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    (Original post by 2776)
    Ive found the value of t when p returns to O. It is 0 or 2.

    So it must be 2.

    Put that into equation, I get, v= -8.

    Does that make sense? as -8 means it is reversing, so it must be true.
    yeah that's right.
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    (Original post by king of swords)
    yeah that's right.
    Also Ive found out that t= 6/4 is the time for p to reach A.

    But how do u find the maximum speed whilst P is travelling from O to A?
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    (Original post by 2776)
    Also Ive found out that t= 6/4 is the time for p to reach A.

    But how do u find the maximum speed whilst P is travelling from O to A?
    for v to be a maximum, dv/dt = 0 and d2v/dt^2 = -ve....for a maximum.
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    (Original post by king of swords)
    for v to be a maximum, dv/dt = 0 and d2v/dt^2 = -ve....for a maximum.
    Imagine a rectangle. At one edge is a pulley. A is on teh rectangle atached to B which is dangled over the pulley. Each is 0.6kg. Now I tried finding the tension in teh string/acceleration as the B falls. But all I found is that 1.2a = 0?

    i.e.

    0.6a= T-0.6g

    0.6a = 0.6g -T

    so when u add them together u get 1.2a = 0.
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    (Original post by 2776)
    Imagine a rectangle. At one edge is a pulley. A is on teh rectangle atached to B which is dangled over the pulley. Each is 0.6kg. Now I tried finding the tension in teh string/acceleration as the B falls. But all I found is that 1.2a = 0?

    i.e.

    0.6a= T-0.6g

    0.6a = 0.6g -T

    so when u add them together u get 1.2a = 0.
    Is the rectangle rough if so what is the value for μ?
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    (Original post by king of swords)
    Is the rectangle rough if so what is the value for μ?
    rectangle is smooth, btw ive given u rep, for being sooooo helpful.
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    (Original post by 2776)
    rectangle is smooth, btw ive given u rep, for being sooooo helpful.
    well:
    0.6a = T - 0.6g --->the vertical bit
    T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g :confused: )
    1.2a = 2T - 0.6g
    are you sure there isn't anymore information?
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    (Original post by king of swords)
    well:
    0.6a = T - 0.6g --->the vertical bit
    T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g :confused: )
    1.2a = 2T - 0.6g
    are you sure there isn't anymore information?
    forget that:
    0.6a = 0.6g - T....better
    T = 0.6a
    so T + 0.6a = 0.6g = 1.2a
    so a = 0.5g so T = 0.3g
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    (Original post by king of swords)
    well:
    0.6a = T - 0.6g --->the vertical bit
    T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g :confused: )
    1.2a = 2T - 0.6g
    are you sure there isn't anymore information?
    Nope, no more,
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    (Original post by king of swords)
    forget that:
    0.6a = 0.6g - T....better
    T = 0.6a
    so T + 0.6a = 0.6g = 1.2a
    so a = 0.5g so T = 0.3g
    How much rep did i give u? check
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    (Original post by 2776)
    Nope, no more,
    yeah i have look at my reply to correct that post a little further up.
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    (Original post by 2776)
    How much rep did i give u? check
    7.
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    (Original post by king of swords)
    7.
    Last ques: Which school do u go to?
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    (Original post by 2776)
    Last ques: Which school do u go to?
    Bradford Grammar School....i'm guessing you are at Leeds.
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    (Original post by king of swords)
    Bradford Grammar School....i'm guessing you are at Leeds.
    Ur right!. U know adshur is at BGS as well
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    Has this whole thread been based on that first post? If so, surely that must be the longest thread to answer 1 question?
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    (Original post by 2776)
    Ur right!. U know adshur is at BGS as well
    Yeah she's my best friend lol .
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    (Original post by king of swords)
    Bradford Grammar School....i'm guessing you are at Leeds.
    Right, last ques: when A has moved a distance of 2 metres it becomes detached from the string. From this instant B takes a further 0.2 seconds to reach the floor. Find h. (i.e. the distance between B and the floor from rest)
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    (Original post by king of swords)
    Yeah she's my best friend lol .
    lol

    I have found that after moving 2 m, the velocity is root 2g.
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    (Original post by 2776)
    Right, last ques: when A has moved a distance of 2 metres it becomes detached from the string. From this instant B takes a further 0.2 seconds to reach the floor. Find h. (i.e. the distance between B and the floor from rest)
    When A moves 2m B also moves 2m. From then on B accelerates only due to gravity since there is no tension left in the string.
    So:
    s = ?, u = ?, a = g, t = 0.2
    so s = tu + ½at^2.....then h = s + 2.
 
 
 
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