Bouncing a lot

The momentum transfer at a collision is $\Delta p=2mv$ . It takes a time of $\Delta t= 2L/v$ for the ball to return for the next collision. The force is $F=\Delta p /\Delta t= mv^2/L$ . We can rewrite this in terms of the kinetic energy of the ball, $E=\frac{1}{2} mv^2$ , as

$F= \frac {2E}{L}$

We have to apply this external force in order to keep the distant $L$ steady. Changing the distance between the plates by $dL$ requires work, $dW=- dL F$ . Since there is no dissipation in the system this work will change the kinetic energy of the bouncing ball:

$dE=- dL F= - \frac{2E}{L}dL$

We can solve this differential equation by separation of variables:

$\frac{dE}{E}=-2\frac{dL}{L} \qquad \rightarrow \qquad \log E/E_0 = - 2\log L/L_0 \qquad \rightarrow \qquad E/E_0=(L/L_0)^{-2}$

Accordingly, the force can be expressed as

$F= \frac {2E}{L}= \frac {2E_0}{L} \left(\frac{L_0}{L}\right)^{2}$

In terms of the original parameters:

$F= \frac {mv_0^2}{L} \left(\frac{L_0}{L}\right)^{2}$

so the answer is A=2.0 .

Note that for a quantum mechanical "particle in a box" one gets an extremely similar result, see here , but the underlying physics is entirely different.

I just l guessed the answer 2 and got it right at the first place itself...!! Then I'm eligible for writing my own solution that's it😅

The ball, moving at speed $v$ between the plates at separation $L$ , bounces with each of the plates $\frac{v}{2L}$ times per second. At each bounce, a momentum of $2mv$ is transferred. The average force is momentum per time, so the force on the each of the plates is $F= \frac{2mv}{\frac{v}{2L}}=\frac{mv^2}{L}$ , pressing outward.

If we bring the plates a bit closer together by an external force, we put an amount of work $dE=-FdL= -mv^2 \frac{dL}{L}$ into the system. The kinetic energy of the system is $\frac{1}{2}mv^2$ , so that we can describe the process just in terms of energy and separation: $\frac{dE}{E} = -2\frac{dL}{L}$ .

Integrating both sides gives the relation $ln(E) = -2 ln(L)+C$ . Exponentiate to find $E = k/L^2$ where $k=e^C$ . Now find k by plugging in the original values $v_0$ and $L_0$ : $k = \frac{1}{2}L0^2mv_0$ so that $E =\frac{1}{2} mv_0^2 \frac{L_0^2}{L^2}$ .

Lastly $F=-dE/dL = mv_0^2\frac{L0^2}{L^3} = \frac{mv_0^2}{L}(\frac{ L0}{L})^2$

So that the asked A equals 2.00

My English is not very good.

This is a one-dimensional adiabatic process，and the adiabatic process equation is $pV^{γ}=const$ .

We know that $γ=\frac{C_{p}}{C_{V}}=\frac{0.5+1}{0.5}=3$ ,so the equation will be $FL^{3}=const$ .(according to the theorem of equipartition of energy)

so the answer is $A=3-1=2$ .

PS:This is not a very rigorous solution,but it is always correct.