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Simple kinematics

  1. Jan 17, 2013 #1
    http://www.examsolutions.net/a-level-maths-papers/Edexcel/Mechanics/Mechanics-M1/2010-June/paper.php [Broken]

    Q. 8

    T = 3.36
    a = 1.4


    for part c)

    First I found the velocity of A after it has moved for 0.5 s, and the displacement it has moved after 0.5 seconds:

    I get that its velocity is 0.7 after 0.5 seconds, and displacement is 0.175 after 0.5 seconds.

    Then for B:

    I try to find how long it takes for it to get to it's maximum height as it moves up for a short period of time before moving back down:

    v = u + at
    0 = 0.7 -gt
    t = 1/14

    then I find the time it takes to go back down to the ground:

    -1.175 = 0.7t - 4.9t^2

    solving this I get t = 0.566...

    so the total time that elapses is 2(1/14) + 0.566.... which is about 0.71 to 2sf

    however in the mark scheme they only use 0.566... where have I gone wrong?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 17, 2013 #2

    TSny

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    I think I'm following your method. In what direction is the initial velocity for the 0.7 factor in your equation above?
     
    Last edited by a moderator: May 6, 2017
  4. Jan 17, 2013 #3
    oh crap, I took it to be going up when it should be going down... works when going down. Thanks!

    Could I ask, why does it work when just using the displacement of -1.175 and velocity of 0.7 directly? I don't understand how that would work right away.
     
  5. Jan 17, 2013 #4

    HallsofIvy

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    You can treat the "AB combination", both masses together, as a whole. The gravitational force, alone, on A is .4g N and on B is .3g N. That means that there is a net force on the "AB combination" of .1g N which will cause AB, as a whole to move at acceleration .1/(.7)= 1/7 m/s^2 in the direction such that A is going downward and B is going upward.
     
  6. Jan 17, 2013 #5
    I'm sorry this has confused me even more - I don't understand why you can just consider the displacement found and get the answer.
     
  7. Jan 17, 2013 #6

    TSny

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    Once the string breaks, block B is in free fall and has a constant acceleration of 9.8 m/s2 downward from the time the string breaks until it strikes the ground. So, the constant acceleration equations hold for the entire flight. So, there is no need to break it up into finding the time to go up and then finding the time coming back down. Even though the velocity switches direction during the free fall, the acceleration remains the same in magnitude and direction.

    At the time the string breaks, the velocity is +0.7 m/s (upward) and the final position is -1.175 m below the starting point. So, if you set up the constant acceleration equation with initial velocity of +7.0 m/s and the final position of -1.175 m, the equation will give you the total time of free fall.
     
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