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Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread

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Is my working out too OCD compared to some of your working out? :lol:

Obviously, I won't do them like this in the real exam I.e free hand drawing but I'd rather
perfect everything neatly drawn so I can get into a perfect pattern etc. :moon:

Also, how are some of your working outs when you answer questions compared to mine? :hmmmm2:

Reply 401

Cryptokyo

Original post by KloppOClock

would there be tension when A and B are back to there original position, or if A and B reach a height of 2L? Also if A gets tension before B does after reaching its original position first, does it rebound?

In regards to your questions.

Original post by KloppOClock

if A and B reach a height of 2L?

1. The question states that

u^{2}<\frac{1}{2}gl

. This inequality is here such that the particle with speed

2u

does not collide with the pulley. Hence neither particle goes above a height

l

from their original position.

Original post by KloppOClock

Also if A gets tension before B does after reaching its original position first, does it rebound?

2. Both particles must get tension at the same time as they are connected by a single string. Tension occurs will the length of the string is of definite length

2l

with no slack. In reality there would be rebound but this does not have to be considered for the question.

Original post by KloppOClock

would there be tension when A and B are back to there original position

3. There would be tension if, and only if both particles were in their original position.

Hints for question
Split the problem into the 3 parts:
- When both A and B are going upwards.
- When A is going upwards and B goes downwards.
- When both A and B are going downwards.

Hope this helps,
Cryptokyo

Reply 402

Cryptokyo

Original post by XxKingSniprxX

Is my working out too OCD compared to some of your working out? :lol:

Obviously, I won't do them like this in the real exam I.e free hand drawing but I'd rather
perfect everything neatly drawn so I can get into a perfect pattern etc. :moon:

Also, how are some of your working outs when you answer questions compared to mine? :hmmmm2:

I pretty much do the same but I make the diagrams as simplistic as possible while still ensuring all the information is there. But a slight difference may be that I:
-represent vector quantities with solid lines and use a little arrow (or arrows!) - but more on that later.
-represent solid surfaces with a solid line with stripes on the opposite side to the surface.
-always represent an object with a single blob of ink.
-use dashed lines for measurements.
Just on arrows...
-use single-headed arrows for forces.
-use double-headed arrows for acceleration.
-arrows halfway along the line for velocity.
-use bracket endings on an arrow for displacement.

But that is much overkill for one to do. But it makes the drawings only have about 4 lines. :biggrin:

How do I do part A?

Original post by Sam1999

How do I do part A?

just from looking at it I would guess that you resolve the weight of P and apply F=ma to it

Reply 405

Sam1999

Original post by KloppOClock

just from looking at it I would guess that you resolve the weight of P and apply F=ma to it

But how do I resolve it as I don't know the hypotenuse

Reply 406

KloppOClock

Original post by Sam1999

But how do I resolve it as I don't know the hypotenuse

im just doing the question now so ill have a look

Reply 407

KloppOClock

Original post by Sam1999

But how do I resolve it as I don't know the hypotenuse

just set the hypotenuse to equal a mass of "P" times G to get a weight. Put that into F=MA and the P should cancel out i think

Edit: just checked the mark scheme, P's cancel out and final answer is 8.49N

(edited 7 years ago)

Reply 408

Sam1999

Original post by KloppOClock

just set the hypotenuse to equal a mass of "P" times G to get a weight. Put that into F=MA and the P should cancel out i think

But doesn't the weight act vertically down?

Reply 409

Sam1999

Original post by Sam1999

But doesn't the weight act vertically down?

Never mind I'm being stupid 😤 I've been stuck on that for 20 mins

Reply 410

KloppOClock

Original post by Sam1999

But doesn't the weight act vertically down?

yes but your trying to find acceleration which isnt vertical.

First resolve the force of P. Use a weight of P*G (mass times gravity). Then taje the horizontal component and sub that into F=MA

Reply 411

KloppOClock

Original post by Cryptokyo

-use bracket endings on an arrow for displacement.

what do you mean by that?

(edited 7 years ago)

Reply 412

_Priyesh_

Stuck on 1b from iygb paper v (dont think there is ms)

Posted from TSR Mobile

1465145988769.jpg43.9KB

Reply 413

ozziew

Can someone please explain to me why for part d we use 2s1 instead of just s1 + s2, taken from Jan 10.

Spoiler

(edited 7 years ago)

Reply 414

Themathgeek

Original post by _Priyesh_

Stuck on 1b from iygb paper v (dont think there is ms)

Posted from TSR Mobile

Which paper? Whats iygb

Posted from TSR Mobile

Reply 415

KloppOClock

Just marking the M1 Solomon Paper D that I did and I have a question about the mark scheme:

Surely it would also depend upon the size of the stone as well? How can you say the tennis ball will have more wind resistance if they don't state the size or shape of the stone?

Would I get any marks for my answer?

"The minimum speed required would be different depending upon the mass of either object as the resultant force of 'F' in F=MA would change which would effect the acceleration as they have different masses, therefore more or less speed may be required."

(edited 7 years ago)

Reply 416

metrize

Original post by KloppOClock

just from looking at it I would guess that you resolve the weight of P and apply F=ma to it

I would have just resolved 9.8 with the angle they gave, would it have been the same answer?

Edit yeah same answer, its because the ball is acellerating vertically down due to gravity so you can resolve