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# Official M4 Edexcel Thread – Wednesday 15th June 2016

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1. (Original post by classic777)
I got (44/3)mlg for the second derivative

me too
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2. (Original post by Ewanclementson)
Unofficial Markscheme - Let me know if you spot any errors!
1)1/root5
2)k=1
3)5root17 256 degrees 48.5m
4)shown then 6
5)shown then when t=pi/12 speed=U and distance travelled when they meet is (U/6)pi+3
6)shown then sin(theta)=5/6 and it is stable as the second derivative equals 48lmg here
All the same apart from on Q4 I kept messing up my constants for the final part and had about 4 different answers, think one of them may have been 6m although it was all an absolute mess so I wrote an apology to the examiner on the paper😂 no idea what I put down in the end!
3. (Original post by oShahpo)
Well this exam wasn't really much about intelligence at all, it i's merely standard stuff that you would have learned if you did a few past papers, so don't blame yourself, it's just that you didn't practice enough. Focus on STEP and show them Cambridge dons who's got the skittles!!
Thank you!

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4. (Original post by physicsmaths)
Damn man how'd you calculate the coefficient of sin(6t), I ended up with x = Bsin(6t) after the solving the aux equation but couldn't figure out how to find B...

Fine for the most part though, I got the right answers except the last one in 5

How many you reckon I have lost in (c)? It was out of 5 marks

I differentiated y w.r.t time and then plugged in for the pi x U/6 term and drew a diagram showing the Upi/6 and 1.5+x (x was 0 of course) and then showed adding them together to give 1.5 + Upi/6...
5. (Original post by Euclidean)

I differentiated y w.r.t time and then plugged in for the pi x U/6 term and drew a diagram showing the Upi/6 and 1.5+x (x was 0 of course) and then showed adding them together to give 1.5 + Upi/6...
I did something similar. I got pi*u/6 and then did distance = 1.5 + that distance. Will probably lose no more than 4 and no less than 2.
6. (Original post by oShahpo)
I did something similar. I got pi*u/6 and then did distance = 1.5 + that distance. Will probably lose no more than 4 and no less than 2.
I reckon I got 3 or 2 marks on that. 1 mark for diff wrt time, 1 mark for subbing dx/dt into it and 1 mark for subbing t = pi/6 I think. Then the other two for the 3 metres part.

Anyone got any predictions on 100 boundary?
7. (Original post by Euclidean)
Damn man how'd you calculate the coefficient of sin(6t), I ended up with x = Bsin(6t) after the solving the aux equation but couldn't figure out how to find B...

Fine for the most part though, I got the right answers except the last one in 5

How many you reckon I have lost in (c)? It was out of 5 marks

I differentiated y w.r.t time and then plugged in for the pi x U/6 term and drew a diagram showing the Upi/6 and 1.5+x (x was 0 of course) and then showed adding them together to give 1.5 + Upi/6...
You lose 1 i think.
Well i showed x=Bsin6t then i diff equated to U at t=0 since xdot=U intially and y dot=0 B=U/6
Then at pi/12 U-ydot=0 hence speed is U.
At pi/5 y dot=2U so travels piu/6 till it salck travels a relative distance of 1.5 and speed is 2U so time taken is 1.5/U times gives 3m in last prt of motion giving final answer.

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8. (Original post by Euclidean)
I reckon I got 3 or 2 marks on that. 1 mark for diff wrt time, 1 mark for subbing dx/dt into it and 1 mark for subbing t = pi/6 I think. Then the other two for the 3 metres part.

Anyone got any predictions on 100 boundary?
You did not need to differentiate as the answer was already found in a previous part of the question where the string was found slack.
9. (Original post by physicsmaths)
You lose 1 i think.
Well i showed x=Bsin6t then i diff equated to U at t=0 since xdot=U intially and y dot=0 B=U/6
Then at pi/12 U-ydot=0 hence speed is U.
At pi/5 y dot=2U so travels piu/6 till it salck travels a relative distance of 1.5 and speed is 2U so time taken is 1.5/U times gives 3m in last prt of motion giving final answer.

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Ah I couldn't figure out what xdot was initially, that's why!

I actually managed to get part(b) and part(c) correct without the coefficient though which is nice
10. (Original post by oShahpo)
You did not need to differentiate as the answer was already found in a previous part of the question where the string was found slack.
Oh maybe not as well as I thought then. Who knows tbh...

Anyway I think I bagged >70 so should be good
11. (Original post by Euclidean)
Oh maybe not as well as I thought then. Who knows tbh...

