Question

A gas molecule of mass \(m\), moving with velocity \(\vec{v}=a\hat{i}+b\hat{j}+c\hat{k}\) hits a container wall that is parallel to the \(xz\)-plane. Assuming a perfectly elastic collision, what is the change in momentum of the molecule?

• A. \(-2ma\hat{i}\)
• B. \(-2mb\hat{j}\)
• C. \(-2mc\hat{k}\)
• D. zero

Hint:

Only the velocity component perpendicular to the wall reverses in a perfectly elastic collision. A wall parallel to the xz-plane has its normal along the y-axis.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a perfectly elastic collision with a fixed wall, the component of velocity along the wall is unchanged and the component along the normal reverses.

Step 2: Identify the normal
A wall parallel to the xz-plane has its normal along the y-axis, so only the y-component \(b\) reverses.

Step 3: Calculation
Initial momentum \(=m(a\hat{i}+b\hat{j}+c\hat{k})\). Final momentum \(=m(a\hat{i}-b\hat{j}+c\hat{k})\).

Step 4: Change in momentum
\(\Delta\vec{p}=-2mb\,\hat{j}\)

Final Answer:
The momentum changes by minus 2mb along the y-direction. \[ \boxed{\text{(B) }-2mb\hat{j}} \]