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 and the second ball to the height h/16 the impulse applied on the floor are i1 and i2 then relation between i1 and i2

Answer 1

The ball dropped ghastly on floor

Answer 2

Hey dear,

● Answer-
I1/I2 = 2/3

● Explaination-
# Given-
h1 = h/4
h2 = h/16

# Solution-
Let v be velocity of both balls when dropped from height h.
v = √(2gh)

Assume v1 and v2 be respective velocities of two balls on 2nd collision.

For first ball,
v1 = √(2gh1)
v1 = √(2gh/4)
v1 = 1/2 √(2gh)
v1 = v/2

Impulse of first ball,
I1 = m∆v = m(v-v1)
I1 = m(v-v/2)
I1 = mv/2

For second ball,
v2 = √(2gh2)
v2 = √(2gh/16)
v2 = 1/4 √(2gh)
v2 = v/4

Impulse of second ball,
I2 = m∆v = m(v-v2)
I2 = m(v-v/4)
I2 = 3mv/4

Ratio of I1/I2,
I1/I2 = (mv/2) /(3mv/4)
I1/I2 = 2/3

Hope it helps...