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# Edexcel(IAL) M1 June 8th 2016 Discussion

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• View Poll Results: Edexcel M1 exam: how was it?
Easy as pie
20.63%
Was OK
42.50%
Ugh, it was hard
25.63%
Failed
11.25%

1. (Original post by SaadKaleem)
Got T=7 and s=73.5

Who else got 77 kg as the integer max mass in Moments question?
Did you get T=7 by taking the final velocity of B as 3T and then equating the areas of the graph under A and B. So that gives some quadratic equation and there were two t values but one was less than 3.5 so we ignored it and took the other value which was 7?
2. (Original post by Becka99)
Did you get T=7 by taking the final velocity of B as 3T and then equating the areas of the graph under A and B. So that gives some quadratic equation and there were two t values but one was less than 3.5 so we ignored it and took the other value which was 7?
Yes, I got a quadratic by equating the distances. Got two values exactly!, the other was less then 3.5 (2.333 something as I recall - don't remember exactly)

I took 7 since it's >3.5 which was obvious.
3. [QUOTE=SaadKaleem;65552301]77.7 isn't an integer though.
do they consider the steps even if the working is wrong?
4. [QUOTE=mr.holmes;65552613]
(Original post by SaadKaleem)
77.7 isn't an integer though.
do they consider the steps even if the working is wrong?
If your working is correct, you'll be awarded method marks accordingly to the point where you've done correctly.. The final answer only carries 1 mark (A1)
5. (Original post by AlvlVictim)
76 or 77 Kg

Posted from TSR Mobile
Me too 78😍😍😍
6. (Original post by AlvlVictim)
I got 4.7 although it's gotta be wrong

Posted from TSR Mobile
I got like 5 something weird too😂😂
7. My Answers: (They could possibly be wrong)
Spoiler:
Show

Horizontal Kinematics: 1) a=3.5 ms^-2 and u=12 ms^-1
Momentum & Impulse: 2) k=1/2 and I=3/2mu
Vertical Kinematics & Forces: 3) h=2.5, R=2250N [probably wrong]
Moments: 4) F(A)=3460, F(C)=4250, m=77kg
Forces & Acceleration as Vectors: 5) b) a=4.47 (2 square root 5) ms^-2
Horizontal Kinematics (Overtaking) 6) a=4, T=7, D(Q)=73.5
Connected Particles: 7) m=2, Force of pulley=22.2N
8. (Original post by SaadKaleem)
My Answers: (They could possibly be wrong)
Spoiler:
Show

1) a=3.5 ms^-2 and u=12 ms^-1
2) k=1/2 and I=3/2mu
3) h=2.5, R=2250N
4) F(A)=3460, F(C)=4250, m=77kg
5) b) a=4.47 (2 square root 5) ms^-2
6) a=4, T=7, D(Q)=73.5
7) m=2, Force of pulley=22.2N

How did u get r?
9. (Original post by Mimiastc)
How did u get r?
The Resistive force?

Basically, I found the deceleration using suvat of the combined mass 10.5 kg and Applied F=ma by considering forces acting on the combined mass to find R.
10. (Original post by SaadKaleem)
The Resistive for
Basically, I found the deceleration using suvat of the combined mass 10.5 kg and Applied F=ma to the equation to find R.
Was ur deceleration -150 ? And was the downward force 10.5g ?
11. (Original post by Mimiastc)
Was ur deceleration -150 ? And was the downward force 10.5g ?
I recall it being -200 something.
12. (Original post by SaadKaleem)
My Answers: (They could possibly be wrong)
Spoiler:
Show

Horizontal Kinematics: 1) a=3.5 ms^-2 and u=12 ms^-1
Momentum & Impulse: 2) k=1/2 and I=3/2mu
Vertical Kinematics & Forces: 3) h=2.5, R=2250N
Moments: 4) F(A)=3460, F(C)=4250, m=77kg
Forces & Acceleration as vectors: 5) b) a=4.47 (2 square root 5) ms^-2
Horizontal Kinematics (Overtaking) 6) a=4, T=7, D(Q)=73.5
Connected Particles: 7) m=2, Force of pulley=22.2N
on Q7 did you round up your m=2? I got m=1.92
13. (Original post by ordu_55)
on Q7 did you round up your m=2? I got m=1.92
Got exactly 2 kg.
14. (Original post by Mimiastc)
Was ur deceleration -150 ? And was the downward force 10.5g ?
yeah i got that too but i think its wrong
15. I got 1575 for R.. Pretty sure it's wrong.
Others seem okay except for the force on pulley..
How much marks was that?
16. (Original post by SaadKaleem)
My Answers: (They could possibly be wrong)
Spoiler:
Show

Horizontal Kinematics: 1) a=3.5 ms^-2 and u=12 ms^-1
Momentum & Impulse: 2) k=1/2 and I=3/2mu
Vertical Kinematics & Forces: 3) h=2.5, R=2250N
Moments: 4) F(A)=3460, F(C)=4250, m=77kg
Forces & Acceleration as Vectors: 5) b) a=4.47 (2 square root 5) ms^-2
Horizontal Kinematics (Overtaking) 6) a=4, T=7, D(Q)=73.5
Connected Particles: 7) m=2, Force of pulley=22.2N
I got 21.952 N for force on pulley by string. Is that wrong....I found it using Tsinalpha + Tsinbeta
17. (Original post by Becka99)
I got 21.952 N for force on pulley by string. Is that wrong....I found it using Tsinalpha + Tsinbeta
I found it using triangle angles, Splitted the angle between pulley and tension as 45 degrees.

Used, 2TCos45 for the resultant then.
18. pretty certain these weren't any of the questions on my paper.. were there two different m1 edexcel papers today??
19. for the last question i attempted the method but i had a mistake in one of the signs i guess and i didnt have time to fix it

so i used the wrong answer for the T for the next part: the one about the forces on the pulley i did the correct method but obviously used the wrong answer for t
do i gain any mark?
that section had 4 marks am i going to loose them all?

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Updated: June 11, 2016
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