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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-on Collisions:

V1 - V2 = V2' - V1'

↓ Scroll for Problem 1 ↓

Giancoli - page 189) 23

m1 = .450 kg

V1 = 3.00 m/s

m2 = .900 kg

V2 = 0

↓ Scroll for Step 1 ↓

V1 - V2 = V2' - V1'

V2' - V1' = 3 m/s

V2' = 3 m/s + V1'

↓ Scroll for Step 2 ↓

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)

↓ Scroll for Final Step ↓

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 degrees ...
Add vectors using ....

a2 + b2 = c2

Textbook - p. 190) 41

Total Momentum Before

↓ Scroll for Step 1 ↓

Eagle 1:

mv = 4.3 kg(7.8 m/s)
= 34 kgm/s

Eagle 2:

mv = 5.6 kg(10.2 m/s)
= 57 kgm/s

↓ Scroll for Step 2 ↓

Total Mom before:

a2 + b2 = c2

342 + 572 = c2

Total mom before = 66 kgm/s

↓ Scroll for Final Answer ↓

66 kgm/s = (5.6kg + 4.3kg)v

V = 6.7 m/s

problem 42)

Gravity, Friction Momentum

Chapter Review
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