Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A wall is hit elastically and normally by ' $n$ ' balls per second. All the balls have the same mass ' $m$ ' and are moving with the same velocity ' $u$ '. The force exerted by the balls on the wall is
PhysicsCenter of Mass Momentum and CollisionMHT CETMHT CET 2022 (05 Aug Shift 1)
Options:
  • A $2 \mathrm{mnu}^2$
  • B $2 \mathrm{mnu}$
  • C $\frac{1}{2} \mathrm{mnu}^2$
  • D $\mathrm{mnu}$
Solution:
1849 Upvotes Verified Answer
The correct answer is: $2 \mathrm{mnu}$
The correct option is (B).
Force is related to momentum change by following relation,
$\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}$
Momentum change per second, $\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{n}(\mathrm{mu}-(-\mathrm{mu}))=2 \mathrm{nmu}$
Therefore,
$\mathrm{F}=2 \mathrm{nmu}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.