1. Linear Momentum

2. Linear momentum describes motion in which the center of mass of an object or system changes position. We call motion where the c.o.m. changes position linear or translational motion.

3. What is center of mass (sometimes called center of gravity)? Center of mass is the point in any object around which the mass of the object is evenly distributed. It’s located somewhere along a vertical line through the “balance point”. Translational motion is a change in position of the c.o.m. of an object. (Rotation occurs when the object moves---but the c.o.m. doesn’t go anywhere!)

4. On the other hand, an object can be moving and not have linear momentum. An example would be an object that is spinning in position. We call this rotational or angular momentum (which we will study in a later chapter).

5. Momentum is a vector quantity, with momentum and velocity in the same direction: p = mv The units for momentum would be kg  m/s.

6. If there is no outside force to affect the momentum, momentum remains constant. That means that the net (total) momentum before an interaction (such as a collision or explosion) is equal to the net (total) momentum afterward if no external force is applied to the system:  p o =  p f  Just means “add all the quantities or vectors”.

7. What is an interaction?

8. A collision of two astronomical objects…. Physicists study the momenta of objects after such interactions to determine what must have been happening before the interactions.

9. A collision of two sub-atomic particles…. Examination of the momenta of the particles and jets after such proton-proton collisions at the Large Hadron Collider in Cern, Switzerland, may lead us to information about dark matter.

10. A collision of two automobiles… Engineers can study the positions of cars after a collision to determine their speeds before the collision---using conservation of momentum principles, of course.

11. An explosion of a supernova…

12. Could this be the result of a perfectly inelastic collision of hot wings with blue cheese…? Possibly not

13. If the momentum before an interaction is zero, then (assuming no interfering force, such as friction or gravitational force) the momentum afterward is equal to zero. In this case, the momentum before the interaction with the fire extinguisher is zero. After the extinguisher is fired, the momentum of gases backward is equal to the momentum of the wagon forward. http://www.youtube.com/watch?v=IlMtJW7reKg

14. Since momentum is a vector, it must have a direction. We use positive and negative signs to denote directions in one- dimensional motion. So if momentum to the right is positive, momentum to the left is negative…..or if momentum to the east is positive, momentum to the west is negative…..etc.

15. Example 1: A cart with mass 1.5 kg moving at 4.0 m/s to the right collides head-on with a cart with mass 3.0 kg moving at 4.0 m/s to the left. After the collision, the 1.5 kg cart is moving to the left at 1.0 m/s. What is the 3.0 kg cart doing after the collision? ?

16. Example 1: A cart with mass 1.5 kg moving at 4.0 m/s to the right collides head-on with a cart with mass 3.0 kg moving at 4.0 m/s to the left. After the collision, the 1.5 kg cart is moving to the left at 1.0 m/s. What is the 3.0 kg cart doing? Solution:  p o =  p f m 1o v 1o + m 2o v 2o = m 1f v 1f + m 2f v 2 (1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(-1.0 m/s) + (3.0 kg)(v) v = -1.5 m/s (Since the answer is -, this cart is also moving to the left after the collision. We will demonstrate examples of such phenomena on the air track in the laboratory.)

18. For most collisions without an intervening force, momentum is conserved but some of the kinetic energy of the colliding objects is “lost”. Where do this energy go? The “lost” kinetic energy is transferred to thermal energy of molecules and to the sound you hear from the collision (which is actually also a form of thermal energy of molecules).

19. In a totally inelastic collision, the objects stick together during the collision and move as one object after the collision.

20. Example 2: A cart with mass 1.5 kg moving at 2.0 m/s to the right collides head-on with a cart with mass 3.0 kg moving at 3.0 m/s to the left. Their bumpers lock. What are they doing after the collision?

21. Example 2: A cart with mass 1.5 kg moving at 2.0 m/s to the right collides head-on with a cart with mass 3.0 kg moving at 3.0 m/s to the left. Their bumpers lock. What are they doing after the collision? Solution:  p o =  p f m 1o v 1o + m 2o v 2o = (m 1 + m 2 )v (1.5 kg)(2.0 m/s) + (3.0 kg)(-3.0 m/s) = (4.5 kg) v v = -1.3 m/s They move as one object to the left.

23. If a collision is elastic---and only if it is elastic---the kinetic energy is also conserved. In this case, we can write both a momentum equation and a kinetic energy equation. (This means that no energy was lost during the collision…..pretty rare…..probably only happens during particle collisions.)

24. Example 3: A cart with mass 1.5 kg moving at 4.0 m/s to the right collides head-on and elastically with a cart with mass 3.0 kg moving at 4.0 m/s to the left. After the collision, what are the carts doing? ??

