M1 June 2016 Model Answers Edexcel (with paper)

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.

I worked out the acceleration, then did the forward force - friction = m x a to get the forward force. Would this be the same as impulse

What forward force? The only force on the particle is friction.

Think of it this way: If there wasn't any friction would you expect the particle to a) accelerate, b) decelerate, or c) carry on at constant speed?

And no, force and impulse aren't the same thing. If you walk into a lamp post at a given speed you will experience a large painful force for a short time.

If you walked at the same speed into a padded rugby goal post you would experience a smaller force for a longer time. In both cases the impulse (force x time) bringing you to a halt would be the same but I know which I would rather do.

When were we told about the cars traveling the same distance?

As the area under the graph equals the distance the graphs would have to intersect otherwise the 30ms graphs area would be fully inside the 40ms graph so area was smaller not the same....although if the info about distance came after we could get away with it

Yeah I guess you're right. But how many marks you think I would get for calculating the friction and acceleration

For question 5 I got a wrong value for R because i used 2gsin20 instead of cos20, I think however I got the right components on my diagram because I got the right value for F, how many marks would I get for using the wrong value of R out of 10

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!

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!

Yes but we're told that they travel the same distance (we're told both start at X end at Y), therefore the area of the two trapeziums must be the same. If car M travels longer than 25 seconds than the trapezium for car N will be inside the trapezium for car N, thus the area will be smaller. So T<25

I'd imagine it would be one mark for drawing the graph with T<25. Second mark for marking the right shapes. Third mark for marking the initial speeds, fourth mark for marking 25 for when car N starts decelerating.

I'm a mathematician. Yes I meant in terms of what Edexcel accepts, 2sf or 3sf are both accepted.

Yes but we're told that they travel the same distance (we're told both start at X end at Y), therefore the area of the two trapeziums must be the same. If car M travels longer than 25 seconds than the trapezium for car N will be inside the trapezium for car N, thus the area will be smaller. So T<25

I'd imagine it would be one mark for drawing the graph with T<25. Second mark for marking the right shapes. Third mark for marking the initial speeds, fourth mark for marking 25 for when car N starts decelerating.

Yes but we're told that they travel the same distance (we're told both start at X end at Y), therefore the area of the two trapeziums must be the same. If car M travels longer than 25 seconds than the trapezium for car N will be inside the trapezium for car N, thus the area will be smaller. So T<25

I'd imagine it would be one mark for drawing the graph with T<25. Second mark for marking the right shapes. Third mark for marking the initial speeds, fourth mark for marking 25 for when car N starts decelerating.

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.

https://drive.google.com/open?id=0B3xSC6XqSy7kR3dhWDRkRGx0eFk

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)

I wrote "45 degrees down and left" for the final question... do you think this will get the marks?

can i just ask, how do you get access to these papers in such short time?

I kinda didn't know that so I put T further on the x axis than 25 but after calculating T I did another graph on the other page below part b. Will I lose marks for putting T further on my first graph or will they use my second graph below part b? :/