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    The june 2010 last question, i was just curious, if the string breaks, and will b still move up for a split second :s wudn't it? but the mark scheme doesn't do this, they just say oh well as string breaks it comes down :s. I don't know why this is, am i missing something
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    So the instant the string breaks, B will be above where it was at the start (obviously), B will move freely under gravity first upwards (you're right) and then all the way back down. The mark scheme uses 1.175m as the displacement due to the fact that B moves up to a point and then back down so displacement cancels out (I'm so bad at explaining). Like lets say B moves a further 0.05m upwards, since displacement is a vector quantity, when B moves back down to the point 1.175m above the ground, it's moving in the NEGATIVE direction so -0.05m. 0.05-0.05=0 so you don't worry about that distance. I'm assuming you're fine with the rest of the question? If not, you should watch this guy's paper solutions. Good luck
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    (Original post by victoria98)
    So the instant the string breaks, B will be above where it was at the start (obviously), B will move freely under gravity first upwards (you're right) and then all the way back down. The mark scheme uses 1.175m as the displacement due to the fact that B moves up to a point and then back down so displacement cancels out (I'm so bad at explaining). Like lets say B moves a further 0.05m upwards, since displacement is a vector quantity, when B moves back down to the point 1.175m above the ground, it's moving in the NEGATIVE direction so -0.05m. 0.05-0.05=0 so you don't worry about that distance. I'm assuming you're fine with the rest of the question? If not, you should watch this guy's paper solutions. Good luck
    I was fine but I considered the whole thing 👍🏻😂 I get ya thanks a lot
 
 
 
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