1. Section 11.2 Conservation of Energy
 Objectives
Solve problems using the Law of Conservation of Energy.
Analyze collisions to find the change in Kinetic Energy.

2. INTRO/CONSERVATION OF ENERGY
As the height of the ball changes, energy is converted from Kinetic Energy to Potential Energy, but the total amount of energy stays the same.
The key to understanding and using the constancy of energy is in selecting the system (collection of object we want to study).
Similar to Momentum, with energy we need a Closed, Isolated System.
Closed Isolated System – collection of objects such that neither matter nor energy can enter or leave the collection.
Closed System – system that does not gain or lose mass (objects neither enter nor leave it).
Isolated System – Closed System on which the net external force is zero (no net external force is exerted on it).

3. CONSERVATION OF ENERGY
Law of Conservation of Energy – states that in a closed isolated system, energy can change form, but the total energy does not change (it is constant). Energy cannot be created or destroyed.
Mechanical Energy – is equal to the sum of kinetic and potential energy ME = KE + PE
The change in Kinetic Energy can be found by the Work Energy Theorem W = KE = KEf – KEi
The decrease in potential energy equals the increase in kinetic energy. Thus the sum of potential and kinetic energy does not change and therefore Mechanical Energy is Constant.

4. CONSERVATION OF ENERGY
Go Over example p
Conservation of Mechanical Energy – when mechanical energy is conserved, the sum of the KE and PE present in the system before the event is equal to the sum of the KE and PE after the event.
KEi + PEi = KEf + PEf
If a system is not isolated, the mechanical energy will not be conserved (stay the same).

5. CONSERVATION OF ENERGY
Do Example 2 p. 296
a) PE = mgh b) KE = ½ mv2
PE = 22(9.8)(13.3 – 6) = ½ (22)v2
PE = J = 11v2
Thus = v2
KE = Joules m/s = v
Do Practice Problems p. 297 # 15-18

6. ANALYZING COLLISIONS
The strategy is to find the motion of the objects just before and just after the collision.
Elastic Collision – a type of collision in which the Kinetic Energy before and after the collision remains the same.
Note: with 2 unknowns, 2 equations are needed to solve the problem.
Inelastic Collision – collision in which the Kinetic Energy after the collision is less than the kinetic energy before the collision.

7. ANALYZING COLLISIONS Do Example 3 p. 299
A) mavai + mbvbi = mavaf + mbvbf
575(15) (5) = 575v v
= 2150v
16500 = 2150v
7.674 m/s = v
B) ΔKE = KEF – KEI
ΔKE = ½ ( )(7.674)2 – [½(575)(15)2 + ½(1575)(5)2 ]
ΔKE = ½ (2150)(58.89) – [½(575)(225) + ½(1575)(25) ]
ΔKE = – [ ]
ΔKE = – 84375
ΔKE = -21, J
C) Skip / = = %

8. ANALYZING COLLISIONS Do Practice Problems p. 300 # 19-21
Read p
Do 11.2 Section Review p. 301 # 22-28