Classical Mechanics
Lecture 12
Today’s Concepts:
a) Elastic Collisions
b) Center-of-Mass Reference Frame
Mechanics Lecture 12, Slide 1
Main Points
Mechanics Lecture 11, Slide 2
Main Points
Mechanics Lecture 11, Slide 3
Elastic Collisions: Analogy using spring

Mechanical Energy of system is
if the form of kinetic energy
moving block.
 Moving block experiences a
negative acceleration due to
spring force.
 Kinetic Energy of block is
transferred to potential energy of
spring.
 Stationary block experiences an
acceleration due to Spring force
acting on formerly stationary
block.
 Potential Energy of Spring is
transferred back to the Kinetic
Energy of blocks after the
collision. Mechanical Energy is
conserved
Mechanics Lecture 11, Slide 4
Elastic Collisions
Mechanics Lecture 11, Slide 5
Elastic Collisions…final state?
Quadratic Equation…
Solvable !...but “tedious”…
Mechanics Lecture 11, Slide 6
A better way!!
Eliminates Quadratic Equation…
Mechanics Lecture 11, Slide 7
Center of Mass Frame
When viewed from reference frame moving with Center of Mass
Mechanics Lecture 11, Slide 8
Center-of-Mass Frame

VCM



m1v1  m2v2  ... Ptot
=
=
m1  m2  ...
M tot
In the CM reference frame, VCM = 0
In the CM reference frame, PTOT = 0
Mechanics Lecture 12, Slide 9
Center of Mass Frame
Mechanics Lecture 11, Slide 10
Elastic Collisions in CM frame
Initial state
Final state
Mechanics Lecture 11, Slide 11
Elastic Collisions in CM frame
Equal speeds but opposite velocity
Mechanics Lecture 11, Slide 12
Elastic Collisions in Lab frame
Mechanics Lecture 11, Slide 13
Elastic Collisions in CM frame
Mechanics Lecture 11, Slide 14
Elastic Collisions in CM frame
Mechanics Lecture 11, Slide 15
Elastic Collisions in CM frame
What are the speeds of the boxes in the Center of Mass frame
after the collision?
In the center of mass frame the solution is TRIVIAL!!!!
The final speeds of each object are the same as their initial
speeds. But in opposite direction…again in the CM frame…
Mechanics Lecture 11, Slide 16
Elastic Collisions in CM frame
Need to transform the velocities back into the Lab frame…
Mechanics Lecture 11, Slide 17
Elastic Collisions in CM frame
Generalized result…
Mechanics Lecture 11, Slide 18
Checkpoint
A.
B.
C.
33%
33%
33%
Mechanics Lecture 11, Slide 19
2-d elastic collisions
and one object at rest
Mechanics Lecture 11, Slide 20
Mechanics Lecture 11, Slide 21
Clicker/Checkpoint
A.
B.
C.
0%
0%
0%
Mechanics Lecture 11, Slide 22
Summary
Mechanics Lecture 11, Slide 23
Summary
In center of mass frame
Speeds remain the same
Even in 2-d!!!
Mechanics Lecture 11, Slide 24
Summary
Mechanics Lecture 11, Slide 25
Center of Mass Frame & Elastic Collisions
The speed on an object is the same before and after an
elastic collision is 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 26
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 27
Example
Four step procedure:
Step 1:
First figure out the velocity of the CM, VCM.

1
 m1  m 2
VCM = 

 (m1v1,i + m2v2,i), but v2,i = 0 so


 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 28
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 29
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 30
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 31
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 32
Collision with Friction
m1v1 = m1  m2 v f
vf =
m1v1
m1  m2 
 m1v1 
1
1

K i = m1  m2 v 2f = m1  m2 
2
2
 m1  m2  
2
W friction = -  k m1  m2 gd = K
 m1v1 
1

K = K f - K i = - m1  m2 


m

m
2
2 
 1
2


1
m1  m2  m1v1 
2
- K
 m1  m2  
=
k =
m1  m2 gd
m1  m2 gd
2
2
 m1v1  1

 k = 


m

m
2  2 gd
 1
Mechanics Lecture 11, Slide 33
vCM


m1v1,i  m2v2,i
=
m1  m2
v1,i = v1,i - vcm
v1, f = -v1,i
v1, f = v1, f  vcm
Mechanics Lecture 12, Slide 34
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 35
Bumper Cars 2



