Solving 2D momentum 4.5 in text

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

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

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. )

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

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]

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°

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]

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

Now solve: X plane: pto = ptf 0 = p1f + p2f 0 = -31.2 + p2f (remember stone #1 went west = -ve) p2f = 31.2 kg·m/s [E]

Now solve: X plane: pto = ptf 0 = p1f + p2f 0 = -31.2 + p2f (remember stone #1 went west = -ve) p2f = 31.2 kg·m/s [E] Y plane: pto = ptf m1v1o + m2v20 = m1v1f + m2v2f 97.5 + 0 = 54.0 + ptf ptf = 43.5 kg·m/s [N] Remember – this is 2nd stone!

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

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…

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

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?

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!