What impulse was given to the ball by the floor?

Question

A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

Answer 1

It seems that we need only the conditions after the impact.

I= F dt= m( dv/ dt)dt=m dv
Where
F – force the floor pushes the ball
dt –time durations the force was applied
dv/dt – acceleration (or change of momentum)
dv – change in velocity during impact

dv= V2-V1=-V1 (since at .6m V2=0)
Kinetic energy at the floor level equal to potential energy at .6m
Pe=mgh
Ke=(mv^2)/2
Then V1=v=sqrt(2gh)

So I=mV1=msqrt(2gh)=
I=.12 sqrt(2x9.81x .6)=
I=.412 kg m/sec

Answer 2

the rate whilst it hits the floor v1 = ?(2gh1) = ?(2(9.8)(a million.25)) = 4.ninety 5 m/s the rate with which it rebounds v2 = ?(2gh2) = ?(2(9.8)(0.ninety six)) = ? 4.34 m/s.. The ? ve sign is used via fact if v1 is +ve, v2 is ? ve. substitute in momentum = very final momentum? preliminary momentum. = m [v2 - v1] = 0.one hundred fifty * [? 4.34?4.ninety 5] = ? a million.3935 Ns or kgm/s ? ve sign shows that the impulse is opposite to the action.