For 1a, the friction force is -20 Newtons. Treat that as given and don't worry about mu.
The push force is 35 N. So the net force is 15 N.
After 4 seconds the pushing stops and the only force is the -20 N friction force. Does that help?
Has anyone done number 2? I got 140m for the height of the tower, and 140m for the highest trajectory, which doesn't make sense. I think I'm calculating the initial y velocity wrong...?
I think I speak for most of the students here and find this method of studying extremely frustrating and inefficient. By not posting an official answer key, we are spending more time lost and confused. Many of us want to engage with the material and learn the interesting concepts behind each question, but we are stuck. It is as if the blind is leading the blind. Please reconsider the format of this class and take into consideration the needs of the hard working students of this course. Most of us will have to take a second physics class in the series and having a strong foundation of fundamental mechanical concepts is crucial. We all want to come out of this more knowledgeable and appreciative of physics, but no answer key is making that very difficult.
To the students, please post what you have gotten for the answers. We got this and I wish you all the best on the final!
Thanks. I think one point of confusion was regarding mu for problem one. I posted a note at the top of this post to clarify that. Please read that.
I posted a solution to problem 2, and some other solutions, in the post Practice Problem Notes (top post). I think this will really help. I think it is important to find a balance in these things. In physics, the focus is on solving problems by thinking about them in the moment, not on remembering a problem solution that you have seen previously as done by an expert. We want to help you develop the tools to tackle problems with thought and intuition and physics problem solving skills.
PJ. I think you understand it really well. Once it stops, it is stopped. But if you are unsure on an exam, you can explain your reasoning, because your reasoning is really good. But basically, follow your intuition, which is that it stops, and trust that over your equation, which is outside its region of validity. Understanding the range of validity of an equation is a key skill; maybe more important than the equations themselves.
For question 3a, is the velocity for t=-2 up right before t=0 20m/s and then the velocity drop to -5m/s (negative for direction change)at t=0 until t=6?
What I got was 40m/s. I got this by using SOHCAHTOA on the triangle made by the 37 degree angle and the Voy found in part a (30m/s). So it's tan(37 degrees) = opposite/adjacent = 30m/s / Voy
Excellent and correct. The method is a little different. Physics help (comment just above, got everything with a more familiar method, (but both are fine). " Sure thing, I started by calculating the velocity after the collision of block 2 (10kg), this will be the velocity at t=0. Also, it says that Potential energy is 0 as the collision occurs, this means that X(0)=0. Because x(0)=0, we can find out that the phase shift=-pi/2 (draw the sin wave). Knowing all this, I set up the velocity equation for v(t)=-A*w*Sin(wt-phi)--> 10=-A*pi/2*Sin(-pi/2), which gives A=6.37. From there you should have everything you need for a regular SHM problem.
Has anyone done number 2? I got 140m for the height of the tower, and 140m for the highest trajectory, which doesn't make sense. I think I'm calculating the initial y velocity wrong...?
ReplyDeleteCatherine. That sounds totally right. i think the launch height is 35 and the top is at 80 m. Does that sound right?
DeleteHey Zach, do you think you could upload the full equation sheet we will be able to use on the final?
ReplyDeleteFor the collision problem in #4, do we assume it is an elastic collision?
ReplyDeleteI think I speak for most of the students here and find this method of studying extremely frustrating and inefficient. By not posting an official answer key, we are spending more time lost and confused. Many of us want to engage with the material and learn the interesting concepts behind each question, but we are stuck. It is as if the blind is leading the blind. Please reconsider the format of this class and take into consideration the needs of the hard working students of this course. Most of us will have to take a second physics class in the series and having a strong foundation of fundamental mechanical concepts is crucial. We all want to come out of this more knowledgeable and appreciative of physics, but no answer key is making that very difficult.
ReplyDeleteTo the students, please post what you have gotten for the answers. We got this and I wish you all the best on the final!
Thanks. I think one point of confusion was regarding mu for problem one. I posted a note at the top of this post to clarify that. Please read that.
DeleteI posted a solution to problem 2, and some other solutions, in the post Practice Problem Notes (top post). I think this will really help. I think it is important to find a balance in these things. In physics, the focus is on solving problems by thinking about them in the moment, not on remembering a problem solution that you have seen previously as done by an expert. We want to help you develop the tools to tackle problems with thought and intuition and physics problem solving skills.
Exactly what I mean, Sara. I think its important we work through the problems step by step, but learning is useless if we cannot check if it is right.
DeleteThanks Sara.
DeleteFor 3, for the 10 kg mass,
v(2)=-10 m/s.
x(1) = 6.3 m
Everything else has been already posted here and in the post "practice problem notes" (top post)
some. see post above.
ReplyDeletePJ. I think you understand it really well. Once it stops, it is stopped. But if you are unsure on an exam, you can explain your reasoning, because your reasoning is really good. But basically, follow your intuition, which is that it stops, and trust that over your equation, which is outside its region of validity. Understanding the range of validity of an equation is a key skill; maybe more important than the equations themselves.
ReplyDeleteShelby, you are not wrong. What force in the problem statement would make it go backwards? Physics should make sense. It is not mysterious.
ReplyDeleteExcellent point anonymous!
ReplyDeleteFor problem 3 does SHM start right when the masses impact or after the spring has fully compressed and is moving back?
ReplyDeleteright after the collision
DeleteWhat is the velocity of the 10 kg mass right after the collision? What is its position right before and right after the collision?
ReplyDeleteFor question 3a, is the velocity for t=-2 up right before t=0 20m/s and then the velocity drop to -5m/s (negative for direction change)at t=0 until t=6?
ReplyDeleteYes, that's what I got. I especially wanted to verify that it's negative due to the direction change.
DeleteThis comment has been removed by the author.
ReplyDeleteWhat I got was 40m/s. I got this by using SOHCAHTOA on the triangle made by the 37 degree angle and the Voy found in part a (30m/s). So it's
Deletetan(37 degrees) = opposite/adjacent = 30m/s / Voy
You can just use 10 m/s^2 for g instead of 9.8
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteExcellent.
ReplyDeleteExcellent and correct. The method is a little different. Physics help (comment just above, got everything with a more familiar method, (but both are fine).
ReplyDelete"
Sure thing, I started by calculating the velocity after the collision of block 2 (10kg), this will be the velocity at t=0. Also, it says that Potential energy is 0 as the collision occurs, this means that X(0)=0. Because x(0)=0, we can find out that the phase shift=-pi/2 (draw the sin wave). Knowing all this, I set up the velocity equation for v(t)=-A*w*Sin(wt-phi)--> 10=-A*pi/2*Sin(-pi/2), which gives A=6.37. From there you should have everything you need for a regular SHM problem.