Your Comments
This entire prelecture/checkpoint confused me, but especially the double
mass but half speed checkpoint.
Hey kid in Seat D23, your shoe is untied!
Switching reference frames makes this way too hard. i'd rather just solve
quadratic equations
I feel like this topic should be relatively simple, but I think I need to go over
more examples to help clear up some confusions. I'd like to talk about the
center of mass velocity affecting the system as a whole, and I want to go over
collisions made at angles. I'm also confused because we say the velocity is
the same after the collision as before the collision, but the example we did
earlier showed the box having a different velocity starting out compared to
finishing.
It's really amazing how good I feel when I realize the homework says "starts
tomorrow" and not "due tomorrow".
Can we call elastic collisions "Bouncy Smacks"? I think this would make
things very interesting.
How in the almighty universe are we supposed to know what direction those
balls will end up going in after looking at measly arrows.
Hour Exam 1
Average 76.5%
(no scaling)
Mechanics Lecture 12, Slide 2
I did poorly on the first exam, do you recommend
dropping the course?
No. Don’t Panic.
Each hour exam is worth 10% of your grade.
Even if you got 50% on the exam, it means you have lost
only 50 points out of 1000.
There are 950 other points you haven't lost yet !
We have posted a spreadsheet to help you estimate your grade
Office Hour Reminder
Physics 211
Lecture 13
Today’s Concepts:
a) Elastic Collisions
b) Center-of-Mass Reference Frame
In high school I was taught that the main difference
between elastic and inelastic collisions is that in inelastic
collisions the blocks stick together and in elastic collisions
they don't. However, after the last few lectures, I'm
starting to think there's a lot more to it. It would be really
helpful if we could just make a list of the key differences
between the two. Also, please play Simon and Garfunkel!
In elastic collisions, E = K + U is the same before & after
Mechanics Lecture 12, Slide 5
A collision is inelastic if non-conservative forces
do work:
- Friction
- Deformation
- Sticking together
- Just about anything
This is usually the case in real life
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 totVcm
If you are in a reference frame
moving along with the CM then the
total momentum you measure is 0.
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
CheckPoint Response
A) Kinitial > Kfinal
B) Kinitial  Kfinal
C) Kinitial < Kfinal
initial
final
A) This is not an elastic collision since they stick together and
therefore the KE isn't conserved after they collide.
B) Because there is no friction, the kinetic
energy is conserved in the system..
C) two boxes are moving after the collision.
Relationship between Momentum & Kinetic Energy
2
1 2
p
1 2 2
K  mv 
mv 
2m
2
2m
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
K Initial
K final
final
p2

2M
p2

2(2M )
same p
Center-of-Mass Frame
I would like more examples and elaboration on problems
that incorporate center of mass and relative motion.
videos
vCM
m1v1  m2v2  ...
Ptot


M tot
m1  m2  ...
..\..\movies\cmel.mpg.
mpg
In the CM reference frame, vCM  0
In the CM reference frame, PTOT  0
..\..\movies\cminel.
mpg.mpg
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
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
Example
Four step procedure:
Step 1:
First figure out the velocity of the CM, vCM.
vCM
vCM
vCM
 1 

  m1v1,i  m2v2,i 
0 in this case
 m1  m2 
m1

v1,i
m1  m2
.2kg

v1,i
.2kg  .8kg
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
Example
Step 2: Calculate the initial velocities in the CM reference frame
(all velocities are in the x direction):
vbox,lab  vbox,cm  vcm,lab
v  v*  vCM
v*  v - vCM
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
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
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!
Summary:
DEMO
v1,i
m1
v1, f
v2,i  0
m2
m1
m1
x
m2
m2
v2, f
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 v1,i* = v1,i – vcm and v2,i* = v2,i – 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: v1,f* = - v1,i* v2,f* = - v2,i*
4. Transform the velocities back to the lab frame using
v1,f = v1,f* + vcm
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
CheckPoint
A) m > M
B) M  m C) M > m
A) m sets M into motion
B) if m > M, then both blocks would move to the right. If m < M, m
would move to the left.
C) The second block must be larger because it makes the first
block stop.
M
m
m
Before Collision
M
After Collision
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
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
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) for cm calculation, |v2i|=|v2f|, so the just swap directions because
of the collision. What was -v before is +v now.
C) If the velocity of the center of mass is 0, neither block can be moving
otherwise the Vcm would be nonzero.
D) In center of mass frame, initial velocity equals
final velocity.