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Question on theory for solving 2015 AP Physics C Free Response Questions

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data

    Here are the free response questions:

    https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_frq_physics_c-m.pdf

    Here are the solutions:
    https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_physics_cm_sg.pdf

    I do not understand how they solved a few of the questions
    2. Relevant equations

    more below
    3. The attempt at a solution
    So for 1. e.
    Why do they have the difference in slopes of the two line segments with the the first segment being steeper than the second? And why is the point before tf/2 and not at tf/2?
    In other words.... what is the explanation behind the way they drew the graph the way they did?

    And then for 3. (c)
    Why do they choose to have velocity squared for the y-axis and the length for the horizontal axis?
    I see how the units of the slope of the line match the acceleration... but I do not understand the theory that allows g to be calculated from it. Could anyone please explain?

    Thanks in advance so much!
     
  2. jcsd
  3. Mar 2, 2016 #2
    For your second question as to why they are plotting velocity squared on the y-axis and length on the horizontal axis. Taking the equation derived in part 3 (b) ##v=\sqrt{3gL}## and squaring it we obtain ##v^2=3gL##, now note that ##3g## is a constant and ##L## is the independent variable. Therefore as ##L## increases linearly it is being multiplied by ##3g##. Perhaps it's easier to see if you write the equation as ##v^2=(3g)L##.

    Ex: Suppose ##L## increases from 0 to 1, then you will have ##v^2(1)=(3g)(1)=3g## and ##v^2(0)=(3g)(0)=0##, your slope is then ##m=\frac{3g-0}{1}=3g##
     
  4. Mar 2, 2016 #3
    Yes!! I see now! Thanks so much.

    Could ypu please take a look at my question regarding 1e? Thanks a lot!
     
  5. Mar 2, 2016 #4

    haruspex

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    What two forces act on the block? Does each have a component parallel to the slope?
    When the block is going up the slope, do those two components act in the same direction or in opposite directions?
    What about when going down the slope?
    Which case will produce the greater net force?
     
  6. Mar 2, 2016 #5
    Forces acting: friction and gravity. Both have such components.

    When block is going up the slope, they act in the same direction.
    When block is going down, they act in opposite direction.

    Going up the slope will produce a greater magnitude of net force.... ... so that is why the segment of the velocity vs time is steeper... more acceleration.

    Yes?

    If I got this right, then I do not understand why the point is before tf/2
     
  7. Mar 3, 2016 #6

    haruspex

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    That's all correct.
    What do you know about distance travelled and features of the velocity-time graph?
     
  8. Mar 5, 2016 #7
    OK. so my distance traveled equals integral of velocity time graph.

    Velocity y axis. time x axis.

    OK... so how does that factor in the solution with the point being before tf/2?
     
  9. Mar 5, 2016 #8

    haruspex

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    you know the graph will look like two lines of different slopes, so the areas will be two triangles. What can you say about the relationship between those two areas?
     
  10. Mar 5, 2016 #9
    These areas must be the same since distance covered is the same.
     
  11. Mar 5, 2016 #10

    haruspex

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    Right. But the slopes are different, and both triangles are based on the x-axis. So what does that tell you about the lengths of those bases?
     
  12. Mar 5, 2016 #11
    Well... I will need to have a larger base for the second triangle..... it will take less time for me to reach the top going up than the bottom going down.
     
  13. Mar 5, 2016 #12
    So thus, the point is less than tf/2.

    Thanks a lot for the help!!
     
  14. Mar 5, 2016 #13

    haruspex

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    You are welcome.
     
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