A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis under

Q: Why is the distance between collisions $2L$ and not $L$?

The particle hits the piston, travels distance $L$ to the closed end, bounces off, and travels another distance $L$ back to hit the piston again. So the total distance covered between two piston hits is $L + L = 2L$.

Q: Why does the speed increase by $2V$ and not just $V$?

In an elastic collision with a heavy moving wall, the relative speed of approach equals the relative speed of separation. The particle approaches at $v+V$ relative to the wall. It must separate at $v+V$ relative to the wall. Since the wall itself is moving at $V$, the particle's absolute speed becomes $(v+V) + V = v+2V$.

Q: Why is $dL = -V dt$?

$V$ is the speed of the piston moving inward, which means the length $L$ of the tube is decreasing. Therefore, the rate of change of length $dL/dt$ is negative, specifically $-V$.

Q: Can we solve this using work-energy theorem?

Yes! The force exerted by the particle on the piston is the rate of change of momentum. $F = \frac{\Delta p}{\Delta t} = \frac{2mv}{2L/v} = \frac{mv^2}{L}$. The work done by the piston is $-F dL$. Equating this to the change in kinetic energy $d(\frac{1}{2}mv^2)$ gives the same differential equation.

Q: What happens if the collision is inelastic?

If the collision were inelastic, the relative speed of separation would be $e$ times the relative speed of approach, where $e$ is the coefficient of restitution. The speed would not increase as much, and the $vL = \text{constant}$ relationship would not hold.

Question

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 $V \ll v_0$, where $dL$ is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct? [JEE(Advanced) 2019]

Options:

(A) The rate at which the particle strikes the piston is $v/L$

(B) After each collision with the piston, the particle speed increases by $2V$

(C) The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from $L_0$ to $\frac{1}{2}L_0$

(D) If the piston moves inward by $dL$, the particle speed increases by $2v\frac{dL}{L}$

Sigma Percentile · Accepted Answer

Imagine a particle trapped in a tube, bouncing back and forth between a fixed wall and a slowly moving piston. This is a classic setup that beautifully marries kinematics, collisions, and calculus. Let's break it down step by step.

### Analyzing the Setup

The particle travels a distance $L$ to the closed end and $L$ back to the piston. The total distance between two successive collisions with the piston is $2L$. Since the particle moves with speed $v$, the time taken for this round trip is $\Delta t = \frac{2L}{v}$.

The rate of collision, or frequency, is simply the reciprocal of this time period.

$$f = \frac{v}{2L}$$

This immediately tells us that Option A is incorrect, as it suggests the rate is $v/L$.

### The Elastic Collision

When the particle strikes the heavy piston, we must consider relative velocities. The piston moves inward with a very small speed $V$, and the particle approaches it with speed $v$.

The relative velocity of approach is $v + V$.

Because the collision is perfectly elastic and the piston is massive, the relative velocity of separation must equal the relative velocity of approach.

$$v_{	ext{sep}} = v_{	ext{app}} = v + V$$

If $v'$ is the particle's new speed after bouncing off, its velocity of separation from the piston is $v' - V$.

$$v' - V = v + V \implies v' = v + 2V$$

The particle's speed increases by exactly **$2V$** after each collision! This makes Option B correct.

### The Master Equation

Now, let's look at the macroscopic change over a small time interval $dt$. How many collisions occur in this time?

$$dn = f \cdot dt = \frac{v}{2L} dt$$

Since each collision increases the speed by $2V$, the total change in speed $dv$ is:

$$dv = dn 	imes 2V = \left(\frac{v}{2L} dtight) (2V) = \frac{v V}{L} dt$$

We know the piston is moving inward, so the length $L$ is decreasing. The infinitesimal change in length $dL$ is related to the piston's speed by $dL = -V dt$.

Substituting $V dt = -dL$ into our equation:

$$dv = \frac{v}{L} (-dL) = -v \frac{dL}{L}$$

This shows that Option D is incorrect, as it misses the negative sign and has an extra factor of 2.

### Final Calculation

To find the relationship between speed and length, we integrate our differential equation.

$$\frac{dv}{v} = -\frac{dL}{L}$$

Integrating from the initial state $(v_0, L_0)$ to the final state $(v, L)$:

$$\int_{v_0}^v \frac{dv}{v} = -\int_{L_0}^L \frac{dL}{L}$$

$$\ln\left(\frac{v}{v_0}ight) = -\ln\left(\frac{L}{L_0}ight) = \ln\left(\frac{L_0}{L}ight)$$

$$v = v_0 \frac{L_0}{L}$$

This elegant result tells us that the product **$vL$ is constant**!

The problem asks about the kinetic energy when the length is halved to $L = \frac{L_0}{2}$.

Using our relation, the new speed is:

$$v = v_0 \frac{L_0}{L_0/2} = 2v_0$$

The speed doubles! Since kinetic energy $K = \frac{1}{2}mv^2$, doubling the speed increases the kinetic energy by a factor of $2^2 = 4$.

$$K_{	ext{final}} = \frac{1}{2}m(2v_0)^2 = 4 \left(\frac{1}{2}mv_0^2ight) = 4 K_0$$

Option C is absolutely correct.

Select Answer:

Visualized Solution

$f = \frac{v}{2 L}$

$Δ v = 2 V$

$d v = \frac{v V}{L} d t$

$d v = - v \frac{d L}{L}$

$v L = v_{0} L_{0}$

$K_{final} = 4 K_{0}$

The Sigma Insight: Head-on Collision

Analyzing the Setup

The Elastic Collision

The Master Equation

Final Calculation

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