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 and the second ball to a height h/16 .The impulse applied by the first and second ball on the floor are I1 and I2 respec, then relation bet I1 and I2 is
Asked by m.nilu
17th September 2018,
8:27 PM
Answered by Expert
Answer:
speed v attained by the first ball when it is dropped from height h is given by, v = 
if first ball reaches a height h/4 after bouncing, speed v1' after bouncing is given by, v1' = 
impulse l1 given by the first ball to floor = m( v - v1' ) = (1/2)v .......................(1)
similarly if v2' is the speed of second ball after bouncing, v2' is given by = 
impulse l2 given by the second ball to floor = m( v - v2' ) = (3/4)v .......................(2)
from (1) and (2), we get the required ratio, l1/l2 = 2/3
Answered by Expert
18th September 2018,
2:43 PM
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