Question

Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces to a height h//4, after the collision & the second ball to a height h//6. The impulse applied by the first & second ball on the floor are I_(1) and I_(2) respectively. Then :-

Answer 1

Answer:

using second equation of motion

for first ball:

velocity at the instant of striking the floor v1=$\sqrt{2gh}$

and velocity after striking v2 =$\sqrt{gh/2}$

for second ball:

velocity at the instant of striking the floor v3=$\sqrt{2gh}$

and velocity after striking v2 =$\sqrt{gh/3}$

now by impulse momentum theorem impulse= change in momentum

$l_1=m\sqrt{gh} (\sqrt{2} -1/\sqrt{2} )$=$m\sqrt{gh} /\sqrt{2}$

$l_2=m\sqrt{gh} (\sqrt{2} -1/\sqrt{3} )=m\sqrt{gh} (\sqrt{6} -1)/\sqrt{3}$