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Math Help - Problem with pulley

  1. #1
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    Problem with pulley

    Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .

    I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as

    v^2=0+2x(9.8x3.15)
    v=7.857 m/2

    so i used V=U+at
    giving t=0.8017

    What is wrong with my method? Would appreciate help, been bugging me all day
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by chartsy View Post
    Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .

    I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as

    v^2=0+2x(9.8x3.15)
    this is wrong. the above equation is true for a particle in free fall. P is not in a free fall
    v=7.857 m/2

    so i used V=U+at
    giving t=0.8017

    What is wrong with my method? Would appreciate help, been bugging me all day
    now try it.
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  3. #3
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    skeeter's Avatar
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    when P hits the ground, Q is still rising at speed v = at = 4.2 \, m/s

    time for it to rise and fall to its original position is

    t = \dfrac{2 \cdot v_0}{g} = \dfrac{6}{7} \, sec
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  4. #4
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    so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time
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  5. #5
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    what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chartsy View Post
    so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time
    You aren't combining them. You are using them one at a time. You need to find the speed that Q is moving upward initially (the speed it gets to after P hits the ground). That gives you the v0 for the problem after P is on the ground.

    -Dan
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  7. #7
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    Quote Originally Posted by chartsy View Post
    what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?
    I get it! it's symmetrical because it decelerates as fast as it accelrates!
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