AP Physics
Conservation of Momentum
Head on Elastic Collisions
Elastic Collision?
No energy lost in collision
(1/2)m1v12 + (1/2)m2v22
= (1/2)m1v1f2 + (1/2)m2v2f2
Combine equations above with
m1v1 + m2v2 = m1v1′ + m2v2′
you get .....
For Head Collisions
V1 - V2 = V2' - V1'
Giancoli - page 189) 23
| m1 = .450 kg | m1 = .450 kg |
| V1 = 3.00 m/s | V1' = |
| m2 = .220 kg | m2 = .900 kg |
| V2 = 0 | V2' = |
V1 - V2 = V2 - V1
V2 - V1 = 3 m/s
V2 = 3 m/s + V1
Total Mom before = Total Mom After
| m1 = .450 kg | m1 = .450 kg |
| V1 = 3.00 m/s | V1' = |
| m2 = .900 kg | m2 = .900 kg |
| V2 = 0 | V2' = |
Total Mom before = Total Mom After
1.35 kgm/s = .45V1 + .90kg(3+V1)
1.35 = 1.35V1' + 2.7
V1'= -1 m/s or 1 m/s West
Since
V2 = 3 m/s + V1
then ...
V2 = 2 m/s East
Collisions in 2 Dimensions
Momentum is conserved
in x and y directions
Total Mom in X direction before =
Total Mom in X direction after
Total Mom in Y direction before =
Total Mom in Y direction after
If angle between = 90 degree ...
Add vectors using ....
a2 + b2 = c2
Textbook - p. 190) 41
Total Momentum Before
a2 + b2 = c2
Eagle 1:
mv = 4.3 kg(7.8m/s) = 34 kgm/s
Eagle 2:
mv = 5.6kg(10.2 m/s) = 57kgm/s
Total Mom before:
a2 + b2 = c2
342 + 572 = c2
Total mom before = 66 kgm/s
66 kgm/s = (5.6kg + 4.3kg)v
V = 6.7 m/s
problem 42)
Gravity, Friction Momentum
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