# Collisions by WLae56

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```									Collisions
The previous example is an example of a collision. Collisions play a
central role in many parts of modern physics and are the basis of our current
understanding of particle physics. The essential effect of collisions is to
redistribute the total momentum of the colliding objects. We can classify all
collisions into one of three categories:

1. Elastic collisions, which conserve kinetic energy,
2. Inelastic collisions, which do not conserve kinetic energy, and
3. Completely inelastic collisions, in which the objects stick together
afterwards.

No matter what type of collision occurs, we can study them all in the same way.
The guiding principle is that of conservation of linear momentum.
A simple way of dealing with collisions is to use the concept of the center
of mass. Suppose we have a collection of n particles, with a mass of m1,
m2,...,mn respectively. Also, let the position of the particles be given by (x1,y1),
(x2,y2),...,(xn,yn). Then recall that the center of mass for this system is at
n


m x        i i
xCM    i 1
n

m
i 1
i

We can now replace all of our equations of motion with equations involving the
center of mass; thus

      x CM
v CM 
t
n               
 mi
i 1
xi
t
        n

m  i 1
i

n
(54)

 mi vi
    i 1
n

m
i 1
i


pCM

m

and
       
Fext  ma CM

v
 m CM
t
n             
m                vi
i t
(55)
m       i 1
n

m  i 1
i

n

  mi a i
i 1

where aCM follows from the above definitions. Notice that the force in (55) is the
external force only. All of the internal forces acting between the particles will
cancel out, leaving only the forces acting on the system as a whole. From this
we see that when a system of particles is acted on by external forces, the center
of mass moves as if all of the mass was centered at the center of mass and all of
the forces acted as a single resultant force at that point.
Elastic Collisions
Now let us look at collisions, starting with elastic collisions. Elastic
collisions are those that also conserve kinetic energy. It can be shown that due
to the conservation of kinetic energy, the relative velocities before the collision
are equal and opposite to the relative velocities after the collision. Thus, we can
write for two particles
           

v2  v1   v2'  v1'                        (56)

In the center of mass frame, this implies that the two particles will approach each
other at some particular speed and after the collision they will recede from each
other with their velocity vector having the same magnitude but opposite sign.

Example:
A ball of mass m1 = 0.1 kg traveling with a velocity v1 = 0.5 m/sec collides
head on with a ball of mass m2 = 0.2 kg which is initially at rest. Calculate the
final velocities, v1' and v2', in the event that the collision is elastic, in the
laboratory frame of reference.
v
1                                               v'   v'
1    2

m                   m                                 m    m
1                   2                                 1    2

Using momentum conservation we have
                     '
m1v1  m2 v2  m1v1'  m2 v2
m1v1  m1v1  m2 v2
'       '

solving for v2', get

v2 
'        m1
m2

v1  v1'      
We still need v1'. So using (56) we get

'

v1   v 2  v1'       
m                 

v1    1 v1  v1'  v1'     
 m2               
 m1          m     
     1 v1   1  1 v1'
 m2           m2   
m1  m2
v1'              v1
m1  m2
-0.1 kg 0.5 m
             s
0.3 kg
 017
.        m
s

and
m1      m  m2 
v2 
'
 v1  1     v1 
m2      m1  m2 
m1  2m2 
                v1
m2  m1  m2 
2m1
           v1
m1  m2
2 0.1 kg
                       0.5 m 
0.3 kg               s

 0.33    m
s

Example:
Solve the same problem as above, but in the center of mass frame.

In this case, the center of mass is moving with a velocity
       
      m1v1  m2 v 2
v CM 
m1  m2

m1v1

m1  m2
0.1 kg0.5 m  

s
i
0.1 kg  0.2 kg
 017
.        m
s    i

so the velocities of the balls relative to the center of mass is
     
u1  v1  v CM
 0.5 m i  0.17
s
m
s   i
 0.33   m
s   i

     
u2  v 2  v CM
 0  0.17               m
s   i
 0.17          m
s   i

Since kinetic energy must be conserved, we get u1' = -u1, u2' = -u2 and thus,
converting back into the lab frame,
      
v1'  u1'  v CM
 0.33           m
s   i  0.17 m i
s

 017
.         m
s    i

       
v 2'  u2'  v CM
 -0.17          m
s    i  0.17 m i
s

 0.33    m
s   i

Completely Inelastic Collisions
Now let us turn to completely inelastic collisions.

Example:
A 15 g bullet is fired into a 10 g wooden block that is mounted on wheels.
If the block moves 180 m in one second, what was the muzzle velocity of the
bullet?

m
b                                                                V
m
v                                        t
b

Using momentum conservation we have

mb vb  mt v t  mb  mt V
mb vb  mb  mt V

or

vb 
mb  mt V
mb
0.010 kg  0.015 kg180 m 
                           s

0.015 kg
 300     m
s

The change in kinetic energy is
K    1
2mb  mt V 2  2 mb vv2
1

 2  0.025 kg180 m   2  0.015 kg300 m 
1                     2    1                  2
s                     s

 270 J

Example:
If the rifle had a mass of 2.5 kg, what was the recoil velocity of the rifle in
the previous example?

The total momentum of the rifle + bullet was zero before the rifle was
fired, so by conservation of momentum,
        
0  mr v r  mb vb
  mr v r  mb vb
mb
vr       vb
mr
0.015 kg
               300 m 
2.5 kg       s

 18
.     m
s

Notice that, for recoils, if we take the ratio of the velocities of the objects we get
(since the initial momentum was zero)

v2 m1
                                   (57)
v1 m2

similarly, the ratio of the kinetic energies is
2
1
m1v12 m1  v1 
2
   
1
2   m2 v 2 m2  v 2 
2

2
m m 
 1  2                            (58)
m2  m1 
m2

m1

```
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