Transcript P221_2008_week7 - Department of Physics | Indiana University

Chapter 7 problems
Chapter 7 problems
(42 (3 votes), 57 (2votes))
Chapter 7 problems
(42 (3 votes), 57 (2votes))
Chapter 8 Problems (cont.)
(36 (5 votes), 53 (3 votes))
Chapter 8 Problems (cont.)
•
The book gives a mathematical expression for the coordinates (X,Y,Z) of
the center of mass (CoM) of a collection of particles. Someone has
suggested that these formulas amount to stating that the CoM is the massweighted average position of the collection of particles. Comment briefly on
the validity/universality of this statement. (13 Yes; 9 No; 6 didn’t understand
or gave confusing answers; 15 no answer)
• I don't think it's valid, because formula doesn't exactly
give us the mass-weighted average position.
• You know, unless I'm understanding the question
incorrectly. It is worded somewhat strangely.
• If by that statement the person means the relative mass
weights the position, I would consider that valid universal
as it basically just restates the formula. The word
average makes me hesitant because that usually implies
dividing by the number of particles which does not
happen in this case. (This is the idea of a weighted
average!)
• This statement appears to be valid. If a m1 is heavier
than m2, then the center of mass (the mass weighted
average position) will be closer to m1 than m2. This
appears to be valid. (Does that help??)
Weighted averages
e.g. Neutron lifetime
If you wanted to average all of these experimental results, would you just add
them up and divide by 8? No not all results are equally precise, so you’d
weight each of them by something like 1/(error bar)
http://www.iucf.indiana.edu/annrep/2004/n-lifetime.pdf
Chapter 9 Problems
Figure 9.6 The Grand Jete’
Figure 9.4
The Negative mass trick’
Figure 9.8,9 Impulse
• A softball (m=0.2kg) approaches a bat at a speed of 25 m/s in a
direction that is roughly parallel to the ground. After making
contact with the bat for roughly 0.1 sec, the ball reverses
direction and is traveling at 30 m/s. What is the average force
exerted on the ball by the bat? Explain briefly how you arrived at
your answer. (12 forgot that v is a vector (|vf-vi|=55m/s not 5m/s;
9 made other mistakes of various sorts; 6 got it right, 17 didn’t
answer).
• Favg=-change in m/change in time exchange in velocity.
Average force is 20 m/s. (units??)
• Favg = -(1(0.20kg)/(0.1s))(30m/s-25m/s); Favg = 10 N
• 11N. I multiplied the mass of the bat with the speed of the
ball for each direction. One direction product is 5N the other
is 6N. I added the two and got 11N. I'm not sure what I am
supposed to do with the .1sec.
• Since momentum equals mv, the initial momentum is 5.0
kgm/s, and the final momentum is -6.0 kgm/s, so the
change in momentum is 11kgm/s, so the average force
exhibited on the ball is 11.0 N.!! (remember to watch units!!)
Chapter 9 problems
b. What was the direction of travel before the explosion?
Chapter 9 problems