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A ball falls vertically downward and bounces off a horizontal floor.The speed of the ball just before reaching the floor $\left(u_1\right)$ is equal to the speed just after leaving contact with the floor $\left(u_2\right), u_1=u_2$. 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
$a_1 < a_2$
$a_1 > a_2$
$a_1=a_2 \neq g$
$a_1=a_2=g$
Solution
(a)
Air resistance is same in both case. When ball is moving down, air resistance is directed away from $g$.
So, acceleration of ball moving downwards is
$a_1=\frac{m g-k v}{m}$
or $a_1=\left(g-\frac{k}{m} v\right)$
where, $k$ is constant.
When ball is moving up, air resistance and $g$ both are directed downwards.
So,acceleration while moving upwards is
$a_2=g+\frac{k}{m} v$
Clearly, $a_2 > a_1 .$
