I understand the concepts, but the math formulas being spat out like machine gun bullets are really confusing me. Is there a way we can see the formulas worked out by hand? I have a hard time connecting steps when shown on the prelecture/lecture slides
I like physics better than chemistry. All they teach you how to do in chemistry is make crystal meth. And that is very bad indeed.
It's 4 AM on Friday and I couldn't sleep. What do I do? Do Tuesday's homework, prelecture and checkpoint. Who knew having insomnia could be so productive!
Conceptual examples would be nice, since that is what the test consists of.
May your mother be with you. Get it? May the F=ma be with you? Oh, please put this on the screen before class!!
I overall understand this concept more than the rest. But please go over the Double Mass but
Half Speed problem in detail!
This all seems to make sense - momentum/elastic/inelastic tie together well.
Why isn't the answer to the first question in the pre-lecture elastic? Don't both of the boxes move?
I liked the Hitchhiker's Guide themed lecture a few weeks ago, can we have another?
Preferably Doctor Who, I know there are plenty of Whovians in this class. You could have a collision problem where the TARDIS collides with a Dalek traveling through the Time Vortex and see where they end up.
Mechanics Lecture 12, Slide 1

Question:
“why is there no curve on the test we just took?”
Answer:
The average was 80% 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% without any curve.
Mechanics Lecture 12, Slide 2

a) Elastic Collisions b) Center-of-Mass Reference Frame
Mechanics Lecture 12, Slide 3

F

M A tot cm
 dP tot dt
The CM behaves just like a point particle
If F

0 then momentum is conserved
P tot

M V tot cm
If you are in a reference frame moving along with the CM then the total momentum you measure is 0 .
What is meant by "velocity of the center of mass of this system is zero"?
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

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) K initial
B) K initial
C) K initial
>
K final

K final
<
K final initial final
Mechanics Lecture 12, Slide 6

A) K initial
B) K initial
C) K initial
>
K final

K final
<
K final 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

1
K mv
2
2 
1
2 m p
2

2 m since p
 mv
This is often a handy way to figure out the kinetic energy before and after a collision since p is conserved.
initial final
K
Initial p
2

2 M
K final p
2

2(2 M ) same p
Mechanics Lecture 12, Slide 8

v
CM
 m v
1 1
 m v
2 2
 m
1
 m
2

...
...

P tot
M tot
CM
v
CM

0
CM
P
TOT

0
“If two objects collide in opposite directions with the same momentum, DO THEY MOVE AFTER THE
COLLISION??? WHAAAAA???”
Mechanics Lecture 12, Slide 9

CM
m
1 v *
1, i m
2 v *
2, i m
1 m
2 v *
1, f m
1 m
2 v *
2, f
Mechanics Lecture 12, Slide 10

A glider of mass m
1 initial velocity v
1, i m
2


0.2 kg
1.5 m
/ s slides on a frictionless track with
. It hits a stationary glider of mass

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?
m
1 v
1, i v
1, f m
1 m
2 m
1
CM
* m
2 v
2, i

0 x m
2 v
2, f
Mechanics Lecture 12, Slide 11

Four step procedure :
Step 1: First figure out the velocity of the CM , v
CM
.
v
CM v
CM



1 m
1
 m
2


 m v
 m v
1 1, i 2 2, i

0 in this case
 m
1 m
1
 m
2 v
1, i v
CM

.2
kg
.2
kg

.8
kg v
1, i
Mechanics Lecture 12, Slide 12

CM
V
CM
v *
1, f m
1 m
1 v *
1, i m
2 v *
2, i m
2 m
1 m
2 v *
2, f
Mechanics Lecture 12, Slide 13

Step 2: Calculate the initial velocities in the CM reference frame
(all velocities are in the x direction): v
 v *
 v
CM v v *
 v
v
CM v
CM v * v *
1 ,i
 v
1 ,i
v
CM

1.5 m/s
-
0.3 m/s

1.2 m/s v *
2 ,i
 v
2 ,i
v
CM

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

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 v *
2 , f
 v *
2 ,i m
1
V *
1, i m
2
V *
2, i v *
1 , f
 v *
1 ,i
 -
1.2m/s
V *
1, f m
1 m
1 m
2 v *
2 , f
 v *
2 ,i

.3 m/s m
2 V *
2, f x
Mechanics Lecture 12, Slide 15

Step 4: Calculate the final velocities back in the lab reference frame: v
 v *

V
CM v v
CM v * v
1 , f
 v *
1 , f
 v
CM
 -
1.2 m/s

0.3 m/s
 -
0.9 m/s v
2 , f
 v *
2 , f
 v
CM

0.3 m/s

0.3 m/s

0.6 m/s v
1 , f
 -
0.9 m/s v
2 , 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?”
Mechanics Lecture 12, Slide 17

m
1 v
1, i v
1, f m
1 m
1 m
2 v
2, i

0 m
2 x m
2 v
2, f
Mechanics Lecture 12, Slide 18

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
Before Collision m M
After Collision
Mechanics Lecture 12, Slide 19

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
Before Collision m M
After Collision
Mechanics Lecture 12, Slide 20

Mechanics Lecture 12, Slide 21

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 B) Toward the right C) zero
2 v m v
2 m
Before Collision
This is the CM frame
Mechanics Lecture 12, Slide 22

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)
2 v B) v C) 0 D)
v E)
-2 v
2 v m v
2 m
Before Collision
+
Mechanics Lecture 12, Slide 23

Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place?
A)
2 v B) v C) 0 D)
v E)
-2 v v +
2v m
2 m
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 24

do a problem were we have to go between the enter of mass reference frame and the labs reference frame v cm
 m v
1 1, i
 m v
2 2, i m
1
 m
2 v
*
1, i
 v
1, i
v cm v
1,
* f
 v
1,
* i v v v
1, f
 *
1, f
 cm
Mechanics Lecture 12, Slide 25

A
B
C
Do same steps for car 2 v cm is the same = final speed
Compare 1 /
2 mv 2 before and after for both cases
Mechanics Lecture 12, Slide 26