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 ?

detailed solution

Correct option is B

Let say after collision velocity of particle becomes v′ and piston keeps moving with same speed V Velocity of separation = Velocity of approach ⇒v′−V=v+V⇒v′=v+2V Thus after every collision with piston, speed of particle increases by 2V .  Time between two successive collision =T= Total dist to and fro  velocity =2Lv Rate of collision =f=1T=V2L Let say piston move by distance dL in time dt⇒dL=V×dt⇒dt=dLV In this dt time, No of collision occured with particle =n=dt×f=dLV×v2L After these n collisions, the increament in speed of particle =dv=n(2V)=dLV×V2L×2V⇒dv=dLL×v Since, L is decresing, we will put a negative sign, ⇒∫V0v dvV=−∫L0L0/2 dLL⇒v=2v0 Since, KE∝v2 ,  KE becomes 4 times.