1. In almost all of the problems we will consider, the total momentum will be conserved
The total mechanical energy may or may not be conserved
Two kinds of collisions: Elastic – energy conserved (special case) 2. Inelastic – energy not conserved (general)
Example – Problem 7.38
A ball is dropped from rest at the top of a 6.10-m tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses 10.0% of its kinetic energy every

2. time it collides with the ground
time it collides with the ground. How many bounces can the ball make and still reach a window sill that is 2.44 m above the ground?
Solution:
Method: since the ball bounces on the ground, there is an external force. Therefore, we can not use conservation of linear momentum.
An inelastic collision means the total energy is not conserved, but we know by how much is not conserved. On every bounce 10% of the KE is lost:
Given: h0 = 6.10 m, hf = 2.44 m

3. Since energy is conserved from point 0 to point 1. 3 6
Since energy is conserved from point 0 to point 1.
However, between point 1 and 2, energy is lost 1 2 4 5
Total energy after one bounce

4. By the same reasoning
Total energy after two bounces
The total energy after n bounces is then
Answer is 8 bounces

5. Center of Mass
The average location for the total mass of a collection of particles (system)
Consider two particles located along a straight line
If m1=m2=m x1
xcm m2 x m1  cm x2

6. Also can define the center of mass for multiple particles - N number of particles
Where M is the total mass of the system
Can also define the center of mass along other coordinate axes - ycm

7. Can also define the velocity of the center of mass
If linear momentum is conserved, then
The velocity of the center of mass is constant. It is the same before and after a collision.

8. Return to earlier Example
m1 = 45 kg, m2 = 12 kg, vo1 = +5.1 m/s, vo2 = 0
We found vf1 = m/s and vf2 = m/s
Now, compute the velocity of the center of mass
Taking x1=0 and x=x2-x1=x2