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Edexcel(IAL) M1 June 8th 2016 Discussion

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Reply 60
Avatar for Riddzy
Riddzy
You guys where are the unofficial mark schemes posted?
Reply 61
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Riddzy
Original post by SaadKaleem
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.


How did you find the deceleration? Like I used a suvat equation, taking u as 7 I guess.
Original post by Riddzy
How did you find the deceleration? Like I used a suvat equation, taking u as 7 I guess.


Messed it up >.< Mine answer is wrong probably.
Reply 63
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Riddzy
Original post by SaadKaleem
Messed it up >.< Mine answer is wrong probably.


I am soooooooooo scared, because most people in my school found it nice, but I thought some questions were really hard.

BTW I got m as 2 as well for the last question. Did you do F=ma, and then combine it for both particles, and T cancels out?
What values did you guys use for your equation to find the acceleration? Did you use s=.12 or s=12 as the distance was 12cm not 12 m
Original post by queenoflean
What values did you guys use for your equation to find the acceleration? Did you use s=.12 or s=12 as the distance was 12cm not 12 m


ya 0.12 m
Original post by SaadKaleem
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.


I bymistake under pressure used the 2.3333 value 😭😭 how many marks do u think i will loose?
Original post by Mimiastc
I bymistake under pressure used the 2.3333 value 😭😭 how many marks do u think i will loose?


1 mark only.
Original post by SaadKaleem
1 mark only.


Really? But i got the wrong distance too then?... Yay!!!🎉
Original post by Mimiastc
Really? But i got the wrong distance too then?... Yay!!!🎉


There would (probably) be ECF for the next distance question.
Original post by SaadKaleem
1 mark only.

Oh what did ur graphs looks like?
Original post by SaadKaleem
There would (probably) be ECF for the next distance question.


Okayy what a relief thank you haha☺️
Original post by Mimiastc
Oh what did ur graphs looks like?



Something like this: Do you remember how many marks was that Acceleration time graph?

Spoiler

Original post by wenogk
I think I got 78 :K:


Round down :tongue:
Reply 74
Avatar for EUSHA
EUSHA
2
any unofficial mark scheme posted?
Is anyone going to post the model answers for the paper?
Original post by SaadKaleem
Something like this: Do you remember how many marks was that Acceleration time graph?

Spoiler



Yayyy I did that... The first was 3 ans the second 3 or 2😂
Original post by SaadKaleem
Something like this: Do you remember how many marks was that Acceleration time graph?

Spoiler



for the second graph, why did you draw the line going down?
i did the same thing but without this line
i think the first graph was for 4 marks and the second one for 3 marks
These are all the questions and parts I remember and how I did them. Please let me know If I made any errors and please add to this list because I dont remember some of the questions!


Question 1
Not sure what it was exactly but I remember finding it easy overall

Question 2
Related to the momentum of to masses before and after collision. We had to find the impulse and the mass of one of the two moving objects. I totally skipped this one.

Question 3
The two bricks falling.
So we knew that A weighed 9kg and that B weighed 1.5 kg.
A started from rest so initial velocity was 0 and its final velocity was 7m/s. The acceleration was 9.8 and using these three values we could calculate the fallen height required in the first part.
In the second part I modeled A and B together as one particle of mass 10.5 kg. This gave a downward force of 10.5g. I also used the fact that the two bricks finally came to rest so the final velocity was 0m/s and I made the assumption that there was no resistive force such that all off brick A's downward velocity as transferred to Brick B meaning that the initial velocity was 7m/s. We also knew that the bricks travelled a distance of 12cm or 0.12m. using this data i applied the formula v^2=u^2 +2as to calculate the upward acceleration or the deceleration applied to the bricks as a whole. Then using this acceleration I found the upward resistive force applied by the ground on the bricks by resolving the system in an upward direction.

Question 4
This one was about Vectors. seemed Easy but I hadn't studied this ;-;

Question 5
Diver and Springboard.
For part (a) (i) I did:
50 x 4 - 30 x 2
For Part (a)(ii) I did
50 x 3.4 - 30 x 1.4

For part (b) I used
4x - 30x2 = 5000N (I now realise I shoul have subtracted The normal reaction at C from this as well)

Question 6
The two Cars A and B.
I found T=7
So I started by using the fact that Car A had travelled (1/2 x 3,5 x 14)m in the first 3.5 seconds and then 14m each second thereafter. I then drew up a table of values of distance that A had traveled for each second after that point. I also used the formula s=ut +1/2at^2 to find an equation of motion for B becasue we knew it started from rest and had a constant acceleration of 3m/s^2. The formula simplified to s=1.5t^2. Using this answer I then calculated the distances B travelled for different values of time. At T=7 both A and B had travelled 73.5 m and so that was the point that B passed A.
I found the distance travelled by B to be= 73.5m
Drawing the graphs for this question was pretty easy.


Question 7
The two particles over a pulley and a string.
We knew that the mass of the particle on the left was 4kg and that the system was in static equilibrium.
Using this knowledge I first found the normal reaction of the particle on the left by resolving perpendicular to the slope and using R - 4gcos(a) = 0
Then using my answer I found the F(max) by using the coefficient of friction they had provided us.
From there I resolved in the direction of the tension to find the value of the tension for the particle on the left side.
I then argued that since the particle is in static equilibrium that the tension on both sides of the pulley would be the same.
Then by resolving in the direction of the tension for the particle on the right I was able to find the mass of the particle on the right which I found to be
m= 2KG!

I did the other parts too but I want to know what you guys got for it

Well thats all I remember.
Original post by mr.holmes
for the second graph, why did you draw the line going down?
i did the same thing but without this line
i think the first graph was for 4 marks and the second one for 3 marks


The vertical line, was meant to be dotted sorry, to show it's changing acceleration.

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