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Aqa as physics 2018 unofficial mark scheme

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Original post by kishxx01
No but momentum must always be conserved so velocity has to decrease.


you are right, but would conservation still apply considering the truck is on a frictionless surface since added mass with same surface area causes a decrease in velocity due to the normal force and weight increasing friction? so when taken out of a real life situation like it has on the frictionless surface would it still be conserved?
Original post by Matt Barnes
i also said this, i also said how the added mass would not decrease velocity since there was no friction/resistive force, therefore the added mass woud have no effect on the velocity.


That's what I was thinking.
Original post by kishxx01
No but momentum must always be conserved so velocity has to decrease.


yeh, youre right. however if velocity decreased it would counter the rise in momentum so the answer would be contradicted? im not sure but would momentu be conserved in this scenario with the frictionless floor, potentially the air resistence may still conserve momentum by decreasing velocity due to newtons third law??
Original post by Matt Barnes
yeh, youre right. however if velocity decreased it would counter the rise in momentum so the answer would be contradicted? im not sure but would momentu be conserved in this scenario with the frictionless floor, potentially the air resistence may still conserve momentum by decreasing velocity due to newtons third law??


i think you are correct, my bad
How many marks do you think you will have to get in each paper to get a B?
Original post by NaomiWilso
How many marks do you think you will have to get in each paper to get a B?


82/140 if we calculate the average from previous papers.

85/140 if you want to gaurantee.
Reply 26
Original post by randint
82/140 if we calculate the average from previous papers.

85/140 if you want to gaurantee.


I hope Its low. I need an A and I got like 40/70 in paper 1
Reply 27
Original post by MissFattyFatty

PLEASE CHECK OUT MY QUESTION ABOUT STRICT PARENTS AND BRAS AND GOING OUT -----I POSTED IT MONTHS AGO AND NO-ONE REPLIED------- IM NEW TO THIS

https://www.thestudentroom.co.uk/showthread.php?t=5251710#primary_content


You are aware this is a PHYSICS thread?
Reply 28
Original post by randint
82/140 if we calculate the average from previous papers.

85/140 if you want to gaurantee.


WHAT??? I would expect this to be the boundary for the grade A not B
Original post by kishxx01
WHAT??? I would expect this to be the boundary for the grade A not B


It's roughly 50% for a b. As much as I would love the grade boundaries to be incredibly low, you have to think realistically here.
Reply 30
Original post by randint
It's roughly 50% for a b. As much as I would love the grade boundaries to be incredibly low, you have to think realistically here.


Last yr was 83/140 for A
Original post by kishxx01
Last yr was 83/140 for A


i looked at the wrong grade boundaries.

around (70 to 75)/140 for a b.
(edited 5 years ago)
Original post by kishxx01
I hope Its low. I need an A and I got like 40/70 in paper 1


Are you carrying on physics to year 2?
Reply 33
Original post by randint
i looked at the wrong grade boundaries.

around (70 to 75)/140 for a b.


And I would say last yrs paper was easier than this yrs paper 1 so it should be around 80/140 for A
Reply 34
Original post by NaomiWilso
Are you carrying on physics to year 2?


No
I'm also dropping physics - or, well, have dropped it now. I ended up with values of 7500N and 10000N for 4.4 to 2.s.f., and my friend got the same.
Considering the weight of the beam was 12000N and was balanced evenly, I assumed that the vertical component of tension in each cable was 6000N, then using this value with SOH CAH TOA and the angles to resolve in the direction of tension.

Picture shows this (I wasn't sure which side was which angle but these are the two solutions I had)

(edited 5 years ago)
anybody got an edexcel unofficial mark scheme ?
Original post by Electravess
I'm also dropping physics - or, well, have dropped it now. I ended up with values of 7500N and 10000N for 4.4 to 2.s.f., and my friend got the same.
Considering the weight of the beam was 12000N and was balanced evenly, I assumed that the vertical component of tension in each cable was 6000N, then using this value with SOH CAH TOA and the angles to resolve in the direction of tension.

Picture shows this (I wasn't sure which side was which angle but these are the two solutions I had)



Why does each string only support half the weight? The weight is supported by both, and in ratios relating to the angle they make with the vertical.

In other words, both vertical components of tension when added together equal 12000N. You then find that the horizontal component for each equal each other - it is in equilbirium.

Simplen simultaneous equation and then poof.
I've also looked at the volume of the iron question. I even ran the numbers on wolfram alpha to check my arithemetic and I get my answer: 3.2*10^-8

How did you do it?
Reply 39
Original post by DrSebWilkes
Why does each string only support half the weight? The weight is supported by both, and in ratios relating to the angle they make with the vertical.

In other words, both vertical components of tension when added together equal 12000N. You then find that the horizontal component for each equal each other - it is in equilbirium.

Simplen simultaneous equation and then poof.


There is an easier way, that revolves around the fact that the sum of the two angles is 90 degrees. Because the three forces are in equilibrium and therefore sum to zero, they can be represented as the three sides of a vector triangle. And then;
beam.jpg

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