1. COLLISIONS We use the Conservation of Momentum Law for Collisions: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

2. COLLISIONS Limitations of the Conservation of Momentum External forces - forces caused by agents external to the "system“. The “system” is only mass 1 and mass 2. Examples 1) ‘external’ friction (the felt of a pool table) 2) air resistance 3) gravity

3. COLLISIONS Internal forces - forces that only act BETWEEN the objects in the system; such as the 'push' given by one skater to the other, the explosive force between the cannon and a cannon ball. These forces are actually necessary for the exchange of momentum.

4. COLLISIONS Significance? Internal forces occur between the masses; they act on both masses. They provide the impulse that changes the momentum. External forces occur between the individual mass and its surroundings. So the object loses its momentum to the surroundings and not to the other object.

5. COLLISIONS Two Types of Collisions: 1. Elastic (ideal case)2. Inelastic No Energy Is LostEnergy Is Lost Objects rebound:a) objects pass through one another b) objects stick together c) objects may rebound but too much Energy is lost (see next page) m1 m2 Nearly all ordinary collisions are inelastic

6. COLLISIONS 1. Objects pass through 2. Objects that stick “Perfect Inelastic” Friction on block of wood from bullet Friction on bullet from block of wood The friction that occurs between BOTH objects is an “internal” force. Internal forces are absolutely needed to exchange momentum, but energy is still lost. INELASTIC COLLISIONS

7. COLLISIONS Inelastic Analysis CASE 1: The objects STICK TOGETHER PERFECT inelastic collision Maximum loss of KE. If the two objects are stuck together at the end, what can we say about their final velocities?

8. COLLISIONS Conservation of Momentum Note that the final velocities are now the same The above equation is the PERFECT INELASTIC form of the Cons. Of Momen. We can alter the Conservation of Momentum Equation to account for that:

9. COLLISIONS Example: A 0.053 kg piece of clay is thrown at 21 m/s at a 1.3 kg wood cube sitting motionless on ice (frictionless). It hits the cube and sticks to it and they both slide on the ice together. Find their final velocity.

11. COLLISIONS

28. Example: A 0.053 kg piece of clay is thrown at 21 m/s at a 1.3 kg wood cube sitting motionless on ice (frictionless). It hits the cube and sticks to it and they both slide on the ice together. Find their final velocity.

29. COLLISIONS Example: A 0.053 kg piece of clay is thrown at 21 m/s at a 1.3 kg wood cube sitting motionless on ice (frictionless). It hits the cube and sticks to it and they both slide on the ice together. Find their final velocity.

30. COLLISIONS Example: A 0.053 kg piece of clay is thrown at 21 m/s at a 1.3 kg wood cube sitting motionless on ice (frictionless). It hits the cube and sticks to it and they both slide on the ice together. Find their final velocity.

31. COLLISIONS Inelastic Analysis CASE 2: The objects DON’T stick together Use the regular Conservation of Momentum Law.

32. COLLISIONS Example: A 0.123 kg arrow is moving at 42 m/s. It passes through a 5.2 kg piece of balsa wood sitting motionless on a frictionless surface and the arrow emerges moving at 33 m/s. How fast is the piece of balsa wood moving?

34. COLLISIONS

42. Example: A 0.123 kg arrow is moving at 42 m/s. It passes through a 5.2 kg piece of balsa wood sitting motionless on a frictionless surface and the arrow emerges moving at 33 m/s. How fast is the piece of balsa wood moving?

43. COLLISIONS Example: A 0.123 kg arrow is moving at 42 m/s. It passes through a 5.2 kg piece of balsa wood sitting motionless on a frictionless surface and the arrow emerges moving at 33 m/s. How fast is the piece of balsa wood moving?

44. COLLISIONS Elastic Analysis: For a “perfect” elastic collision NO ENERGY IS LOST

45. COLLISIONS That means: With the Conservation of Momentum: We have two equations and can solve for two unknowns!