Q. A $100\ g$ iron ball having velocity $10\ m/s$ collides with a wall at an angle $30{^\circ}$ and rebounds with the same angle. If the period of contact between the ball and wall is $0.1\ second$ , then the force experienced by the wall is
A) $10\ N$
B) $100\ N$
C) $1.0\ N$
D) $0.1\ N$
Explanation
(a) During collision of ball with the wall horizontal momentum changes (vertical momentum remains constant)
$\therefore F = \frac{Change\ in\ horizontal\ momentum}{Time\ of\ contact}$
$= \frac{2P\cos\theta}{0.1} = \frac{2mv\cos\theta}{0.1}$
$= \frac{2 \times 0.1 \times 10 \times \cos{60{^\circ}}}{0.1} = 10\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.