1. Collision with the moving piston

Let the piston move inward with speed $V$ (towards the particle). Since the piston is very heavy, treat it as an infinite-mass wall moving with speed $V$.

For an elastic collision with a wall moving toward the particle:

• velocity before collision: $+v$ toward the piston,
• wall velocity: $-V$ if we take rightward positive and piston moving left.

Using the standard result for elastic collision with a massive moving wall, $v' = 2u - v,$ where $u$ is wall velocity.

Here, just before collision the particle velocity is toward the piston, and after collision its speed increases by $\Delta v = 2V.$ So after each collision with the piston, the speed increases by $2V$.

Therefore, A is correct.

2. Relating change in speed to inward displacement $dL$

Suppose between two successive collisions with the piston, the piston moves inward by a small amount $dL$.

The time between two successive hits on the piston is approximately the round-trip time of the particle: $\Delta t \approx \frac{2L}{v}$ because $V \ll v$.

In this time, the piston moves inward by $dL = V\Delta t = V\cdot \frac{2L}{v}.$ Hence, $V = \frac{v\,dL}{2L}.$

Since speed gain per piston collision is $2V$, $dv = 2V = 2\cdot \frac{v\,dL}{2L} = \frac{v\,dL}{L}.$

So the increase in speed is $dv = \frac{v\,dL}{L},$ not $2v\dfrac{dL}{L}$.

Therefore, B is incorrect.

3. Dependence of speed on tube length

From above, $dv = \frac{v}{L}(-dL_{\text{actual}}),$ if $dL_{\text{actual}}$ denotes the decrease in length. More cleanly, using signed differential of length $L$: $\frac{dv}{v} = -\frac{dL}{L}.$

Integrating, $\ln v = -\ln L + \text{constant}$ so $vL = \text{constant}.$

Thus, $v \propto \frac{1}{L}.$

If length changes from $L_0$ to $L_0/2$, $v_f = v_0\frac{L_0}{L_0/2} = 2v_0.$

Therefore kinetic energy becomes $K_f = \frac12 m(2v_0)^2 = 4\left(\frac12 mv_0^2\right).$

So kinetic energy increases by a factor of $4$.

Therefore, C is correct.

4. Rate of striking the piston

The particle hits the piston once every round trip, so the time between successive hits is approximately $T = \frac{2L}{v}.$

Hence strike rate is $f = \frac{1}{T} = \frac{v}{2L}.$

So it is not $v/L$.

Therefore, D is incorrect.

5. Final conclusion

Correct options are: $\boxed{A,\ C}$

This matches the stored correct answer.