Physics 2111
Unit 12
Today’s Concepts:
a) Elastic Collisions
b) Center-of-Mass Reference Frame
Mechanics Lecture 12, Slide 1
Perfectly Elastic Collisions
WNC = 0 = DME
If DPE = 0 (and it normally is) then
DKE = 0
 KEo = KEf
What does this imply?
(v1o – v2o) = - (v1f – v2f)
Mechanics Lecture 12, Slide 2
Example 12.1 (Two boxes elastically collide)
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?
m2
m1
m1
m1
x
m2
m2
Mechanics Lecture 12, Slide 3
Special Case
What if v2o = 0?
v2f = 2m1/(m1+m2)
v1f = (m1 – m2)/(m1+m2)
Mechanics Lecture 12, Slide 4
Special Case
What if v2o = 0?
m1*v10 = m1*v1f + m2*v2f
v1o = v2f – v1f
v1f =v2f-v1o
v2f=v1o+v1f
Mechanics Lecture 12, Slide 5
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
Mechanics Lecture 12, Slide 6
Center of Mass Frame & Elastic Collisions
When viewed in the CM frame, the speed of both
objects is the same before and after an elastic
collision. They just reverse directions.
v*
1,i
m1
m1
v*1, f
m1
m2
v*2,i
m2
v*2, f
m2
Mechanics Lecture 12, Slide 7
Example 12.2 (Center of Mass Calculation)
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?
m2
m1
m1
m1
x
m2
m2
Mechanics Lecture 12, Slide 8
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 9
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 10
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 11
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 12
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 13
Example 12.3 (Spring compression)
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 with k = 20N/m is attached to the first
glider compresses and relaxes during the collision, but there is
no friction (i.e. energy is conserved). What is the maximum
compression of the spring?
m2
m1
m1
m1
x
m2
m2
Mechanics Lecture 12, Slide 14
Example 12.4 (2D inelastic blocks)
2m/sec
30o
5m/sec
m1
m2
Two blocks, m1=3kg and m2 = 2kg,
are sliding across a frictionless
floor as shown. They collide and
stick together.
What is the direction and
magnitude of their final velocity?
Mechanics Lecture 12, Slide 15
Example 12.5 (2D elastic pucks)
f
m1
m2
a
v1o = 3m/sec
Two punk, m1=2kg and m2 = 1.2kg, collide elastically
while sitting on a frictionless table. They collide
slightly off center, so that m1 is set upwards while m2
is sent downwards.
What is the values of angles f and a?
Mechanics Lecture 12, Slide 16