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    Hey, please can anyone explain why the particle move beyond the top blue line in this video ? Watch the first 2 minutes of this. https://www.youtube.com/watch?v=uP7RBWsoQ44 Is this is maximum height reached by the particle under free gravity after the other particle has hit the ground? So, when the particle moves back down to the blue line, is that when the string becomes taut again? Thank you.
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    This is because when the particle hits the ground the other particle continues for a little bit before returning to the rest point at the blue line. That is just showing you what happens in reality, you won't be asked about the bounce thing which he shows, thats just showing you what happens, you only care about when it returns to rest, not the little bounce.

    you would be asked about the distance it travels when one particle hits the ground, this would be using suvat resolving vertically for one of the particles to find the distance, then times it by 2 and thats the height the other particle is at.
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    (Original post by Sayless)
    This is because when the particle hits the ground the other particle continues for a little bit before returning to the rest point at the blue line. That is just showing you what happens in reality, you won't be asked about the bounce thing which he shows, thats just showing you what happens, you only care about when it returns to rest, not the little bounce.

    you would be asked about the distance it travels when one particle hits the ground, this would be using suvat resolving vertically for one of the particles to find the distance, then times it by 2 and thats the height the other particle is at.
    Oh I get what you mean now. That's why when the questions asks for the height moved by the particle when string is taut would be the distance *2 because it has to move back down again like the video suggests. Right? Thanks.
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    (Original post by coconut64)
    Oh I get what you mean now. That's why when the questions asks for the height moved by the particle when string is taut would be the distance *2 because it has to move back down again like the video suggests. Right? Thanks.
    no.

    the quesiton won't asked for the height moved they will ask for the height of the particle. sec i'll draw a pretty picture


    to find this u use suvat vertically
    e.g it will say particle starts at rest so u = 0
    then it'll say like, the particle travels 8 seconds, t= 8
    then you prob already know accelration since you usually work it out at part (a)

    so you do a suvat to get the distance s
    then times it by 2

    if the string is slack, then you do same but do acceleration as 9.8, and the initial velocity will be the final velocity of the taut suvat
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    (Original post by coconut64)
    Oh I get what you mean now. That's why when the questions asks for the height moved by the particle when string is taut would be the distance *2 because it has to move back down again like the video suggests. Right? Thanks.
    My bad, I described the wrong thing, I see now what you mean, my explanation before was for another question, I think this question is what you was talking about, see 5(c)

    I believe this is the little bounce thing that occurs when the string becomes slack, the question is looking for the maximum height.

    I'll take you through how I would answer a question like this:
    for 5a)
    We need to find the acceleration from the diagram so to do this we know that the lightest particle is already touching the ground and the heavy particle is mid air therefore the heavy particle is going to move downwards when released.

    Therefore the first equation is for the 3kg particle (using F=ma) resolving downwards
    3g - T = 3a

    now for the second equation we use the 1kg particle (using F=ma) resolving downwards
    T -g = a

    now we put them together and add them so T + -T cancels out the T so we can find acceleration like so:

    2g = 4a
    4a = 19.6
    a = 4.90ms^-2

    5b) Here we use suvat vertically. We are told the 1kg particle starts at rest so u=0, the distance it travels when the 3kg particle touches the ground is 0.4m so s=0.4, and the acceleration we already worked out as 4.9ms^-2. We use v^2 = u^2 - 2as.

    Therefore v^2 = 0^2 - 2x4.9x0.4
    v^2 = 3.92
    v = 1.98 ms^-1

    5c) Here is the question relating to what you was asking, i'll demonstrate how to answer these types of questions.
    We need to find the maximum height of the 1kg particle, the string is slack at this point. The final velocity from the previous question is 1.98 ms^-1, this occurs when the particles have being released and the string has finished being taut. At this point it is the beginning of the string being slack so we use the inital velocity as u=1.98, since this is our new starting point. Since the string is slack our acceleration will be a=-9.8ms^-2, since we are resolving upwards. The maximum height occurs when the final velocity is 0, as the particle will stop at 0ms^-1 before it comes sback down, therefore final velocity v=0. Lets call the displacement, s = h, for height.

    we can use v^2 = u^2 - 2as.

    0^2 = 1.98^2 - 2x-9.8xh
    0=3.92 -19.6h
    19.6h = 3.92
    h = 0.2m

    that is the distance travelled when the string is slack.
    We add this to the distance travelled when the string was taut which is 40cm, so 0.4m.

    max height = 0.2m + 0.4m
    =0.6m

    I hope this helps you, part c is what you are looking for I believe, disregard what I said in the previous post.
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    (Original post by Sayless)
    My bad, I described the wrong thing, I see now what you mean, my explanation before was for another question, I think this question is what you was talking about, see 5(c)

    I believe this is the little bounce thing that occurs when the string becomes slack, the question is looking for the maximum height.

    I'll take you through how I would answer a question like this:
    for 5a)
    We need to find the acceleration from the diagram so to do this we know that the lightest particle is already touching the ground and the heavy particle is mid air therefore the heavy particle is going to move downwards when released.

    Therefore the first equation is for the 3kg particle (using F=ma) resolving downwards
    3g - T = 3a

    now for the second equation we use the 1kg particle (using F=ma) resolving downwards
    T -g = a

    now we put them together and add them so T + -T cancels out the T so we can find acceleration like so:

    2g = 4a
    4a = 19.6
    a = 4.90ms^-2

    5b) Here we use suvat vertically. We are told the 1kg particle starts at rest so u=0, the distance it travels when the 3kg particle touches the ground is 0.4m so s=0.4, and the acceleration we already worked out as 4.9ms^-2. We use v^2 = u^2 - 2as.

    Therefore v^2 = 0^2 - 2x4.9x0.4
    v^2 = 3.92
    v = 1.98 ms^-1

    5c) Here is the question relating to what you was asking, i'll demonstrate how to answer these types of questions.
    We need to find the maximum height of the 1kg particle, the string is slack at this point. The final velocity from the previous question is 1.98 ms^-1, this occurs when the particles have being released and the string has finished being taut. At this point it is the beginning of the string being slack so we use the inital velocity as u=1.98, since this is our new starting point. Since the string is slack our acceleration will be a=-9.8ms^-2, since we are resolving upwards. The maximum height occurs when the final velocity is 0, as the particle will stop at 0ms^-1 before it comes sback down, therefore final velocity v=0. Lets call the displacement, s = h, for height.

    we can use v^2 = u^2 - 2as.

    0^2 = 1.98^2 - 2x-9.8xh
    0=3.92 -19.6h
    19.6h = 3.92
    h = 0.2m

    that is the distance travelled when the string is slack.
    We add this to the distance travelled when the string was taut which is 40cm, so 0.4m.

    max height = 0.2m + 0.4m
    =0.6m

    I hope this helps you, part c is what you are looking for I believe, disregard what I said in the previous post.
    Thanks for getting back to this, I understand this. This is not the exam board that I am doing but it still helps me. I am doing Edexcel and their question is slightly different to this.
 
 
 
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