JEE Advanced2019PhysicsWork, Power and EnergyActual
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 l
Options
- AThe rate at which the particle strikes the piston is v L
- BAfter each collision with the piston, the particle speed increases by 2 V
- CThe particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L 0 to 1 2 L 0
- DIf the piston moves inward by d L , the particle speed increases by 2 v d L L
Correct answer
B. After each collision with the piston, the particle speed increases by 2 V
Step-by-step solution
Therefore change in speed = 2 V + V 0 - V 0 = 2 V In every collision it acquires 2 V ⇒ B is correct Now, frequency of collision of particle with piston, ⇒ f = V 2 x (as it has to travel “ 2 x ” distance with speed “ V ” ) ⇒ A is incorrect Acceleration ⇒ d v d t = f × 2 V ⇒ d v = V 2 x 2 V d t d v = V 2 x 2 - d x a s V d t = - d x ∫ V 0 V d v v = ∫ l x - d x x ⇒ l n V V 0 = - l n x l ⇒ V = V 0 l x where x = l 2 , V = 2 V 0 ∴ K i = 1 2 m V 0 2 and K f = 1 2 m 2 V 0 2 = 4 . K i ⇒ K f K i = 4 ⇒ C is correct.