Question

A molecule of mass $m$ moving with velocity $v$ makes 5 elastic collisions with a wall of container per second. The change in momentum of the wall per second in 5 collisions will be

• A. $10\ mv$
• B. $5\ mv$
• C. $15\ mv$
• D. $\frac{1}{10}\ mv$

Hint:

Always remember that momentum is a vector quantity! A common trap is calculating the change in speed as $v - v = 0$, or forgetting the factor of 2 and choosing $5\ mv$ (option B). Because the direction reverses, the magnitude of the change per collision is always doubled ($2mv$).

Answer 1

The Correct Option is A

Step 1: Understand the question.
A molecule of mass $m$ and speed $v$ makes 5 elastic collisions with a wall each second. We find the total change in momentum of the wall per second.
Step 2: Find the momentum change in one collision.
In an elastic collision with a wall, the molecule bounces straight back with the same speed. Its velocity goes from $+v$ to $-v$, so its momentum changes by \[ \Delta p = mv - (-mv) = 2mv. \]
Step 3: Apply Newton's third law to the wall.
By Newton's third law, the wall gains an equal and opposite momentum of $2mv$ in each collision.
Step 4: Count the collisions per second.
There are 5 collisions every second.
Step 5: Add up over all collisions.
Total change per second \[ = 5\times 2mv = 10\,mv. \]
Step 6: State the answer.
The change in momentum of the wall per second is $10\,mv$. \[ \boxed{10\,mv} \]