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QJEEAdv19Ph6511: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph6511 Solution: A small particle of mass m moving inside a heavy, hollow and…

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.

JEE Advanced 2019 Paper 2 Offline Physics - Work Power & Energy Question 9 English

When the distance of the piston from closed end is L = L0, the particle speed is v = v0. The piston is moved inward at a very low speed V such that V < < {{dL} \over L}{v_0} , where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?

Multiple Choice Options

  • A After each collision with the piston, the particle speed increases by 2V.
  • B If the piston moved inward by dL, the particle speed increases 2vdLL2v{{dL} \over L} .
  • C The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to 1/2 L0.
  • D The rate at which the particle strikes the piston is v/L.

Step-by-Step Verified Solution

JEE Advanced 2019 Paper 2 Offline Physics - Work Power & Energy Question 9 English Explanation 1

Change in speed = (2v+v0−v0)=2v(2v + {v_0} - {v_0}) = 2v In every collision it acquires 2v,

JEE Advanced 2019 Paper 2 Offline Physics - Work Power & Energy Question 9 English Explanation 2

f=v2xf = {v \over {2x}}

dvdt=f×2v{{dv} \over {dt}} = f \times 2v

dv=v2x2vdtdv = {v \over {2x}}2vdt

dv=v2x2(−dx)dv = {v \over {2x}}2( - dx)

Integration on both sides limits v0 to v, we get

∫v0vdvv=∫lx−dxx\int\limits_{{v_0}}^v {{{dv} \over v}} = \int\limits_l^x {{{ - dx} \over x}}

⇒ln⁡vv0=−ln⁡xl\Rightarrow \ln {v \over {{v_0}}} = - \ln {x \over l}

⇒v=v0lx\Rightarrow v = {{{v_0}l} \over x}

where, x=l2,v=2v0x = {l \over 2},v = 2{v_0}

so, f=2v02l2=2v0lf = {{2{v_0}} \over {2{l \over 2}}} = {{2{v_0}} \over l}

∴\therefore kfki=4{{{k_f}} \over {{k_i}}} = 4

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