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# M1 question

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1. 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
2. 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
3. (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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