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.

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).

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.

CONSERVATION OF ENERGY Go Over example p. 293-294   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).

CONSERVATION OF ENERGY Do Example 2 p. 296 a) PE = mgh b) KE = ½ mv2 PE = 22(9.8)(13.3 – 6) 1573.88 = ½ (22)v2 PE = 1573.88 J 1573.88 = 11v2 Thus 143.08 = v2 KE = 1573.88 Joules 11.96 m/s = v   Do Practice Problems p. 297 # 15-18

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.

ANALYZING COLLISIONS Do Example 3 p. 299 A) mavai + mbvbi = mavaf + mbvbf 575(15) + 1575(5) = 575v + 1575v 8625 + 7875 = 2150v 16500 = 2150v 7.674 m/s = v B) ΔKE = KEF – KEI ΔKE = ½ (575 + 1575)(7.674)2 – [½(575)(15)2 + ½(1575)(5)2 ] ΔKE = ½ (2150)(58.89) – [½(575)(225) + ½(1575)(25) ] ΔKE = 63306.75 – [64687.5 + 19687.5] ΔKE = 63306.75 – 84375 ΔKE = -21,068.25 J C) Skip -21068.25 / 84375 = .2497 = 24.97%

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