Someone has to write this...

Since nobody is doing the blog and I am contemplating what to do tomorrow, I'll write a quick version of what happened today.

Since Professor Mason is off in Arizona somewhere doing something Physics? related, Professor Christensen was kind enough to sub in until his 7pm Physics 2A class. After the short introduction, we began our wonderful "mini-celebration of knowledge" as it covers over the center of mass momentum.

Paraphrasing from my memory, the first question asked how far would the fisherman move if he tied a string to his cabin's door knob and pulled on a frictionless ice lake.

• Fisherman - 80 kg
• Log Cabin - 250 kg
• Distance in between - 10 m

Using the center of mass of a system defined as the average of their positions , weighted their masses m_i:

Basically you take ((80*0)+(250*10)) / (250 + 80) and we should get around 7.58 m since we treat the fisherman as the origin.

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The second problem had 6 parts to it involving a desperate fisherman applying his momentum force onto the cabin to save his hide.

A frustrated ice-fisherman with a mass of 80.0 kg and his 250 kg fishing cabin are sliding together across a frozen, frictionless lake towards thin ice at a speed of 1.60 m/s to the left. The cabin is ahead of the fisherman. Upon seeing the "thin ice" sign, the frustrated person pushes against the cabin for .300 seconds, giving himself a velocity of 32.0 cm/s in the opposite direction.

1. What is the magnitude and direction of the total momentum of the person and the cabin after they separate?
2. What is the speed of the cabin after they separate?
3. How much average force does the person exert on the cabin?
4. How much work did the person do on himself and the cabin in order to give himself a velocity of 32.0 cm/s in the opposite direction?
5. What does a fishing cabin become when it is surrounded by molecules of H₂O at 10.0^o C?

So we have our givens:

mp	= 80.0 kg	(Mass of the person.)
mc	= 250 kg	(Mass of the cabin.)
vo	= 1.60 m/s	(Initial velocity of the person and cabin.)
vp	= .320 m/s	(Final velocity of the person.)
vc	= ?	(Final velocity of the cabin.)
tp	= .300 s	(Time the person pushes.)
Wp	= ?	(Work done by the person.)
Fave	= ?	(Average force exerted by the person.)

Quick rundown:
1. Conservation of momentum - Final momentum must equal to Initial momentum
(250 kg+ 80 kg)(1.6 m/s) = 528 kg*m/s or N*s

2. Again, conservation of momentum
((250 kg + 80 kg)(1.6 m/s) + (80 kg)(0.320 m/s)) / 250 kg = 2.21 m/s

3. The impulse exerted on the cabin is equal to the average force multiplied by the time
(80 kg)(0.320 m/s +1.6 m/s) = 153.60 m/s

Force avg = Impulse / Time so 153.60 m/s / 0.300s = 512 N

4. Avg Work = Final KE - Initial KE or
(0.5)(250 kg)(2.21)^2 + (0.5)(80 kg)(0.320 m/s)^2 - (0.5)(250 kg + 80 kg)(1.6 m/s)^2
612.94 J + 4.096 J - 422.40 J = 194.64 J or 195 J

5. ......
Wet!

Or refer to Mason's Momentum notes.

After the quiz, Christensen gave us an explanation of kinematics in class about rotational kinematics which uses the same kinematics that we learned in 2A. Or refer also to Mason's Rotational Kinematics notes.

Note: Tomorrow - 4/18.... sigh.