Saturday, November 12, 2011
HW8 #8
Is this question referring to the initial speed of the point or after the time interval given in #5?
HW8 #10
I seem to be stuck on this problem. I converted the rotational velocity so that the units would be rad/s instead of rev/min. I calculated the radius (in m) using the diameter (given in m). I plugged these values into the KE =(1/2)(m)(wr)^2 equation [w = lower case omega], but my answer was incorrect. What am I doing wrong?
Sunday, November 6, 2011
HW7 #7
Student question:
I tried to do #7, but it doesn't seem to work. I have the following values given:
A projectile of mass 5 kg is fired with an initial speed of 198 m/s at an angle of 25◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.2 kg and 1.8 kg . The 1.8 kg fragment lands on the ground directly below the point of explosion 2.8 s after the explosion.
The acceleration due to gravity is 9.81 m/s2 .
Find the magnitude of the velocity of the 3.2 kg fragment immediatedly after the explosion.
I know that the 1.8 kg has no horizontal velocity because it falls straight down, so I found the horizontal velocity of the 3.2 kg fragment by doing (5 kg)(198cos25 m/s)/(3.2 kg). I tried to find the vertical velocity by finding the initial vertical velocity of the 1.8 kg. I used X=V0t+1/2at^2 and used conservation of energy in the vertical direction to find the height, or X. I then put this velocity into conservation of momentum in the vertical direction and found the vertical velocity for the 3.2 kg fragment. Finally, I used the pythagorean triangle to find the resulting velocity. I've done this five times and it hasn't worked. Am I doing something wrong or missing some important step? If you could reply as soon as possible, I'd greatly appreciate it. Thanks!
Dr. Winters' response:
Very nice start. This is a fairly complex problem, where you need to look at kinematics and momentum. I like the way you found the horizontal component of the 3.2kg fragment using conservation of momentum in the horizontal direction.
I didn't follow how you found the vertical velocity of the 1.8kg fragment. It sounds as though you first found the vertical velocity and then found the height. I think that's backwards. You can find the height using kinematics (the explosion takes place at the top of the trajectory). Knowing the height and the time to fall, you can find the initial vertical velocity of the 1.8kg fragment.
After that, use conservation of momentum in the vertical direction, as you described, to find the vertical component of velocity of the 3.2kg fragment.
It sounds as though you are doing everything right except attacking the middle calculation backwards. Hope that helps. Good luck.
I tried to do #7, but it doesn't seem to work. I have the following values given:
A projectile of mass 5 kg is fired with an initial speed of 198 m/s at an angle of 25◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.2 kg and 1.8 kg . The 1.8 kg fragment lands on the ground directly below the point of explosion 2.8 s after the explosion.
The acceleration due to gravity is 9.81 m/s2 .
Find the magnitude of the velocity of the 3.2 kg fragment immediatedly after the explosion.
I know that the 1.8 kg has no horizontal velocity because it falls straight down, so I found the horizontal velocity of the 3.2 kg fragment by doing (5 kg)(198cos25 m/s)/(3.2 kg). I tried to find the vertical velocity by finding the initial vertical velocity of the 1.8 kg. I used X=V0t+1/2at^2 and used conservation of energy in the vertical direction to find the height, or X. I then put this velocity into conservation of momentum in the vertical direction and found the vertical velocity for the 3.2 kg fragment. Finally, I used the pythagorean triangle to find the resulting velocity. I've done this five times and it hasn't worked. Am I doing something wrong or missing some important step? If you could reply as soon as possible, I'd greatly appreciate it. Thanks!
Dr. Winters' response:
Very nice start. This is a fairly complex problem, where you need to look at kinematics and momentum. I like the way you found the horizontal component of the 3.2kg fragment using conservation of momentum in the horizontal direction.
I didn't follow how you found the vertical velocity of the 1.8kg fragment. It sounds as though you first found the vertical velocity and then found the height. I think that's backwards. You can find the height using kinematics (the explosion takes place at the top of the trajectory). Knowing the height and the time to fall, you can find the initial vertical velocity of the 1.8kg fragment.
After that, use conservation of momentum in the vertical direction, as you described, to find the vertical component of velocity of the 3.2kg fragment.
It sounds as though you are doing everything right except attacking the middle calculation backwards. Hope that helps. Good luck.
Tuesday, November 1, 2011
Adopt-a-Physicist open until Nov 22
We have a unique opportunity to talk to 5 different people who have physics backgrounds. They would like to share their knowledge and opinions. Please take advantage of this offer.
To log on: go to www.adoptaphysicist.org
Log in by scrolling down to our Class: NY: Smithtown AP-C Physics
and our Class PIN (supplied in class).
Read about our physicists and write to them. Write thoughtfully. Find out about their careers, their education, and ask for their suggestions for you. They are happy to answer just about any relevant question, often at length. Keep up a conversation.
Our physicists are:
Dr. Bill Freeman. Manager for a $278M particle accelerator project at Fermilab.
Dr. Bryan Gorman. Mathematical modeling (statistical prediction). One example was for the US Coast Guard for improved search and rescue and drug interdiction.
Dr. Michaela Kleinert. Physics professor at Willamette University in OR. Her interests are in atomic/molecular physics and optics.
Mr. Scott Dodd. Medical Physicist, works in radiation oncology at a cancer treatment center.
Dr. von Foerster. Retired; former university professor and publisher.
To log on: go to www.adoptaphysicist.org
Log in by scrolling down to our Class: NY: Smithtown AP-C Physics
and our Class PIN (supplied in class).
Read about our physicists and write to them. Write thoughtfully. Find out about their careers, their education, and ask for their suggestions for you. They are happy to answer just about any relevant question, often at length. Keep up a conversation.
Our physicists are:
Dr. Bill Freeman. Manager for a $278M particle accelerator project at Fermilab.
Dr. Bryan Gorman. Mathematical modeling (statistical prediction). One example was for the US Coast Guard for improved search and rescue and drug interdiction.
Dr. Michaela Kleinert. Physics professor at Willamette University in OR. Her interests are in atomic/molecular physics and optics.
Mr. Scott Dodd. Medical Physicist, works in radiation oncology at a cancer treatment center.
Dr. von Foerster. Retired; former university professor and publisher.
Sunday, October 23, 2011
HW6 #1
Does anyone know how to solve #1? I have already tried change in energy over time, P=Fv, etc...
HW6 #17
Hey I am having a problem getting #17 to work out on the homework with my numbers. When I tried solving it with other people's numbers the same way, I got the right answer, but mine arent working out. I have an 18 kg child on a playground swing moving with a speed of 4.7 m/s at the lowest point, and the swing is 1 meter long. I need to find the angle the string makes with the vertical. First I tried using conservation of energy with K=U, so .5mv^2=mgh. The m's cancel out and I am trying to solve for h, height above the lowest point. I end up with .5(4.7^2)=9.81h, and when I solve for h, I get 1.12, which doesn't work because it is higher than the length of the string. I also tried solving it by setting cos(theta)=(centripital acceleration)/g, and the acceleration is greater than g so it doesnt work. Please let me know if I am doing something wrong! Thanks!
Sunday, October 16, 2011
HW5 #18
Hey everyone, I just wanted to share a helpful tip for #18. Be careful when converting cubic kilometers to cubic meters. It's not a 1:1,000 ratio because the value is cubed.
Subscribe to:
Posts (Atom)