Physics 211
Lecture 12
Today’s Concepts:
a) Elastic Collisions
b) Center-of-Mass Reference Frame
I'm still a little unsure about the difference between elastic
and inelastic collisions. How can you tell the difference?
Hang on, . . . Here we go!
Mechanics Lecture 12, Slide 1
Hour Exam 1
High Score of 100%
Charles Huh
Andrew Yu
Mean score of 82.1%
(Yes, the KEY got an A - )
Change in my office hours
2:30 to 3:30 on M, Tu, W
10:30 to 11:30 on Thursday
Mechanics Lecture 11, Slide 2
iClicker and “bonus points”
The Course Description mentions “bonus points” for
correct answers with your iClickers.
That feature is not implemented this summer.
I am using the iClickers for feedback. Do well. But
there are no add’l points for getting answers correct
with your iClickers.
Mechanics Lecture 11, Slide 3
Summary / Paraphrase
What are we doing today?
I need someone to summarize or restate or
paraphrase what you learned in today’s Prelecture.
This is important and the Prelecture seems very
well done – to me. But it is somewhat different.
What did YOU hear? What did you learn?
Why are we doing this?
Help me out. I need some feedback.
Mechanics Lecture 11, Slide 4
Center of Mass & Collisions so far:
Fnet ,ext  M tot Acm
If
Fnet ,ext  0
Ptot  M totVcm
dPtot

dt
The CM behaves just like
a point particle
then momentum is conserved
If you are in a reference frame
moving along with the CM then the
total momentum you measure is 0.
Mechanics Lecture 12, Slide 5
Inelastic / Elastic Collisions
Inelastic collisions are those in which the objects stick
together. Think of balls of clay or gummy
marshmellows. ( S p l a t ! ) KE will not be conserved.
Elastic collisions are like billiard balls or air hockey
pucks or air track gliders with springs or rubber bands.
( B o i n g ! ) KE will be conserved.
Inelastic and Elastic collisions may also be called
Splat – Boing problems.
Mechanics Lecture 12, Slide 6
CheckPoint/ACT
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 7
CheckPoint Response
A) Kinitial > Kfinal
B) Kinitial  Kfinal
C) Kinitial < Kfinal
initial
final
A) The boxes stick so it is not an elastic colllision
B) Because there are no external forces acting, the energy
will be conserved.
C) There are two boxes moving instead of 1, so there is
more energy.
Mechanics Lecture 12, Slide 8
Your Comments
Because this is not an elastic collision KE is not conserved. Due to this when the mass
increases the KE increases. If KE increases, you need to have a source for that in mind.
So the collision is not elastic and some K lost during the collision
if the boxes stick that means there is an inelastic collision happening, thus there is some
energy lost due to the internal forces of the boxes.
energy is conserved in an elastic collision True; but this is an INELASTIC collision.
since they are identical boxes, their kinetic energies will be the same Hmmm, . . . That
sounds like you’re answering a different question. How do the initial and final KE’s compare?
The momentum will we the same before and after the boxes collide and stick together,
however the mass of the object moving doubles after the collision, making the initial kinetic
energy smaller than the final kinetic energy. Good observation.
kinetic energy is lost with inelastic collisions
thats how it works
lost energy due to object material ? ? ?
Should be equal Tell me more.
The forces from the initial is the reason why the final force is moving I don’t understand.
The collision is inelastic so kinetic energy is not conserved. Good
Be careful That is often my response.
Inelastic collision ! ! !
Mechanics Lecture 12, Slide 9
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
same p
p2

2(2M )
Mechanics Lecture 12, Slide 10
CheckPoint/ACT
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.
90% correct. Must conserve
Energy and Momenturm
How does M compare to m?
A) m > M
B) M  m C) M > m D) Need more information
M
m
m
Before Collision
M
After Collision
Mechanics Lecture 12, Slide 11
CheckPoint/ACT
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 D) Need more information
M
m
m
Before Collision
M
After Collision
Mechanics Lecture 12, Slide 12
Newton’s Cradle
Demos: Newton's cradle,
Air track elastic collisions
Mechanics Lecture 12, Slide 13
Center-of-Mass Frame
VCM
m1v1  m2v2  ...
Ptot


M tot
m1  m2  ...
In the CM reference frame, VCM  0
In the CM reference frame, PTOT  0
Films: cmel, cminel
Mechanics Lecture 12, Slide 14
Center of Mass Frame & Elastic Collisions
In the prelecture when finding the velocities of the objects in the center of mass,
they did 5-2=3 and 0-2= -2 but then switched the signs to be -3 and 2, why?
The speed on an object is the same before and after an
elastic collision as 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 15
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?
v*1,i
m1

m2
v*2,i  0
VCM
  CM
m1
v*1, f
m1

x
m2

m2
v*2, f
Mechanics Lecture 12, Slide 16
Example
Four step procedure:
Step 1:
First figure out the velocity of the CM, VCM.
 1 
 (m1v1,i + m2v2,i), but v2,i  0 so
VCM  
 m1  m2 
 m1
VCM =  m  m
2
 1

 v
1,i

(for v2,i  0 only)
So VCM = 1/5 (1.5 m/s) = 0.3 m/s
Mechanics Lecture 12, Slide 17
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 18
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 19
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 20
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 21
CheckPoint/ACT
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
Mechanics Lecture 12, Slide 22
CheckPoint Response
The velocity of the center of mass of this system before the collision is
A) Toward the left B) Toward the right C) zero
2v
v
m
2m
This is the CM frame
Before Collision
A) because it has twice the mass
B) the net velocity is towards the right.
C) The total momentum of the system is zero.
This means that the velocity of the center of
mass must be zero.
Mechanics Lecture 12, Slide 23
CheckPoint/ACT
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) v B) - v C) 2v D) - 2v E) zero
2v
m
v
2m
+
Before Collision
This is the CM frame
Mechanics Lecture 12, Slide 24
CheckPoint Response
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) v B) - v C) 2v D) - 2v E) zero
2V
m
+
V
2m
Before Collision
This is the CM frame
A) Since the collision is elastic, and the velocity of the
center of mass is zero then the blocks simply travel
backwards at their same speeds, (or opposite velocities).
B) the magnitude of the velocity stays the same but the
direction changes
E) in order to conserve momentum, the blocks will
both stop.
Mechanics Lecture 12, Slide 25
Do a problem where we have to go between the enter of mass reference
frame and the labs reference frame
Vcm =
m1v1,i + m2 v2,i
m1 + m2
v1,i* = v1,i -Vcm
v1,* f  - v1,* i
v1, f = v1,* f +Vcm
Mechanics Lecture 12, Slide 26
Do same steps for car 2
vcm is the same = final speed
Compare 1/2 mv 2 before
and after for both cases
Mechanics Lecture 12, Slide 27
Special Case
Consider m1 coming in with velocity v1i
and colliding with mass m2 at rest.
What will be their final velocities?
Mechanics Lecture 12, Slide 28
Special Case
Mechanics Lecture 12, Slide 29
Final Results for this Special Case
Mechanics Lecture 12, Slide 30
2D collisions
Now conserve Px, Py (& K if elastic).
Scattering angle depends on impact parameter (glancing blow vs
head-on)
Demo: pool table
Mechanics Lecture 12, Slide 31