-1 
-4 
m1v1, f - m1v1,i =
m1v1,i - m1v1,i =
m1v1,i
3
3
ptotal = 0
ptotal = p1  p2 = 0
p2 = -p1
Mechanics Lecture 11, Slide 36
v2, f = p2, f / m2 = ( p2,i  p2 ) / m2 = p2 / m2
K1 = K1, f - K1,i =
K1 =
p12, f
2m1
-
p12,i
2m1

=
( p1,i  p1 ) 2
2m1
1
p12,i  2 p1,i p1  p 2 - p12,i
2m1
-
p12,i
2m1

2
1
1 
-4
 -4
 
2
2 p1,i p1  p1 =
2 p1,i 
p1,i   
p1,i  
2m1
2m1 
 3
  3
 
1   - 24 2   16 2  
1  8 2 
K1 =
 
p1,i    p1,i   =
  p1,i  
2m1   9
 9
  2m1   9

K1 =


Elastic collision 
Ktot = 0
Ktot = K1  K 2 = 0
 K 2 = -K1
Mechanics Lecture 11, Slide 37
v 2 w ,i = v 2 w , f  v 2 b , f
vw,i = vw, f cos w  vb , f cos b
0 = vw, f sin  w  vb , f sin  b




mwvw,i  mb vb ,i = mwvw, f  mb vb , f
vw, f sin  w = -vb , f sin  b
1
1
1
1
mwv 2 w,i  mb v 2 b ,i = mw v 2 w, f  mb v 2 b , f
2
2
2
2




vw,i  vb ,i = vw, f  vb , f
vb , f = -vw, f
v 2 w ,i  v 2 b ,i = v 2 w , f  v 2 b , f
vw,i  vb,i x = vw, f  vb, f x
vw,i  vb,i y = vw, f  vb, f y
v
2
w ,i
=v
vw, f =
2
w, f
sin  w
sin  b
2
  sin   2 

sin  w 
2
w
 = v w, f 1  
 
  vw, f


sin

sin

b 
b 

 

v 2 w ,i
  sin   2 
w
1  
 
  sin  b  


Mechanics Lecture 11, Slide 38
v 2 w ,i = v 2 w , f  v 2 b , f
v 2 b , f = v 2 w ,i - v 2 w , f
vb , f = v 2 w,i - v 2 w, f
ptotal, f = ptotal,i = mwvw,i
K total, f = K total,i =
1
1 2
mwv 2 w,i =
p total,i
2
2m
Mechanics Lecture 11, Slide 39
Spring Loaded Collision
vCM =
1
m1v1  m2v2  = m1v1
m1  m2 
m1  m2 


m1

v1*,i = v1,i - vCM = v1,i 1 

m

m
1
2 



m1

v1*, f = -v1*,i = -v1,i 1 

m

m
1
2 

Mechanics Lecture 11, Slide 40
B.
Checkpoint
C.
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
33%
33%
initial
33%
final
Mechanics Lecture 11, Slide 41
CheckPoint Response
A) Kinitial > Kfinal
B) Kinitial = Kfinal
C) Kinitial < Kfinal
initial
final
A) Since the two boxes stick together, this collision is
inelastic, and therefore energy is not conserved.
Mechanics Lecture 12, Slide 42
Relationship between Momentum & Kinetic Energy
1 2 2
p2
1 2
K = mv = m v =
2
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
p2
K Initial =
2M
K final
final
same p
p2
=
2(2M )
Mechanics Lecture 12, Slide 43
Checkpoint
A.
B.
C.
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
33%
33%
Before Collision
33%
m
M
After Collision
Mechanics Lecture 11, Slide 44
CheckPoint
M
m
m
A) m > M
Before Collision
M
After Collision
B) M = m C) M > m D) Need more information
B) In order to stop m in the collision, M must be
the same as m. If m were smaller, it would
bounce back and if m were bigger, it would
continue to move forward.
Mechanics Lecture 12, Slide 45
Newton’s Cradle
Mechanics Lecture 12, Slide 46
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
Mechanics Lecture 12, Slide 47
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
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 48
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) v B) -v C) 2v D) -2v E) zero
2v
m
v
2m
+
Before Collision
This is the CM frame
Mechanics Lecture 12, Slide 49
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 50