JEE Challenger
More from Laws of Motion
PhyHardAdv 2019 • Paper 2

Particle Collisions in a Tube with Moving Piston

Consider a small particle of mass mm moving inside a heavy, hollow, straight, frictionless tube along its axis. It undergoes elastic collisions at both ends. One end is permanently closed by a flat surface, while the other end is fitted with a heavy movable flat piston as depicted in the figure. Initially, when the separation between the piston and the closed end is L=L0L = L_0, the speed of the particle is v=v0v = v_0. The piston is then shifted inward at a very small constant velocity VV obeying V≪dLLv0V \ll \frac{dL}{L} v_0, where dLdL represents the infinitesimal displacement of the piston. Select the correct statement(s):

Question Diagram 1

Options

A

The rate at which the particle hits the piston is given by v/Lv/L

B

Following every collision with the piston, the particle's speed increases by 2V2V

Correct
C

When the piston advances inward by an infinitesimal distance dLdL, the particle speed increases by 2vdLL2v \frac{dL}{L}

D

The kinetic energy of the particle increases by a factor of 4 as the piston is pushed inward from L0L_0 to 12L0\frac{1}{2} L_0

Correct

Topics & Concepts

Laws of MotionCollisionKinematicsNewton's Laws

Step-by-Step Solution

To determine the correct statement(s), let us analyze each option individually:

Analysis of Option (A):

Between two consecutive collisions with the moving piston, the particle travels to the fixed closed end (a distance LL) and returns to the piston (a distance LL). The total distance traveled by the particle between two successive collisions with the piston is 2L2L.

The time interval Δt\Delta t between consecutive collisions with the piston is: Δt=2Lv\Delta t = \frac{2L}{v}

Therefore, the rate (frequency) at which the particle strikes the piston is: f=1Δt=v2Lf = \frac{1}{\Delta t} = \frac{v}{2L}

Hence, Option (A) is incorrect.

Analysis of Option (B):

Consider a collision between the particle of mass mm moving at speed vv and the heavy piston moving inward at speed VV.

  • In the reference frame of the heavy piston, the velocity of the particle before collision is v+Vv + V towards the piston.
  • Since the collision is elastic and the piston is heavy (M≫mM \gg m), the particle rebounds with the same relative speed v+Vv + V relative to the piston.
  • In the laboratory frame, the speed of the particle after the collision is: v′=(v+V)+V=v+2Vv' = (v + V) + V = v + 2V

The increase in the speed of the particle after a single collision with the piston is: Δv=v′−v=2V\Delta v = v' - v = 2V

Hence, Option (B) is correct.

Analysis of Option (C):

Let the piston move inward by an infinitesimal displacement dLdL in time dt=dLVdt = \frac{dL}{V}.

The number of collisions dNdN the particle undergoes with the piston during this time dtdt is: dN=dtΔt=dL/V2L/v=v dL2VLdN = \frac{dt}{\Delta t} = \frac{dL / V}{2L / v} = \frac{v \, dL}{2 V L}

Since the particle's speed increases by 2V2V during each collision, the total change in speed dvdv as the piston moves inward by dLdL is: dv=dN×(2V)=(v dL2VL)(2V)=vdLLdv = dN \times (2V) = \left( \frac{v \, dL}{2 V L} \right) (2V) = v \frac{dL}{L}

Thus, the speed increases by vdLLv \frac{dL}{L} (and not 2vdLL2v \frac{dL}{L}).

Hence, Option (C) is incorrect.

Analysis of Option (D):

From the differential relation derived in Option (C), where dLdL represents the reduction in length LL: dvv=−dLL\frac{dv}{v} = -\frac{dL}{L}

Integrating both sides from the initial state (L0,v0)(L_0, v_0) to the final state (L02,v)\left(\frac{L_0}{2}, v\right): ∫v0vdvv=−∫L0L0/2dLL\int_{v_0}^{v} \frac{dv}{v} = -\int_{L_0}^{L_0/2} \frac{dL}{L}

ln⁡(vv0)=−ln⁡(12)=ln⁡(2)\ln\left(\frac{v}{v_0}\right) = -\ln\left(\frac{1}{2}\right) = \ln(2)

v=2v0v = 2v_0

The final kinetic energy KfK_f compared to the initial kinetic energy KiK_i is: Ki=12mv02K_i = \frac{1}{2} m v_0^2 Kf=12mv2=12m(2v0)2=4(12mv02)=4KiK_f = \frac{1}{2} m v^2 = \frac{1}{2} m (2v_0)^2 = 4 \left( \frac{1}{2} m v_0^2 \right) = 4 K_i

Thus, the kinetic energy increases by a factor of 44.

Hence, Option (D) is correct.

Conclusion:

The correct statements are B and D.

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