Question

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_ the particle speed is v = v_. The piston is moved inward at a very low speed v such that V << dL/L v_,where dL is the infinitesimal displacement of the piston. which of the following statement(s) is/are correct ?
A. The particleÕs kinetic energy increases by a factor of 4 when the piston is moved inward from L_ to 1/2 L_
B. After each collision with the piston, the particle speed increases by 2V.
C. If the piston moves inward by dL, the particle speed increases by 2v dL/L
D. The rate at which the particle strikes the piston is v/L

Answer 1

A). The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L_ to 1/2 $L_{0}$

B). After each collision with the piston, the particle speed increases by 2V.

Explanation:

Given that,

The piston strike the particle at the rate:

f = v/2x

If x is equal to L, then f = v/2L

The rate at which the speed of the particle changes:

dv/dt = f * 2V

Since, dv = $(\frac{v}{2x})^{2Vdt}$

dv = $(\frac{v}{2x}) ^{2(-dx)}$

∵ v = $\frac{v0l}{x}$

If x = $L_{0} /2$, then

v = $(v_{0} L_{0} * 2)/ L_{0}$ = $2v_{0}$

Thus, Kinetic Energy at x = L_{0} /2 = $k_{f}$ = 1/2 * m * $4v^{2}_{0}$

∵ $k_{f}/k_{i}$ = 4

Learn more: Kinetic energy

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