1. Solving 2D momentum
4.5 in text

2. Note: We have the Law of Conservation of Momentum.
So we solve conserving momentum (not velocity or anything else)

3. (FYI no need to write what I include in brackets)

4. Example #7 page 203 in text. Read 1st.
(We are looking at just before and just after collisions so we do not need to worry about loss of speed due to friction. )

5. Draw this because it will go away on next slide!
Example #7 page 203 in text.
E +
Before
After
30° m2
V20 = 0 m/s m2 m1
V2f = ?
V1f = 3.2 m/s [N 30° W] m1
V10 = 5.0 m/s [N]
Draw this because it will go away on next slide!
m1 = m2 = 19.5 kg

6. Before the collision
X-Plane Pto = 0 in x plane (no W/E movement)
Y-Plane Pto is all in y plane due to 1st curling stone.
p = mv
p = (19.5 kg)(5 m/s) = 97.5 kg·m/s [N]

7. Now break oblique angles into x & y planes. After the collision:
( We only have information on 1st stone. Calculate the momentum and then break momentum into components.)
30°

8. Now break oblique angles into x & y planes. After the collision:
( We only have information on 1st stone. Calculate the momentum and then break momentum into components.)
30°
p = mv
p = (19.5 kg) (3.2 m/s)
= 62.4 kg·m/s [N 30° W]

9. Now break oblique angles into x & y planes. After the collision:
We only have information on 1st stone. Calculate the momentum and then break momentum into components.
px = sin 30° (62.4) = 31.2 kg·m/s
30°
p = mv
p = (19.5 kg) (3.2 m/s)
= 62.4 kg·m/s [N 30° W]
py = cos 30° (62.4) = 54.0 kg·m/s

10. Now solve:
X plane:
pto = ptf
0 = p1f + p2f
0 = p2f (remember stone #1 went west = -ve)
p2f = kg·m/s [E]

11. Now solve:
X plane:
pto = ptf
0 = p1f + p2f
0 = p2f (remember stone #1 went west = -ve)
p2f = kg·m/s [E]
Y plane: pto = ptf
m1v1o + m2v20 = m1v1f + m2v2f
= ptf
ptf = kg·m/s [N]
Remember – this is 2nd stone!

12. Now add the x & y components
Px = 31.2 kg·m/s [E]
Py = 43.5 kg·m/s [N]

13. Now add the x & y components
Px = 31.2 kg·m/s [E]
Py = 43.5 kg·m/s [N]
Final momentum = solve by Pythagorean (a2 + b2 = c2)
Θ = tan-1 (31.2/43.5)
Final answer…next slide…

14. The 2nd curling stone has a momentum of 53.5 kg·m/s [N 35.6° E]
But remember, we wanted the velocity!

15. The 2nd curling stone has a momentum of 53.5 kg·m/s [N 35.6° E]
But remember, we wanted the velocity!
So p = mv
53.5 = 19.5 v
v = 2.7 m/s [N 36° E]
Look at 1st diagram. Does this answer make sense?

16. Homework:
Do page 210
# 1,2,3,4
Hints:
#1. Watch +/- signs
#2. What is the mass after if they are locked together?
#3. is a grenade problem. It is at rest initially so Momentum = 0 in both x & y plane.
#4. No hint needed.
Check answers in back binder!