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Question

(A) The particle moves between two walls separated by distance $L$. The time taken for one round trip is $\Delta t = 2L/v$. The frequency of collision

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(A) The particle moves between two walls separated by distance $L$. The time taken for one round trip is $\Delta t = 2L/v$. The frequency of collision with the piston is $f = 1/\Delta t = v/(2L)$. Thus, statement $(1)$ is incorrect as it states $v/L$.In an elastic collision with a piston moving at speed $V$ towards the particle, the velocity of the particle changes from $v$ to $v + 2V$. The change in speed is $(v + 2V) - v = 2V$. Thus, statement $(2)$ is correct.For a small displacement $dL$ of the piston, the change in speed is $dv = 2V 	imes (	ext{number of collisions in time } dt)$. Since $dt = dL/V$, the number of collisions is $dt / (2L/v) = (dL/V) \cdot (v/2L) = v dL / (2LV)$.Therefore, $dv = 2V \cdot (v dL / 2LV) = v dL / L$. Thus, statement $(4)$ is incorrect as it states $2v dL/L$.Integrating $dv/v = dL/L$ (with $L$ decreasing, $dv/v = -dL/L$), we get $\ln(v/v_0) = -\ln(L/L_0) = \ln(L_0/L)$, so $vL = v_0 L_0$. If $L$ becomes $L_0/2$, then $v = 2v_0$. The kinetic energy $KE = \frac{1}{2}mv^2$ becomes $\frac{1}{2}m(2v_0)^2 = 4 	imes (\frac{1}{2}mv_0^2) = 4 KE_0$. Thus, statement $(3)$ is correct.The correct statements are $(2)$ and $(3)$.

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