Question

In a cylinder a heavy piston is moving with speed $v$ as shown in the diagram and gas is filled inside it. A gas molecule is moving with speed $v_0$ towards the moving piston. Then which of the following is correct. (Assume $v<<<<v_0\frac{∆l}{l}$ and collision is elastic).

• A. Change in speed after collision is $2v$
• B. Change in speed after collision is $2v_0\frac{∆l}{l}$
• C. Rate of collision is $\frac{v}{I}$
• D. When piston is at $\frac{I}{2}$ its kinetic energy will be four times

Answer 1

The correct option is A

Change in speed after collision is $2v$

Step 1: Given

Speed of piston: $v$

Speed of gas molecule: $v_0$

Assume the velocity of the gas molecule after the collision: $v\text{'}$

Step 2: Formula used

Coefficient of restitution: $e=\frac{Velocityofseparation}{Velocityofapproach}$

Step 3: Find the change in speed after the collision

Find the velocity of separation by subtracting the velocity of the piston from the velocity of the gas molecule after the collision.

$Velocityofseparation=v\text{'}-v$

Find the velocity of the approach by adding the velocity of the piston and the velocity of the gas molecule before the collision.

$Velocityofapproach=v_0-v$

Since the collision is elastic, $e=1$. Substitute the values in the formula to find the new velocity of the gas molecule.

$$\begin{gathered} 1=\frac{v\text{'}-v}{v_0+v} \\ \Rightarrow v\text{'}-v=v_0+v \\ \Rightarrow v\text{'}=v_0+v+v \\ \Rightarrow v\text{'}=v_0+2v \end{gathered}$$

Find the change in velocity by subtracting the velocity of the gas molecule before collision from the velocity of the gas molecule after collision.

$\begin{array}{rcl}∆v& =& v\text{'}-v_0\\ & =& \left(v_0+2v\right)-v_0\\ & =& 2v\end{array}$

Hence, option A is correct.