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

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 $V << \frac{dL}{L}v_{0}$, where $dL$. is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?

  1. The rate at which the particle strikes the piston is vL
  2. After each collision with the piston, the particle speed increases by 2V
  3. The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to 12L0
  4. If the piston moves inward by dL, the particle speed increases by 2vdLL

Solution


Therefore change in speed =2V+V0-V0
=2V
In every collision it acquires 2V⇒B is correct

Now, frequency of collision of particle with piston,
⇒f=V2x (as it has to travel “2x” distance with speed “V” ) ⇒A is incorrect
Acceleration ⇒dvdt=f×2V
⇒dv=V2x2 Vdt
dv=V2x2-dx as Vdt=-dx
∫V0Vdvv=∫lx-dxx
⇒ln⁡VV0=-ln⁡xl
⇒V=V0lx
where x=l2, V=2V0
∴Ki=12mV02 and Kf=12m2V02=4.Ki
⇒KfKi=4⇒C is correct.

Asked in: JEE Advanced 2019 (Paper 2)

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