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    A block P of mass 0.6kg resting on a smooth surface of a horizontal table. inextensible light strings connect P to blocks A and B which hang freely over light smooth pulleys placed at opposite parallel edges of the table. the masses of A and B are 0.3kg and 0.5kg respectively. all portions of the string are taut and perpendicular to their respective edges of the table. the system is released from rest. calculate:

    a) the common magnitude of the acceleration of the blocks
    b) the tension in the strings

    I've tried working this out and came to this, however my answer was wrong, any help would be appreciated.
    Block A: t - 0.3g = 0.3a
    Block B: 0.5g - t = 0.5a
    therefore 0.2g = 0.8a and a = 2.45
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    (Original post by javari112)
    A block P of mass 0.6kg resting on a smooth surface of a horizontal table. inextensible light strings connect P to blocks A and B which hang freely over light smooth pulleys placed at opposite parallel edges of the table. the masses of A and B are 0.3kg and 0.5kg respectively. all portions of the string are taut and perpendicular to their respective edges of the table. the system is released from rest. calculate:

    a) the common magnitude of the acceleration of the blocks
    b) the tension in the strings

    I've tried working this out and came to this, however my answer was wrong, any help would be appreciated.
    Block A: t - 0.3g = 0.3a
    Block B: 0.5g - t = 0.5a
    therefore 0.2g = 0.8a and a = 2.45
    You've assumed that the tension is the same in each of the strings. This is not the case, since the masses are accelerating.


    Easiest method is to treat the whole setup as one thing to start, and use F=ma

    F= (0.5-0.3)g = (0.5+0.6+0.3)a
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    (Original post by RKM21)
    Hi, as much as I appreciate your method, the OP has done pretty much what the markscheme normally dictates in M1 modules. Resolving using two equations is how most of us are taught to answer these questions. The only problem with the OP's working is that they have done it the wrong way round. I have attached an image with my working.Attachment 542181
    You have forgotten the mass of the 0.6kg block in your calculations.
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    (Original post by RKM21)
    Hi, as much as I appreciate your method, the OP has done pretty much what the markscheme normally dictates in M1 modules. Resolving using two equations is how most of us are taught to answer these questions. The only problem with the OP's working is that they have done it the wrong way round. I have attached an image with my working.Attachment 542181
    I appreciate you may be expected to do it via multiple equations, however your diagram and working don't match the question. It's a different arrangement.
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    Here is diagramName:  ImageUploadedByStudent Room1464875473.050085.jpg
Views: 63
Size:  113.9 KB

    I apologise for the abysmal lines.
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    (Original post by ghostwalker)
    I appreciate you may be expected to do it via multiple equations, however your diagram and working don't match the question. It's a different arrangement.
    So sorry, I am currently doing an M1 past paper and mixed up two questions.
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    (Original post by ghostwalker)
    You've assumed that the tension is the same in each of the strings. This is not the case, since the masses are accelerating.


    Easiest method is to treat the whole setup as one thing to start, and use F=ma

    F= (0.5-0.3)g = (0.5+0.6+0.3)a
    Thank you! I reread the question and realised the particles are attached using two seperate/different strings, which made the question a lot easier to answer
 
 
 
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