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 inside the tube and how it interacts with the moving piston. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
- The particle of mass \( m \) is moving inside a hollow tube.
- The tube is closed at one end and has a movable piston at the other end.
- Initially, when the distance of the piston from the closed end is \( L = L_0 \), the speed of the particle is \( v = v_0 \).

### Step 2: Analyze the Motion and Collisions
- The particle undergoes elastic collisions with the piston and the closed end of the tube.
- When the particle collides elastically with the piston, its speed will change based on the conservation of momentum and kinetic energy.
- After colliding with the piston, the speed of the particle will increase.

### Step 3: Determine the Speed After Collision
- The speed of the particle after the first collision with the piston can be calculated. Since it is an elastic collision, the speed of the particle after the collision becomes:
\[ v' = 2v_0 \]
This is because the piston is heavy and does not move significantly compared to the particle.

### Step 4: Calculate the Time Taken for Piston Movement
- The piston moves inward at a very low speed \( V \) such that \( V \ll \frac{dL}{L} v_0 \).
- If the piston moves by an infinitesimal distance \( dL \), the time taken for this movement is:
\[ T = \frac{dL}{V} \]

### Step 5: Calculate the Number of Collisions
- The particle travels a total length of \( 2L \) (to the closed end and back) in time \( T \).
- The number of collisions in time \( T \) can be calculated as:
\[ \text{Number of Collisions} = \frac{v_0}{2L} \cdot T = \frac{v_0}{2L} \cdot \frac{dL}{V} \]

### Step 6: Increment in Velocity After Collisions
- The increment in velocity after \( n \) collisions can be expressed as:
\[ \Delta v = \frac{v_0 \cdot dL}{L} \]
Thus, the new speed of the particle becomes:
\[ v_f = v_0 + \Delta v = v_0 + \frac{v_0 \cdot dL}{L} \]

### Step 7: Relate Velocity to Kinetic Energy
- The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
- If the speed doubles, the kinetic energy increases by a factor of 4:
\[ KE_f = \frac{1}{2} m (2v_0)^2 = 4 \cdot \frac{1}{2} mv_0^2 \]

### Conclusion
- After analyzing the situation, we conclude:
1. After each collision with the piston, the particle speed increases by a factor of 2.
2. The kinetic energy increases by a factor of 4 when the piston is moved inward from \( L_0 \) to \( L_0/2 \).

### Final Statements
- The correct statements are:
- After each collision with the piston, the particle speed increases by a factor of 2.
- The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from \( L_0 \) to \( L_0/2 \).