( m \) moving inside a hollow tube. The tube has one end closed and the other end has a movable piston. The particle undergoes elastic collisions with both ends of the tube. Initially, when the distance of the piston from the closed end is \( L = L_0 \), the particle's speed is \( v = v_0 \).

### Step 2: Analyze the Movement of the Piston
The piston moves inward at a very low speed \( V \) such that \( V \ll \frac{dL}{L} v_0 \). This means the piston is moving slowly compared to the speed of the particle and the changes in length \( L \).

### Step 3: Calculate the Time for Collisions
The time taken for the particle to collide with the piston can be calculated as follows:
- The total length of the tube is \( 2L \) (from one end to the other).
- The time \( T \) for one complete round trip (collision with both ends) is given by:
\[ T = \frac{2L}{v_0} \]

### Step 4: Determine the Number of Collisions
The number of collisions \( N \) in a time interval can be calculated by:
- The number of collisions in one second is:
\[ N = \frac{v_0}{2L} \]
- In a small time interval \( dt \), the number of collisions is:
\[ dN = \frac{v_0}{2L} dt \]

### Step 5: Effect of Piston Movement on Particle Speed
As the piston moves inward by an infinitesimal distance \( dL \), the speed of the particle after each collision changes. The change in speed \( dv \) after \( N \) collisions can be calculated as:
- The increment in velocity after \( N \) collisions:
\[ \Delta v = 2v_0 \cdot dN = 2v_0 \cdot \frac{v_0}{2L} dt = \frac{v_0^2}{L} dt \]

### Step 6: Integrate to Find Final Speed
Integrating the change in speed gives us the final speed \( v_f \):
\[ v_f = v_0 + \int \frac{v_0^2}{L} dt \]
As the piston moves inward, the effective length \( L \) decreases, leading to an increase in speed.

### Step 7: Kinetic Energy Calculation
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} m v^2 \]
Initially, the kinetic energy is:
\[ K_0 = \frac{1}{2} m v_0^2 \]
After the piston moves and the speed increases to \( v_f \), the new kinetic energy becomes:
\[ K_f = \frac{1}{2} m (2v_0)^2 = 2m v_0^2 \]
Thus, the kinetic energy increases by a factor of 4.

### Conclusion
After analyzing the system, we can conclude:
1. After each collision with the piston, the particle's speed increases.
2. The particle's kinetic energy increases by a factor of 4 when the piston moves inward.