A ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor (u1) is equal to the speed just after leaving contact with the floor (u2), u1=u2. The corresponding magnitudes of accelerations are denoted respectively by a1 and a2. The air resistance during motion is proportional to speed and is not negligible. If g is acceleration due to gravity, then

Question

A ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor (u1) is equal to the speed just after leaving contact with the floor (u2), u1=u2. The corresponding magnitudes of accelerations are denoted respectively by a1 and a2. The air resistance during motion is proportional to speed and is not negligible. If g is acceleration due to gravity, then (A) a1<a2 (B) a1>a2 (C) a1=a2 ≠ g (D) a1=a2=g

Tardigrade Admin · Accepted Answer

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 a1=(m g-k v/m) a1=(g-(k/m) v) where, k is constant. When ball is moving up, air resistance and g both are directed downwards.

So, acceleration while moving upwards is a2=g+(k/m) v Clearly, a2>a1 .

$a_{1} < a_{2}$

$a_{1} > a_{2}$

$a_{1} = a_{2} \neq = g$

$a_{1} = a_{2} = g$

a1​<a2​

a1​>a2​

a1​=a2​=g

a1​=a2​=g

$a_{1} < a_{2}$

$a_{1} > a_{2}$

$a_{1} = a_{2} \neq = g$

$a_{1} = a_{2} = g$