Q. A 1.5 kg ball drops vertically on a floor hitting with a speed of $\ 25{ms}^{- 1}$ . It rebounds with an initial speed of$\ 15{\ ms}^{- 1}$. If the ball was in contact for only 0.03 seconds, the force exerted on the floor by the ball is
A) 2000 N
B) 3000 N
C) 3500 N
D) 4000 N
Explanation
(a) Mass of ball $m = 1.5$ kg
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Speed of ball at the time of hitting,
$u = 25\ {ms}^{- 1}$
Speed of ball while rebounding
$v = 15\ {ms}^{- 1}$
Duration of contact with floor $t = 0.03\ s$
Let force exerted by the ball on floor $= F$
Applying Newton’s II law of motion
$F = \frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}$
$F = \frac{mv - mu}{t} = \frac{1.5 \times 15 \times ( - 25)}{0.03}$
$= \frac{1.5(15 + 25)}{0.03} = 50 \times 40N$
$F = 2000$ N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.