25. Example 2: A cart with mass 1.5 kg moving at 4.0 m/s to the right collides head-on and elastically with a cart with mass 3.0 kg moving at 4.0 m/s to the left. After the collision, what are the carts doing? Solution: Since we don’t know the velocity of either cart after the collision, there are two unknowns—so we’ll need two equations. Since the collision is elastic, we have two equations---conservation of momentum and conservation of kinetic energy.  p o =  p f m 1o v 1o + m 2o v 2o = m 1f v 1f + m 2f v 2 (1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(v 1 ) + (3.0 kg)(v 2 ) also: KE 10 + KE 2o = KE 1f + KE 2f ½ (1.5 kg)(4.0 m/s) 2 + ½ (3.0 kg)(-4.0 m/s) 2 = ½ (1.5 kg)(v 1 ) 2 + ½ (3.0 kg)(v 2 ) 2 Remember: We can only write that energy equation b/c the collision is elastic!

26. Now solve the two equations simultaneously for the two unknown variables: Your Algebra II teacher would be proud of you…but don’t forget to F-O-I-L! 1. (1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(v 1 ) + (3.0 kg)(v 2 ) 2. ½ (1.5 kg)(4.0 m/s) 2 + ½ (3.0 kg)(-4.0 m/s) 2 = ½ (1.5 kg)(v 1 ) 2 + ½ (3.0 kg)(v 2 ) 2 1.-6 = 1.5 v 1 + 3 v 2 2.First cancel all the ½ ‘s in equation 2: 72 = 1.5 v 1 2 + 3v 2 2 1. Solve for one variable in equation 1: v 1 = -4 – 2v 2 2.Substitute into equation 2: 72 = 1.5 (-4 – 2v 2 ) 2 + 3v 2 2 72 = 1.5 (16 + 16v 2 + 4v 2 2 ) + 3v 2 2 72 = 24 + 24 v 2 + 6v 2 2 + 3v 2 2 0 = 9v 2 2 + 24v 2 – 48 0 = 3v 2 2 + 8v 2 - 16 (go on)

27. Now solve this quadratic: 0 = 3v 2 2 + 8v 2 - 16 v 2 = -1.95 m/s or 0.177 m/s Then, by subsitution: v 1 = 0.1 m/s when v 2 = -1.95 m/s or v 1 = -4.35 m/s when v 2 = 0.177 m/s That’s not possible! How can car 1 move to the right and car 2 move to the left after the collision--- when they can’t pass through each other! Here’s the answer!

29. Impulse, which is also a vector (sometimes labeled J or I ), is defined as “change in momentum of an object or system”. J =  p =  m  v Since changing both m and v at the same time will require calculus, all of our problems will have a constant mass, so: J = m(v f – v o ).

30. Example: What is the impulse of a cart of mass 2.0 kg moving at 3.0 m/s if the cart is moving at -2.5 m/s after hitting the block?

31. Example: What is the impulse of a cart of mass 2.0 kg moving at 3.0 m/s if the cart is moving at -2.5 m/s after hitting the block? Answer: J = m(v f – v o ) J = (2.0 kg)(-2.5 m/s – 3.0 m/s) = -11 kg  m/s [Notice that impulse has the same units as momentum.]

32. Since we already know that an external force is required to change the momentum of an object, impulse can be related to force. As Isaac Newton formulated in his second law of motion: “Force is proportional to the rate in change of momentum.” F =  p/  t Newton wrote it in the above form, but we have simplified it to F = ma, which is equal to the above if you substitute.

33. Example: What is the force exerted on a cart of mass 2.0 kg moving at 3.0 m/s if the cart is moving at -2.5 m/s after hitting the block and being in contact with it for 1.0 ms?

34. Example: What is the force exerted on a cart of mass 2.0 kg moving at 3.0 m/s if the cart is moving at -2.5 m/s after hitting the block and being in contact with it for 1.0 ms? Answer: F = m(v f – v o )/t F = (2.0 kg)(-2.5 m/s – 3.0 m/s)/(1.0 x 10 -3 s) = -11,000 N Q: What is the force exerted by the cart on the block? Answer? The same---but in the opposite direction. The cart exerts an 11,000 N force on the block. Don’t forget Newtons’ third law of motion…action and reaction.

35. A corollary to Conservation of Momentum is “Conservation of the Velocity of Center of Mass”: The original motion of each shell is a parabola. After the shell explodes, the individual pieces move in such a way that if you add their positive and negative momenta, they still follow the original parabolic path.

36. The Grand Finale During this display, watch for conservation of momentum in the form of “conservation of the velocity of center of mass of the system”. http://www.youtube.com/watch?v=K3THhUGBNJE