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Official M4 Edexcel Thread – Wednesday 15th June 2016

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



me too
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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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Original post by physicsmaths
U-y^=Ucos6t=-u leading to y dot=2U


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 :biggrin:

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...
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.
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?
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 :biggrin:

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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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.
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 :biggrin:
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 :biggrin:
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 :biggrin:


I did exactly the same as you so high five ma man!
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. :s-smilie:
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 :lol:
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. :s-smilie:


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 :eek:
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 :eek:


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!😂
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?
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 :biggrin:

They'll cater for alternative solutions
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
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?
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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K=1 had me tripping for 10 minutes cause it was like 2 lines.


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