10/21 Momentum Intro

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

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

What do you think? v = 0 v = 0 Velocities before collision are the same in both cases X A A Y v = 0 A rebounds off of X but does not rebound off of Y A X A Y Student 1: I say Y since block A stops dead in that case, giving all its momentum to 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.

A B NB,A NA,B What do you think? v = 0 v = 0 Velocities before collision are the same in both cases X A A Y v = 0 A rebounds off of X but does not rebound off of Y A X A Y Let’s check into Student 2’s idea. In either collision, the FBD’s look like: 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!

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

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

pf,A pi,A pf,A pi,A pA pA pY pX Student 2’s revenge v = 0 v = 0 Velocities before collision are the same in both cases X A A Y v = 0 A rebounds off of X but does not rebound off of Y A X A 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! X is faster than Y!

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

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!!! r The momentum of the “system” does not change and the initial momentum equals the final momentum, for the “system.”

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

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

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

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

Two cars collides with a parked truck on glare ice, (frictionless). Fnett = p v = 0 before v v, same as A A B C After A and B collide, C collides with B and B comes to rest again. mA = m mB = 6m mC = m after AB, before BC v = v/2 v, same as A What happens to C? A B C mB = 6m mC = m mA = m after BC collision v = 0 Summary: vf,C = v/2 v/2 A B C Momentum mA = m mB = 6m mC = m 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!

10/21 Momentum Intro