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    The first bit of the queston is two particles of mass 5KG an 7KG are attached to cables of a ligh inextensible string, passing over a smooth pulley. Initially the particles are at rest, and at same level, when the particles are released find the acceleration of them, and the time when the difference in the height of the particles are 1m.

    I got the acceleration as 2.67 which is correct

    for time i used suvat using
    S = 1
    U = 0
    V = ?
    A = 2.67
    T = X
    I get T as 0.86 but this is the wrong answer, any ideas?
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    (Original post by SunDun111)
    The first bit of the queston is two particles of mass 5KG an 7KG are attached to cables of a ligh inextensible string, passing over a smooth pulley. Initially the particles are at rest, and at same level, when the particles are released find the acceleration of them, and the time when the difference in the height of the particles are 1m.

    I got the acceleration as 2.67 which is correct

    for time i used suvat using
    S = 1
    U = 0
    V = ?
    A = 2.67
    T = X
    I get T as 0.86 but this is the wrong answer, any ideas?
    are you sure about that?
    i got a=\dfrac{g}{6}
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    (Original post by SunDun111)
    The first bit of the queston is two particles of mass 5KG an 7KG are attached to cables of a ligh inextensible string, passing over a smooth pulley. Initially the particles are at rest, and at same level, when the particles are released find the acceleration of them, and the time when the difference in the height of the particles are 1m.

    I got the acceleration as 2.67 which is correct

    for time i used suvat using
    S = 1
    U = 0
    V = ?
    A = 2.67
    T = X
    I get T as 0.86 but this is the wrong answer, any ideas?
    Shouldn't the acceleration be g/6?=1.63?
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    (Original post by Steelmeat)
    are you sure about that?
    i got a=\dfrac{g}{6}
    Just beat me to it
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    (Original post by Middriver)
    Shouldn't the acceleration be g/6?=1.63?
    (Original post by Steelmeat)
    are you sure about that?
    i got a=\dfrac{g}{6}
    its hundred per cent 2.67, according to the back of the book and what I got from my working?
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    (Original post by Middriver)
    Just beat me to it
    soory
    (Original post by SunDun111)
    its hundred per cent 2.67, according to the back of the book and what I got from my working?
    nah it can't be here's me working
    resolving vertically down 7g-T=7a
    resolving vertically up T-5g=5a

    adding them together gets you 7g-5g=12a
    hence 2g=12a
    g=6a
    a=\dfrac{g}{6}

    did you do something different?
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    (Original post by SunDun111)
    its hundred per cent 2.67, according to the back of the book and what I got from my working?
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    (Original post by Steelmeat)
    soory


    nah it can't be here's me working
    resolving vertically down 7g-T=7a
    resolving vertically up T-5g=5a

    adding them together gets you 7g-5g=12a
    hence 2g=12a
    g=6a
    a=\dfrac{g}{6}

    did you do something different?
    O

    (Original post by Middriver)
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    I think im very stupid, I wrote the question wrong its 4KG and 7KG? does that make a difference to your answers?
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    (Original post by SunDun111)
    O



    I think im very stupid, I wrote the question wrong its 4KG and 7KG? does that make a difference to your answers?
    so when you add them together you get 3g=11a which does get you 2.67
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    (Original post by SunDun111)
    O



    I think im very stupid, I wrote the question wrong its 4KG and 7KG? does that make a difference to your answers?
    Oh ok, it's probably something to do with your displacement then, you put 1 maybe it's 0.5(since they both move 0.5 would be 1m apart) I'll do it now.
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    (Original post by SunDun111)
    The first bit of the queston is two particles of mass 5KG an 7KG are attached to cables of a ligh inextensible string, passing over a smooth pulley. Initially the particles are at rest, and at same level, when the particles are released find the acceleration of them, and the time when the difference in the height of the particles are 1m.

    I got the acceleration as 2.67 which is correct

    for time i used suvat using
    S = 1
    U = 0
    V = ?
    A = 2.67
    T = X
    I get T as 0.86 but this is the wrong answer, any ideas?
    Does using displacement as 0.5 work?
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    (Original post by Middriver)
    Does using displacement as 0.5 work?
    Yes it does got it right thanks!
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    (Original post by Middriver)
    Does using displacement as 0.5 work?
    Sorry got another question, The question is a diagram shows a mass of A 5kg intially at rest on a horizontal table, a resistance force of 10N acts against the motion of A which is connected to Mass of B of 3kg by a light inextensible string which passes over the smooth pulley. The sysstem is released from rest, Calculate the acceleration of A, and the tension in the string, when B hits the ground calculate the Acceleration of A.

    I got the initial acceleration as 2.43 and tension as 22.15N how do i find the other acceleration when B has hit the ground?
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    There is no longer any tension in the string so acceleration is matched only by the resistive force of 10N

    --> -10 = 5a
    a = -2 m s^-2
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    (Original post by SunDun111)
    Sorry got another question, The question is a diagram shows a mass of A 5kg intially at rest on a horizontal table, a resistance force of 10N acts against the motion of A which is connected to Mass of B of 3kg by a light inextensible string which passes over the smooth pulley. The sysstem is released from rest, Calculate the acceleration of A, and the tension in the string, when B hits the ground calculate the Acceleration of A.

    I got the initial acceleration as 2.43 and tension as 22.15N how do i find the other acceleration when B has hit the ground?
    Resolve, for A alone. There will be no tension since the rope will be slack when it hits the ground.
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    (Original post by Middriver)
    Resolve, for A alone. There will be no tension since the rope will be slack when it hits the ground.
    Thanks, stuck on another question, basically its one over a smooth pulley, two childs have a mass of 40 and 60 kg. I had to find the acceleration and tension in the strings, over the pulley which i did. Now it says one of them stood 2 metres above the ground level, and steps of the two metres and falls, find the total distance that the other travels upwards?
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    (Original post by SunDun111)
    Thanks, stuck on another question, basically its one over a smooth pulley, two childs have a mass of 40 and 60 kg. I had to find the acceleration and tension in the strings, over the pulley which i did. Now it says one of them stood 2 metres above the ground level, and steps of the two metres and falls, find the total distance that the other travels upwards?
    Do you have a diagram that sounds very confusing, does it say which one steps off?
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    (Original post by Middriver)
    Do you have a diagram that sounds very confusing, does it say which one steps off?
    Sorry worded it terrible, the one that weights 60kg stepped of a 2m vertical height. so the other one (40kg) came up, its asking how far the total distance the other one came up
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    (Original post by SunDun111)
    Sorry worded it terrible, the one that weights 60kg stepped of a 2m vertical height. so the other one (40kg) came up, its asking how far the total distance the other one came up
    Find the speed at which B(60kg) hits the floor, this will be the initial speed for A(40kg) then resolve for A new acceleration since tension will be 0. Then Suvat to find displacement (may need to divide this by 2 if the child falls back down since they want the max height? Or if it just says distance then don't divide by 2) then add the 2m from from the fall of B. Posting a photo of the question would be useful.
 
 
 
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