Your Comments Everything seemed to make sense What is the average for the Exam 1? Answer: 80.5 I'm very confused as to how you analyze the center of mass frame and why. Also how do you transform it back to the normal frame reference. The velocities you are adding are not clear Professor Selen, PLEASE PLEASE PLEASE differentiate between center of mass reference frame and lab reference frame. … the prelecture says that in a CM reference frame Vi=Vf, but I didn't see that in the picture. Thanks! I really wish for a universe where a cannon firing propels the cannon backwards a long distance while the ball falls to the ground I was confused by the checkpoint section "Double Mass but Half Speed." Can you explain this further in lecture today? I'd like to see more problems using the center of mass frame We never learned about these tricks using the center of mass reference frame, I found them pretty neat. Teach center of mass again! That was a pain in the center of my @$$ JOSEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Mechanics Lecture 12, Slide 1

Question: “why is there no curve on the test we just took?”

Answer: The average was 80.5% and we only scale an exam up if the average is less than 75%, we never scale it down. The only time there is ever any curving in this class is if we mess up and make something too hard. This is extremely rare. Question: “That test was nothing like the previous midterms. Why did you have to make it harder?” Answer: It wasn’t harder. The average was 80.5% without any curve. Mechanics Lecture 12, Slide 2

Physics 211 Lecture 12

Today’s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame

Mechanics Lecture 12, Slide 3

Review: Center of Mass & Collisions so far

Fnet , ext  M tot Acm 

dPtot dt

The CM behaves just like a point particle

If Fnet , ext  0 then momentum is conserved

Ptot  M totV cm

If you are in a reference frame moving along with the CM then the total momentum you measure is 0.

Mechanics Lecture 12, Slide 4

”…And can we go over more on how to tell the difference between an inelastic and elastic collision and the kinetic energies related to them.?” “How are elastic and nonelastic collision different??”

Inelastic if nonconservatve forces do work: - Friction - Deformation - Sticking together - Just about anything

Mechanics Lecture 12, Slide 5

CheckPoint A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. Compare the initial and final kinetic energies of the system.

A) Kinitial > Kfinal B) Kinitial  Kfinal C) Kinitial < Kfinal

initial

final Mechanics Lecture 12, Slide 6

CheckPoint Response A) Kinitial > Kfinal B) Kinitial  Kfinal C) Kinitial < Kfinal

initial

final A) Since the boxes stuck together, this is not an elastic collision, therefore Kinetic Energy is not maintained. B) Since it sticks, all kinetic energy is conserved. C) This is not an elastic collision, therefore some Ke is lost.

Mechanics Lecture 12, Slide 7

Relationship between Momentum & Kinetic Energy

K 

1 2

mv

2



1 2m

2

m v

2



p

2

since p  m v

2m

This is often a handy way to figure out the kinetic energy before and after a collision since p is conserved.

initial

K Initial 

K

final

final



p

2

2M p

same p

2

2 (2 M )

Mechanics Lecture 12, Slide 8

Center-of-Mass Frame

videos

vC M 

m1 v1  m 2 v 2  ... m1  m 2  ...



..\..\movies\cmel.mpg. mpg

Pto t M

..\..\movies\cminel. mpg.mpg

to t

In the CM reference frame, vCM  0 In the CM reference frame, PTOT  0

Mechanics Lecture 12, Slide 9

Center of Mass Frame & Elastic Collisions The speed of an object is the same before and after an elastic collision if viewed in the CM frame: v*1,i

m1 m1 v*1, f

m1

m2

v*2,i

m2

v*2, f

m2

Mechanics Lecture 12, Slide 10

Example: Using CM Reference Frame A glider of mass m1  0.2 kg slides on a frictionless track with initial velocity v1,i  1.5 m/s. It hits a stationary glider of mass m2  0.8 kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What are the final velocities? v1,i

CM

*

m1

v2,i  0

m2 CM

m1 v1, f

m1

*

x

m2 CM

*

m2

v2, f Mechanics Lecture 12, Slide 11

Example Four step procedure: Step 1:

First figure out the velocity of the CM, vCM. vC M

  1    m 1 v1, i  m 2 v 2 , i   m1  m 2  0 in this case

vC M 

vC M 

m1 m1  m 2

v1, i

.2 kg .2 kg  .8 kg

v1, i

Mechanics Lecture 12, Slide 12

Example Now consider the collision viewed from a frame moving with the CM velocity VCM.

