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 mov

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 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 $V ≪ \frac{d L}{L} v_{0}$ , where $d L$ is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct [2019]

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(1, 4)

When the small particle moving with velocity $v_{0}$ undergoes an elastic collision with the heavy movable piston moving with velocity $v$ , it acquires a new velocity $v_{0} + 2 v$ . So, the increase in velocity after every collision is $2 v$ .

Time period of collision when the piston is at a distance $' L'$ from the closed end

$T = \frac{distance}{speed} = \frac{2 L}{v'}$

Where $v'$ is the speed of the particle at that time.

$∴$ Frequency or rate at which the particle strikes the piston $= \frac{v'}{2 L}$

The rate of change of speed of the particle

$= \frac{d v'}{d t} = (frequency) \times 2 v$ $∴$ $d v' = \frac{v'}{2 L} 2 v d t$

$∴$ $\frac{d v'}{v'} = \frac{v d t}{L} = \frac{- d L}{L}$

Where $d L$ is the distance travelled by the piston in time dt. The minus sign indicates decrease in $' L'$ with time.

$∴$ $\int_{V_{0}}^{V} \frac{d v'}{v'} = - \int_{L_{0}}^{x} \frac{d L}{L}$

$∴$ $l n (\frac{v'}{v_{0}}) = - l n (\frac{L}{L_{0}}) or | v' | = \frac{v_{0} L_{0}}{L}$

$When L = \frac{L_{0}}{2}, we have | v' | = \frac{v_{0} L_{0}}{L_{0} / 2} = 2 V_{0}$

$∴$ ${K.E}_{L_{0} / 2} = \frac{1}{2} m {(2 v_{0})}^{2}$

$∴$ $K . E_{L_{0}} = \frac{1}{2} m v_{0}^{2}$

$∴$ $\frac{{K.E}_{L_{0} / 2}}{{K.E}_{L_{0}}} = 4$

Solution

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

2 If the piston moves inward by $d L$ , the particle speed increases by $2 v \frac{d L}{L}$

3 The rate at which the particle strikes the piston is $\frac{v}{L}$

4 After each collision with the piston, the particle speed increases by $2 V$

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Solution

1 The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to 12L0

2 If the piston moves inward by dL, the particle speed increases by 2vdLL

3 The rate at which the particle strikes the piston is vL

4 After each collision with the piston, the particle speed increases by 2V

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

2 If the piston moves inward by $d L$ , the particle speed increases by $2 v \frac{d L}{L}$

3 The rate at which the particle strikes the piston is $\frac{v}{L}$

4 After each collision with the piston, the particle speed increases by $2 V$