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    Name:  uptr.png
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Size:  45.6 KBHello there

    I am having trouble understanding the point I have circled in green in the picture. Why are they adding this to the value it takes the 6kg mass to hit the ground and then multiplying the other value of t by 2?

    Thanks
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    (Original post by upthegunners)
    Name:  uptr.png
Views: 75
Size:  45.6 KBHello there

    I am having trouble understanding the point I have circled in green in the picture. Why are they adding this to the value it takes the 6kg mass to hit the ground and then multiplying the other value of t by 2?

    Thanks
    It's the time taken for the 4kg weight to go up and come back down again.
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    When the 6kg mass hits the ground, the 4kg shares the velocity at which that mass collides. The 4kg mass will reach a maximum height (you are told it does not hit the pulley) before changing direction and reaching the position it was at when the other mass hit ground. This is when the string is again taut.

    You multiply the time for the mass to come to rest by 2 because the time for it to come down again is the same by symmetry (constant acceleration).
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    (Original post by Brister)
    When the 6kg mass hits the ground, the 4kg shares the velocity at which that mass collides. The 4kg mass will reach a maximum height (you are told it does not hit the pulley) before changing direction and reaching the position it was at when the other mass hit ground. This is when the string is again taut.

    You multiply the time for the mass to come to rest by 2 because the time for it to come down again is the same by symmetry (constant acceleration).
    but surely it will go down faster as it is not going against gravity but it is going against gravity on the way up?
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    (Original post by upthegunners)
    but surely it will go down at a different speed as gravity is acting in an opposite direction?
    Okay, let's be a little more quantitative.

    With constant acceleration, in general you will have x = x_0 + v_0 t + \frac{1}{2}at^2. This is a parabola, which is symmetric about an axis through the local maximum/minimum depending on the sign of a.

    In this case, if you take the upward direction as being positive, we have a local maximum for the mass. Now since velocity is the time derivative of displacement, velocity is clearly zero at the local maximum. We can therefore use v = u + at with v = 0 to find the time for the mass to go from the initial displacement to the maximum easily and double that to get the time when it returns to the initial displacement by symmetry.

    Regarding your comment, the direction of gravity does not change, but yes the velocity is obviously not the same. However, there is an interesting result concerning speed. With constant acceleration you will notice that the time for an object to go from a velocity of +k to 0 is the same as the time for it to go from 0 to -k. If you draw the velocity-time graph, this is an obvious point. Again, the symmetry of the displacement equation means that for any displacement x = m, the object will have a certain speed (not velocity, but speed) whether it is on the way up or on the way down.

    In other words, if you fire a an object up from horizontal ground with some speed, it will hit the ground again with the same speed. If you think about conservation of energy, this makes sense.
 
 
 
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