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Question: Answered & Verified by Expert
A ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor $\left(\mathrm{u}_{1}\right)$ is equal to the speed just after having contact with the floor $\left(\mathrm{u}_{2}\right) ; \mathrm{u}_{1}=\mathrm{u}_{2} .$ The corresponding magnitudes of accelerations are denoted respectively by $a_{1}$ and $a_{2} .$ The air resistance during motion is proportional to speed and is not negligible. If $g$ is acceleration due to gravity, then-
PhysicsMotion In Two DimensionsKVPYKVPY 2012 (SA)
Options:
  • A $a_{1} < a_{2}$
  • B $a_{1}>a_{2}$
  • C $\mathrm{a}_{1}=\mathrm{a}_{2} \neq \mathrm{g}$
  • D $\mathrm{a}_{1}=\mathrm{a}_{2}=\mathrm{g}$
Solution:
2164 Upvotes Verified Answer
The correct answer is: $a_{1} < a_{2}$


Downward- $a_{1}=g-\frac{F_{R}}{m}$
Upward:- $\quad a_{2}=g+\frac{F_{R}}{m}$
$\mathrm{a}_{2} < \mathrm{a}_{1}$

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