Question

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. $a_{1}
• B. $a_{1}>a_{2}$
• C. $a_{1}=a_{2} \neq g$
• D. $a_{1}=a_{2}=g$

Answer 1

Answer: (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} $
$a_{1}=\left(g-\frac{k}{m} v\right)$
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} .$