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Find the Acceleration

  1. Oct 4, 2014 #1
    I'm working on my University Physics 1 homework and I need someone to coach me through the problems. I only have until 10/05/14 (Tomorrow at 11p.m.) and I need to get this done, but I have been struggling with what equations to use and how to solve it (The 2 books that came with the course do a poor job of telling me every equation that I need to know, so for the past month I have been trying to get help and figure the equations out on my own without being shown the proper steps to work the problems-- its gone rather unsuccessfully and only half the class is remaining this semester).
    Final edit: Please note that my College does not have a Physics tutor, as the only one available has transferred to UTD. I have been collaborating with the teacher and the students as best I can, but nothing seems to 'click'.

    1. The problem statement, all variables and given/known data

    Two children on ice skates pull toward each other on a rope held taut between them. The inertia of one child is 34kg , and the inertia of the other is 26kg .

    2. Relevant equations
    If at one instant the 34-kg child is accelerating at 1.3m/s2 to the left, what is the acceleration of the 26-kg child at that instant?
    Express your answer with the appropriate units.

    3. The attempt at a solution
    For the solution, I thought that it would be an equation that only involved the inertia and acceleration of child 1 to somehow obtain the acceleration of child 2. But that is exactly where I'm getting stuck.
     
  2. jcsd
  3. Oct 4, 2014 #2

    SteamKing

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    Homework Helper

    Well, right off the bat, your text is using the term 'inertia' in a sloppy fashion here. What they are really describing is the 'mass' of the children, which is measured in units of kilograms. Later on, you will learn that the term 'inertia' is used in a different context altogether, and is measured in different units.
     
  4. Oct 4, 2014 #3
    I thought as much. So for starters, how would I write this equation? I'll check back if I continue to have issues.
     
  5. Oct 5, 2014 #4
    I just was looking for this question and found your thread so I made an account to help you out on this one.
    It is conservation of momentum, so m1v1 = m2v2
    Hence (34)(1.3) = (26)(v2)
    When you solve you get the second child's acceleration at 1.6 m/s^2.

    Hope I could help!
     
  6. Oct 5, 2014 #5
    Thank you! However, I've just discovered that answer is incorrect both when I use your answer and try to make it negative. The correct answer (which I just barely managed to get with 2-3 attempts left) was 1.7m/s^2.
     
  7. Oct 5, 2014 #6
    That's strange, because I just had that question on masteringphysics for me and it accepted my answer.
     
  8. Oct 5, 2014 #7
    That's really weird then, I think something is obviously up with this class. At least now I'll have this in my notes for future reference for when I eventually have to retake the course. Again, thank you for your help.
     
  9. Oct 5, 2014 #8
    Have you drawn a free body diagram on one of the skaters yet? If T is the tension on the rope, what does Newton's 2nd law tell you about the relationship between the tension T and the mass times acceleration of the first skater? What is the tension in the rope? This same tension is also causing the other skater to accelerate. What does Newton's 2nd law tell you about the relationship between the tension T and the mass times acceleration of this other skater? What is the acceleration of this skater?

    Chet
     
  10. Oct 5, 2014 #9
    It's not asking for tension. also, this problem has been completed already. I see what I did wrong now. When I went to calculate the answer again, I got it right.
     
  11. Oct 5, 2014 #10
    Well excuse me. I was merely trying to provide an alternate, and, in my best judgement, a better way of analyzing this problem.

    Chet
     
  12. Oct 5, 2014 #11
    I think you read my tone wrong, all I said was that it wasn't asking for the tension, I know that you were trying to help. I'm sorry for any misunderstanding that may have occurred.
     
  13. Oct 5, 2014 #12
    Thank you. No hard feelings. I hope that you consider trying the problem the way I suggested. It should only take a couple of minutes, and I think it would be worth the time.

    Chet
     
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