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# M1 June 2016 Model Answers Edexcel (with paper) watch

1. (Original post by kprime2)
15N to 2sf is still correct. Answers involving g should in fact be rounded to 2sf but they also accept 3sf. Full raw well done
Ah - that's great. Thanks for the solutions again - very much appreciated. Will you be doing M2 btw?
2. (Original post by Marxist)
I hope so - Edexcel probably going to make it 75 though. :/
Nah last year boundary was 59 for an A and 100UMS was 71. 74 should definitely be 100 ums
3. (Original post by Marxist)
Ah - that's great. Thanks for the solutions again - very much appreciated. Will you be doing M2 btw?
Yes
4. (Original post by kprime2)
Yes
Ah thanks - hope that goes well too.
5. Special thanks to TeeEm for his booklets. Made things much better.
6. Got everything right except for the graph, didnt make it insersect, i also didnt write the full bearing for the last question, other than that i thibk i got it all, how many marks do you think i dropped and the ums with it?
7. (Original post by kprime2)
I think they should be under 60 for an A. 65 should be 90 at least. 60 should be a low A
What would happen if say you got 50 on the first paper and 70 on the second paper? Do they come together to create one grade or can you count the grade on the one that got higher marks?
8. (Original post by thegeorgesmith)
What would happen if say you got 50 on the first paper and 70 on the second paper? Do they come together to create one grade or can you count the grade on the one that got higher marks?
Are you referring to retakes?
9. Do you reckon 73 is 100UMS, kprime2? Also where do you find the 100UMS boundary for previous papers?
10. (Original post by TheRandomGenius)
Do you reckon 73 is 100UMS, kprime2? Also where do you find the 100UMS boundary for previous papers?
Yeah should be. This paper should be below 60 for an A, so even if it's 59 then 71+ would be full UMS. I found it from the Edexcel mark converted on their website.
11. (Original post by metrize)
Got everything right except for the graph, didnt make it insersect, i also didnt write the full bearing for the last question, other than that i thibk i got it all, how many marks do you think i dropped and the ums with it?
The graph has to intersect twice, the second intersection is at (40,0). If you missed out the first intersection I'd guess that's one mark lost.

For the last question, you don't HAVE to write the direction as a bearing. Anything along the lines of '45 degrees below the horizontal' or 'south west' or a diagram with an arrow and the angle would be accepted. It's one mark for direction. So probably dropped 2 marks. 73/75 100 UMS. Even if at worse you dropped 3 marks then 72/75 should still be 100 UMS
12. pretty sure i got 55.....would this be enough to scrape a B?
13. (Original post by kprime2)
Trickier than usual. Hell lot of simultaneous equations seems to be the trend in some of these recent papers.

q3, q6 and 7a will probably be answered poorly.

My answers using g have been given to 3sf. 2sf of course is also accepted.

(just a heads up, I will be posting answers to C3, C4 and some of the other modules too)
kprime2 the great, thanks.
14. Oh my goooooooood. For the uniform rod question, I picked the wrong reaction force to be 0.... How many marks will i get for that
15. Also, for the impulse question, I worked out the resultant force after working out the friction and acceleration. Would that be right
16. (Original post by kprime2)
The graph has to intersect twice, the second intersection is at (40,0). If you missed out the first intersection I'd guess that's one mark lost.

For the last question, you don't HAVE to write the direction as a bearing. Anything along the lines of '45 degrees below the horizontal' or 'south west' or a diagram with an arrow and the angle would be accepted. It's one mark for direction. So probably dropped 2 marks. 73/75 100 UMS. Even if at worse you dropped 3 marks then 72/75 should still be 100 UMS
Great, I hope I didn't make some other little error with sig figs or something, thanks a lot for the model answers
17. ****!!!!!! I just realized that I did everything right for the slope question, but because I forgot the component of weight acting downwards, my answer at the end is wrong. How many marks do you think I will get for a perfect method, and component diagram? :/
18. (Original post by kprime2)
15N to 2sf is still correct. Answers involving g should in fact be rounded to 2sf but they also accept 3sf. Full raw well done
I take it you're a physicist then.

In my role as a mathematician I'd stick with 3sf. This is because in maths we talk about 30° and 45° but we don't mean the former to 1sf and the latter to 2sf. We have this concept of exactness. So if I'm told that g = 9.8 (when I was young, it was always 9.81) then that's 9.8 precisely. The exam paper doesn't say 9.8 to 2sf.

However, as an engineer (which is what I actually am - CEng, MIET) I fully understand that the real world is imprecise. And I've had some great conversations in the staff room with my physics colleagues about how we might normalise how we teach things like mechanics in physics v in maths.

I personally simply warn my maths students that there is a difference in approach and explain why!

Edit: Sorry I should make clear that I totally agree with you that Edexcel are quite relaxed about 2sf v 3sf on calculations involving g - but always remember to round at the end!
19. (Original post by fpmaniac)
Also, for the impulse question, I worked out the resultant force after working out the friction and acceleration. Would that be right
Yes, that's how I would have done it.

v² = u² + 2as --> 0 = u² - 2a x 5 (as a is deceleration) to get u² = 10a

F = μR = mg/8 (where m = 0.4 as I know it's going to cancel out in a moment)

F = ma --> mg/8 = ma, so a = g/8

Hence u² = 10g/8 = 49/4 --> u = 7/2 = 3.5

And finally I = (4 + 3.5) x 0.4 = 3Ns.
20. (Original post by kprime2)
The graph has to intersect twice, the second intersection is at (40,0). If you missed out the first intersection I'd guess that's one mark lost.

For the last question, you don't HAVE to write the direction as a bearing. Anything along the lines of '45 degrees below the horizontal' or 'south west' or a diagram with an arrow and the angle would be accepted. It's one mark for direction. So probably dropped 2 marks. 73/75 100 UMS. Even if at worse you dropped 3 marks then 72/75 should still be 100 UMS
I'm not sure I agree. The sketch question (part (a)) comes before the information given for part (b) so technically it's perfectly possible for the graphs not to intersect. Eg T could be very close to 25 secs. But I'm happy for you to prove me wrong!

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