v*1,i

m1 m1 v*1, f

m1

m2

v*2,i

m2

v*2, f

m2

Mechanics Lecture 12, Slide 13

Example Step 2: Calculate the initial velocities in the CM reference frame (all velocities are in the x direction): v

v  v* vCM vCM

v*  v - vCM v*

v*1,i  v1,i - vCM  1.5 m/s - 0.3 m/s  1.2 m/s v*2,i  v2,i - vCM  0 m/s - 0.3 m/s  -0.3 m/s v*1,i  1.2 m/s v*2,i  -0.3 m/s Mechanics Lecture 12, Slide 14

Example Step 3:

Use the fact that the speed of each block is the same before and after the collision in the CM frame.

v*1, f  -v* 1,i

m1

V*1, f

m1

m2

V*1,i

m1

v*1, f  - v*1,i  -1.2m/s

v*2, f  -v*2,i V*2,i x

m2 v*2, f  - v*2,i .3 m/s

m2

V*2, f Mechanics Lecture 12, Slide 15

Example Step 4:

Calculate the final velocities back in the lab reference frame: v

v  v* VCM vCM

v*

v1, f  v*1, f  vCM  -1.2 m/s  0.3 m/s  -0.9 m/s

v2, f  v*2, f  vCM  0.3 m/s  0.3 m/s  0.6 m/s v1, f  -0.9 m/s

v2, f  0.6 m/s Four easy steps! No need to solve a quadratic equation! Mechanics Lecture 12, Slide 16

I don't understand the purpose of switching reference frames. Isn't it enough just to say that momentum and KE are conserved in elastic collisions? “I get why switching to center of mass makes things easier. But at the same time, I feel like quadratic equations would be easier than trying to keep track of switching and switching back”

“Can we just play a few rounds of pool for this week's lab?”

Yes

Try it !

2D demo Mechanics Lecture 12, Slide 17

Summary:

DEMO v1,i

m1 v1, f

v2,i  0

m2

m1

m1

x

m2 m2

v2, f Mechanics Lecture 12, Slide 18

4-step summary 1. Find velocity of center of mass of the system 2. Transfom the velocity of each object to the center of mass using v  v* vCM 3. Reverse the velocities of each object in the center of mass because the speed of each block is the same before and after the collision in the CM frame 4. Transform the velocities back to the lab frame using v  v* vCM

Mechanics Lecture 12, Slide 19

CheckPoint A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is initially at rest. After the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M

B) M  m C) M > m

m

M

m

Before Collision

M

After Collision Mechanics Lecture 12, Slide 20

CheckPoint A) m > M

B) M  m C) M > m

A) m is has greater momentem, enough to push block M B) if m was greater than M, it would keep rolling, and if m was less than M, it would bounce back. C) If m>M then it wouldnt have stopped moving.

M

m

m

Before Collision

M

After Collision Mechanics Lecture 12, Slide 21

Newton’s Cradle

Mechanics Lecture 12, Slide 22

CheckPoint Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. The velocity of the center of mass of this system before the collision is A) Toward the left

2v

m

B) Toward the right C) zero

v

2m

Before Collision

This is the CM frame Mechanics Lecture 12, Slide 23

CheckPoint Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place? A) 2v B) v C) 0 D) -v E) -2v 2v

m

v

2m

+

Before Collision

Mechanics Lecture 12, Slide 24

CheckPoint Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place? A) 2v B) v C) 0 D) -v E) -2v

2v

m

+

v

2m

Before Collision

This is the CM frame

B) The objects just change direction with the same speed in the cm reference frame. C) The total momentum of the system before and after the collison is zero. D) Initial Velocity and Final Velocity are the same in elastic motion. Mechanics Lecture 12, Slide 25

do a problem were we have to go between the enter of mass reference frame and the labs reference frame

vcm 

m1v1,i  m 2 v 2,i m1  m 2

v1,* i  v1, i - v cm v1,* f  - v1,* i v1, f  v1,* f  v cm Mechanics Lecture 12, Slide 26

Do same steps for car 2

vcm is the same = final speed A B C

Compare 1/2 mv 2 before and after for both cases Mechanics Lecture 12, Slide 27