1. 10/21 Momentum Intro  Text: Chapter7  Lab: Momentum  HW: None Assigned, read Ch 7 instead

2. What do you think? Velocities before collision are the same in both cases X and Y both have the same mass but X must be made of a more “elastic” or more bouncy material. A A A hits block X A hits block Y Y X v = 0 AAYX A rebounds off of X but does not rebound off of Y Which block, X or Y, has the greater velocity after the collision?

3. What do you think? Velocities before collision are the same in both cases Student 1: I say Y since block A stops dead in that case, giving all its momentum to Y. A A AA Y Y X X v = 0 A rebounds off of X but does not rebound off of Y Student 2: I say X since block A turns around so it must have had a greater net force on it. Student 1: But we are talking about X and Y, not A! Student 2: But the 3rd law says that the net force on A equals the net force on the other block.

4. What do you think? Velocities before collision are the same in both cases Let’s check into Student 2’s idea. In either collision, the FBD’s look like: A A AA Y Y X X v = 0 A rebounds off of X but does not rebound off of Y A N B,A B N A,B Student 1: Well, so what. The force only lasts for a millisecond, and without knowing  t you can’t find  v anyway! Pbtbthtbthtbt!!!!! Student 2:Oh yeah? Well…well…I… it’s…but... Darn!

5. What do you think? Velocities before collision are the same in both cases Poor Student 2! Let’s see if we really need  t to find  v. A A AA Y Y X X v = 0 A rebounds off of X but does not rebound off of Y AB N B,A N A,B F net = ma tt vv = m F net  t = m  v The F net ’s are equal and opposite for A and B and the  t’s are the same for A and B (contact time) so... The product F net  t must be equal and opposite for A and B and also the product m  v must be equal and opposite for A and B.

6. Momentum Symbol: p p = mv  p = m  v “change in momentum”--works just like  v We work with the initial momentum, final momentum, and the change in momentum. p i +  p = p f momentum definition--unlike energy, it’s a vector, and points the same direction as v

7. Student 2’s revenge Velocities before collision are the same in both cases A A AA Y Y X X v = 0 A rebounds off of X but does not rebound off of Y Student 2:By golly I’ve got it now! The “change in momentum” of A is equal and opposite to the “change in momentum” of a block! We don’t need to know  t! neener neener neener! p i,A p f,A pApA p i,A p f,A pApA pXpX pYpY X is faster than Y!

8. Lab: Linear Momentum (1-D) cart flag air trough photogate velocity from flag length and gate time

9. Momentum of the “System” The momentum of the system is the sum (vector) of the momenta. The change in momentum of the system is the sum (vector) of the changes in momenta. The FBD of the system looks like... And the net force is zero!!! The momentum of the “system” does not change and the initial momentum equals the final momentum, for the “system.” r

10. Two cars collides with a parked truck on glare ice, (frictionless). AB m A = mm B = 6m v v = 0 C m C = m v, same as A After A and B collide, C collides with B and B comes to rest again. After A and B collide, A is moving left at 1 / 2 v What happens to B? Could we use 2nd law? Energy? Momentum? F net  t =  p 3rd law pairs AB During collision A and B N B,A N A,B before A B m A = mm B = 6m v/2 v = ? C m C = m v, same as A after

11. Two cars collides with a parked truck on glare ice, (frictionless). AB m A = mm B = 6m v v = 0 C m C = m v, same as A A B m A = mm B = 6m v/2 v f,B = ? C m C = m v, same as A After A and B collide, A is moving left at 1 / 2 v F net  t =  p AB 3rd law pairs During collision Same, but opposite Same for each block Same for each, but opposite direction  p A = ? = 3 / 2 p A,i = 3 / 2 mv  p B = ? = 3 / 2 p A,i = 3 / 2 mvDirection? pBpB What happens to B? = p f,B p A,i p A,f before after After A and B collide, C collides with B and B comes to rest again. v f,B = 1 / 4 v = 6mv f,B = 3 / 2 mv f,B

12. Two cars collides with a parked truck on glare ice, (frictionless). BC 3rd law pairs During collision F net  t =  p Same, but opposite Same for each block Same for each, but opposite direction  p C = ?  p B = ? = 3 / 2 p A,i = 3 / 2 mv left p i,B What happens to C? p i,C = ? p f,C = ? = 1 / 2 mv right A B m A = m m B = 6m v/2 1/4v1/4v C m C = m after AB, before BC A B m A = mm B = 6m v/2 v = 0 C m C = m v f,C = ? after BC collision v, same as A because p f,B = 0 After A and B collide, C collides with B and B comes to rest again.

13. A B m A = mm B = 6m v/2 v = 0 C m C = m v f,C = v/2 F net  t =  p What happens to C? Summary: ABC initial change final B after BC collision A B m A = m m B = 6m v/2 v = C m C = m after AB, before BC v, same as A AB m A = mm B = 6m v v = 0 C m C = m v, same as A before Momentum After A and B collide, C collides with B and B comes to rest again. Two cars collides with a parked truck on glare ice, (frictionless).

14. A B m A = mm B = 6m v/2 v = 0 C m C = m v f,C = v/2 F net  t =  p What happens to C? Summary: after BC collision A B m A = m m B = 6m v/2 v = C m C = m after AB, before BC v, same as A AB m A = mm B = 6m v v = 0 C m C = m v, same as A before Momentum After A and B collide, C collides with B and B comes to rest again. Two cars collides with a parked truck on glare ice, (frictionless). Take a look a the initial and final momenta of the “system.” They are both zero. In facy, the momentum of the “system” is zreo throughout!