OCR M1 - 31st May 2012 Discussion Thread.
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: OCR M1 - 31st May 2012 Discussion Thread.
Hey! Just had a go at the Jan 2012 paper, which I hadn't already done, and there was a question on collisions. Usually I find these ones the easiest on the paper but I couldn't figure this one out and there isn't a detailed mark scheme which has been released yet. Basically: P and Q moving in same direction, P has mass 0.3 and is travelling at 2.2, Q has mass 0.5 and is travelling at 0.8. They collide (obviously) and afterwards 'the speed of P is reduced by 50% and its direction of motion is unchanged'. Find the speed of Q afterwards. Couldn't figure out the right answer, which by the way is apparently 1.46m/s
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Re: OCR M1 - 31st May 2012 Discussion Thread.momentum before = momentum after so you should get:(Original post by Soggy Cherry)
Hey! Just had a go at the Jan 2012 paper, which I hadn't already done, and there was a question on collisions. Usually I find these ones the easiest on the paper but I couldn't figure this one out and there isn't a detailed mark scheme which has been released yet. Basically: P and Q moving in same direction, P has mass 0.3 and is travelling at 2.2, Q has mass 0.5 and is travelling at 0.8. They collide (obviously) and afterwards 'the speed of P is reduced by 50% and its direction of motion is unchanged'. Find the speed of Q afterwards. Couldn't figure out the right answer, which by the way is apparently 1.46m/s
(0.3 x 2.2) + (0.5x0.8) = (0.3x1.1) + 0.5v (1.1 comes from half of 2.2)
0.66 + 0.4 = 0.33 + 0.5v
so 0.5v = 0.73
so v= 0.73/0.5 => 1.46ms^-1
Hope that helps
I haven't done the Jan12 paper yet so it's also practice for me too!
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Re: OCR M1 - 31st May 2012 Discussion Thread.I feel like such an idiot! Did all that but instead of multiplying 1.1 and 0.3, I added them haha, and couldn't even spot it myself! But thanks a lot though(Original post by As_Dust_Dances_)
momentum before = momentum after so you should get:
(0.3 x 2.2) + (0.5x0.8) = (0.3x1.1) + 0.5v (1.1 comes from half of 2.2)
0.66 + 0.4 = 0.33 + 0.5v
so 0.5v = 0.73
so v= 0.73/0.5 => 1.46ms^-1
Hope that helps
I haven't done the Jan12 paper yet so it's also practice for me too!
Should have a go at it if you could find it then! Though I had to get it from my teacher because I don't think it's allowed to be online yet? As for time graphs, I just struggle finding the displacement when it's a shape I've forgotten how to find the area of, which is my fault and can easily be fixed
Last edited by Soggy Cherry; 29-05-2012 at 21:45.
Feel free to post questions and I'll try to answer them. Anyone else finding time graphs annoying?