Q.
A mass of 100 g strikes the wall with speed $5\ {ms}^{- 1}$ at an angle as shown in figure and it rebounds with the same speed. If the contact time is $2 \times 10^{- 3}$ s, what is the force applied?
A) $250\sqrt{3}\ N\ to\ right$
B) 250 N to right
C) $250\sqrt{3}\ N\ to\ left$
D) 250 N to left
Explanation
(a) [image:5_image190.jpeg]
Change in the velocity$\ = v\sin{\theta - ( - v\sin{\theta) = 2\sin\theta}}$
Change in the momentum
$\mathrm{\Delta}p = 2\ m\ v\sin\theta$
$\therefore\ \ \ Force\ applied\ F = \frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}$
$= \frac{2 \times 100 \times 10^{- 3} \times 5\sin{\theta\ 60{^\circ}}}{2 \times 10^{- 3}}$
$= 100 \times 5 \times \frac{\sqrt{3}}{2}$
$= 250\sqrt{3}$ N (To the right)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.