Lab 7: 2-D inelastic collisions
Only 5 more to go!!
Last week we learned that momentum is conserved in collisions. This led us to the
law of conservation of momentum:
 
pi  p f




m1v1i  m2v2i  m1v1 f  m2v2 f
Also remember last week that the collision happened all in one dimension so we
didn’t have to worry much about the fact that momentum is a vector quantity. But
this week the collision will occur in two dimensions so you will need to write the
conservation of momentum in the x and y directions:




x : m1v1i X  m2v2iX  m1v1 f X  m2v2 f X




y : m1v1iY  m2v2iY  m1v1 fY  m2v2 fY
y
How would we write the momentum of this
object?

p  pX xˆ  pY yˆ
V
What is pX and pY ?

x
mass, m
p X  mvX  mv cos 
pY  mvY  mv sin 
Consider this collision. What is the speed of mass 1 and mass 2 after they hit?
v1f
m1
v1i
m1
m2
V2i= 0


m2
v2f
To solve this we first write the conservation of momentum in the x and
y directions:




x : m1v1i X  m2v2iX  m1v1 f X  m2v2 f X




y : m1v1iY  m2v2iY  m1v1 fY  m2v2 fY
Since V2i = 0 the equations for conservation of momentum become:



x : m1v1iX  m1v1 f X  m2v2 f X



y : m1v1iY  m1v1 fY  m2v2 fY
Now substitute in for the velocity components:



x : m1v1i  m1v1 f cos   m2v2 f cos 


y : 0  m1v1 f sin   m2v2 f sin 
From the equation in the y-direction we can get:


y : m2 v2 f sin   m1v1 f sin 
Now we can solve for V2f !!!!

y : v2 f 

m1v1 f sin 
m2 sin 
Now substitute this equation for V2f into the equation in the x-direction:


x : m1v1i  m1v1 f cos   m2
Solve for V1f and you get this big equation:

m1v1 f sin 
m2 sin 
cos 


v1i
v1 f 
cos   sin  cot 
Now that you have V1f you can substitute back in and find V2f

v2 f

v1i
m1 sin 


m2 sin  cos   sin  cot 
Let’s look at another example: What type of collision is this?
BEFORE COLLISION
+x
Cart #1
mass = m1
velocity = v1i
Cart #2
mass = m2
AFTER COLLISION
+x
Cart #1
velocity = v2i=0
Cart #2
mass = m1+m2
velocity = v(1+2)f
What’s the velocity of the combination after the collision?
We can use cons. Of momentum:
m1v1 + m2v2 = (m1 + m2) v(1+2)f
; but v2 = 0 so this becomes:
m1v1 = (m1 + m2) v(1+2)f
1 1
(1 2 ) f
1
2
v
mv

m m
What if I wanted to know how much Kinetic Energy was lost?
All you need to do is calculate the initial KE of the system then calculate the
final KE of the system and take the difference.
Initial KE of system:
KEi = ½ m1v12 + ½ m2v22 = ½ m1v12 + 0 = ½ m1v12
Final KE of system:
KEf = ½ (m1 + m2) (v(1+2)f)2
KE lost:
KEf- KE1 = ½ (m1 + m2) (v(1+2)f)2 - ½ m1v12
What are we going to do today?
We are going to look at a 2-d collision and see if momentum is
conserved and whether of not KE is conserved