Anyway I think I bagged >70 so should be good
I did exactly the same as you so high five ma man!
12. How did everyone do question 1? I used the dot product because the velocity before and after were perpendicular but I feel there was an easier way. Also question 2 I refused to believe the right answer was k=1 for ages, ended up subbing in values for alpha and e to check it was correct, not sure where the 9 marks would be awarded.
13. (Original post by oShahpo)
I did exactly the same as you so high five ma man!
I think quite a few will be in the same boat, it was confusing tbh!

I actually did 5 part (b) and then skipped to Q6 and finished Q6 within the hour, then spent like 20 minutes trying to figure out part (c) and (d) on 5
14. (Original post by Jonooo123)
How did everyone do question 1? I used the dot product because the velocity before and after were perpendicular but I feel there was an easier way. Also question 2 I refused to believe the right answer was k=1 for ages, ended up subbing in values for alpha and e to check it was correct, not sure where the 9 marks would be awarded.
Q1: The deflection by 90 degrees means the velocity of the particle A afterwards is actually perpendicular to the way it was before, so ucos(alpha) ends up being -usin(alpha) afterwards etc. Then I got around 3 equations from linear momentum and NLR and solved simultaneously for tan(alpha)

Q2: I found out that the angle the final velocity makes (after both collisions) with the wall is 90-alpha and then basically found four equations of motion (2 from each collision) and eliminated all speed variables which left me with something like e = tan(alpha)cot(beta) and ke = tan(alpha)cot(beta). Where beta was the angle the velocity of the ball made with the wall after the first collision.

I can see where the marks would be allocated in that one. Hadn't thought to try scalar projection though
15. (Original post by Euclidean)
Q1: The deflection by 90 degrees means the velocity of the particle A afterwards is actually perpendicular to the way it was before, so ucos(alpha) ends up being -usin(alpha) afterwards etc. Then I got around 3 equations from linear momentum and NLR and solved simultaneously for tan(alpha)

Q2: I found out that the angle the final velocity makes (after both collisions) with the wall is 90-alpha and then basically found four equations of motion (2 from each collision) and eliminated all speed variables which left me with something like e = tan(alpha)cot(beta) and ke = tan(alpha)cot(beta). Where beta was the angle the velocity of the ball made with the wall after the first collision.

I can see where the marks would be allocated in that one. Hadn't thought to try scalar projection though
Haha I was just stuck with piles of equations in question 1 and was going nowhere then just had a eureka moment to take the dot product which eliminated everything but terms in sin alpha and cos alpha! Question 2 I got k=1 with hardly any working, worked in vector form and so the initial velocity was (ucosa)i + (usina)j and the final velocity must be (-kecosa)i+(-esina)j and used similar triangles, not sure how I could have got all the method marks!😂
16. For the relative velocity q, for the first part I did the right thing but I got the bearing wrong - found the angle it was coming from, not where it's going to if you understand me. So I got 79 or smth.. how many marks would I lose?
17. (Original post by Jonooo123)
Haha I was just stuck with piles of equations in question 1 and was going nowhere then just had a eureka moment to take the dot product which eliminated everything but terms in sin alpha and cos alpha! Question 2 I got k=1 with hardly any working, worked in vector form and so the initial velocity was (ucosa)i + (usina)j and the final velocity must be (-kecosa)i+(-esina)j and used similar triangles, not sure how I could have got all the method marks!😂
Ah that's a clever approach

They'll cater for alternative solutions
18. (Original post by ExtraTimeOwnGoal)
For the relative velocity q, for the first part I did the right thing but I got the bearing wrong - found the angle it was coming from, not where it's going to if you understand me. So I got 79 or smth.. how many marks would I lose?
Probably just the one
19. Omg 😭😭😭. Guys, for question 1 I got tan(90-theta) = 5tan(theta. I did the double angle formula for tan and thought I was doing something wrong. This was my thought in exam "changing to sin and cos prolly won't do anything". Omfg . The same thing happened in question 2. 😡😡😡. I'm such a retard. How many marks would j lose?
20. (Original post by Pyslocke)
Omg 😭😭😭. Guys, for question 1 I got tan(90-theta) = 5tan(theta. I did the double angle formula for tan and thought I was doing something wrong. This was my thought in exam "changing to sin and cos prolly won't do anything". Omfg . The same thing happened in question 2. 😡😡😡. I'm such a retard. How many marks would j lose?
tan(90-x)=cotx.
You lose 2 I think.

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