A 450g rubber ball is dropped from 2.0m above the floor. If the ball bounces back up to a height of 1.5m.

A 450g rubber ball is dropped from 2.0m above the floor. If the ball bounces back up to a height of 1.5m.
A. Determine the change in momentum.
B. Calculate the amount of heat transferred to the floor.
C. Explain how momentum is conserved

First find initial velocity by finding time. y=y(0)-1/2gt^2
y=0
0=2-1/2(9.8)t^2
2.9=t^2
t=1.7 seconds

v=v(0)+at v(0)=0
v=at
v=9.8(1.7)
v=16.66m/s

Now find final velocity.

1.5=0-1/2(9.8)t^2
6.4=t^2
t=2.52 sec

v=9.8(2.52)

v=24.79 m/s

momentum (p) is equal to mass * velocity

so change in p would be mv(2)-mv(1)

so it .45kg(24.79-16.66)= change in p=3.66

B. Heat will be the change in kinetic energy.

1/2mv^2(2)-1/2mv^2(1)= change of KE

1/2(.45kg)(24.79m/s)^2-1/2(.45kg)(16.66… change in KE

1/2(.45)(8.13^2)=14.87 J

C. Momentum is conserved in such that in the equation mv(1) + mv(2)=mv'(1) + mv'(2).

There can be inelastic collision, which is when there two objects that collide and combine, therefore having a combined velocity.

Then there is elastic collision, in which there is a "bounce" when the objects collide. Each of them now having different velocities.

In a perfect elastic collision, Kinetic energy(KE) is conserved. However, this is untrue most of the time. KE is lost due to sound, heat, etc.

I find this question very interesting, i'll keep here waiting for the answer.