Physics AP - Elastic Collision Lab (#14)

Background:
In a perfect elastic collision the two objects hit each other and
then bounce off each other with no deformation.  In a perfect elastic
collision momentum and kinetic energy are conserved.  In this
collision their well be loss of kinetic energy because the collision
is not a perfect elastic collision, which don't exist. And there will
be a deformation of the balls.
M1V1 + M2V2 = (M1 + M2)Vf. elastic collision
m1v1i + m2v2i = m1v1f + m2v2f. momentum conserved

Problem:
To find out how close from ball one (B1) the collision balls added
together would be.

Procedure:
1. Drop a ball from the bottom of the ramp to get a drop point.
2. Mark the point with a piece carbon paper.  Label the point drop
   dot.
3. From the top of the ramp let go of B1.
4. Repeat step 2 but label it B1.
5. Place B2 at the bottom of the ramp so when B1 is rolled the two
   balls collide the barely touch.
6. Repeat step 2. for B1 and B2 label it B1c and B2c.
7. Then measure the distance from drop dot to B1.
8. After that measure the distance from drop dot to B1c and then add
   the distance from B1c to B2c.
9. then find the subtotal..

Data:
Drop Dot    B1         B1c             B2c
0.0m        44.92m    sin(32)×35.6m    sin(61)×26.4m
                      18.9m            23.1m

B1c + B2c             B1-(B1c + B2c)
18.9m + 23.1m         44.92-(42)
42m                   2.92m

 B1
 |  B2c
 |   \
 |    \
 |    / B1c
 |  /
 |/
 DP

Conclusion/Analysis:
The distance between B1 and B1c + B2c was 2.92m , which shows that
the collision was not perfectly elastic, or the distance would have
been the same.  The energy was probably lost in heat, noise or
deformation.