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Edexcel M2/M3 June 6th/10th 2013

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Reply 360

Exams v__v

Review exercise 1, q46? Thanks. :smile:

Reply 361

Tuya

Original post by RYRK

yeah , but i dont understand why you do that

Well you work out that tan angle is 52.5/50
So, the ratio of vertical velocity to horizontal velocity is 52.5 to 50. But horizontal velocity is constant. So work out the the vertical velocity by drawing a triangle and using this information.

If this went over your head, there is another way of doing this someone suggested on the previous page I think.

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Reply 362

JenniS

Original post by Boy_wonder_95

you'll be fine tomorrow, good luck!

On question 2b June 2012, for some reason I got the inequality sign wrong Velocity of P is less than 0, so why is it not 0>2u(6e-1)/7

Help please :tongue:

S*** ignore that I got a minus sign wrong earlier on. bugger

(edited 10 years ago)

Reply 363

Anonymous1717

Original post by Exams v__v

M2, june 2012 question 3c. Why is the angle 75? :s why Fcos75?
How do we work this angle?

Do you mean another paper or...?

Reply 364

Anonymous1717

Has anyone actually seen a question about toppling in the current specification's papers?

Reply 365

AB07

Original post by christina_rose

Can anyone help with june 2006, question 8b. Its co-effiecients of restitution and i dont know why but i cant get my head around it :mad:

argh

thanks! :smile:

Okay so you've just worked out the two speeds in terms of u so now you use your term for the speed of B after the collision which is 1/5u(e+1). So now you do another collision calculation but with the wall and you use the equation e= speed of rebound/speed of approach so basically 1/5u(e+1) multiplied by 4/5 (the new e value) to get 4/25u(e+1).

Once you've found the returning speed of particle B you need to know if it'll catch back up with the A particle and to do that it must be faster (greater than the speed of A). So you get:
4/25u(e+1) >1/5u(4e-1)
U will cancel and you can rearrange to get e<9/16.
The second limit comes from the fact A's direction reverses after the first collision but as most people take A in the opposite direction when first calculating it (tells you in the question) it must be greater than 0. So you get
(4e-1)>0
e>1/4
Thus, 1/4<e<9/16

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Reply 366

MathsNerd1

Original post by Anonymous1717

Has anyone actually seen a question about toppling in the current specification's papers?

Nope, that's why I feel one might show up tomorrow.

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Reply 367

JenniS

Original post by MathsNerd1

Nope, that's why I feel one might show up tomorrow.

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oh is toppling M2, my M2 and M3 have merged into one :/

Reply 368

MathsNerd1

Original post by JenniS

oh is toppling M2, my M2 and M3 have merged into one :/

Yeah it's in M2 and I'm almost willing to bet money on the fact of it appearing tomorrow morning as its in the spec but hasn't appeared until this day

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Reply 369

fishnets

Hi,
Guys
Could you help me on these two questions. Both part of june 2012.

Original post by Anonymous1717

Do you mean another paper or...?

Oops, my bad. :P

Could you help in the basic idea of toppling? What do we need to know?

Reply 371

Bromers

Does anyone have any tips on when to take moments and when to just equate?

Reply 372

Boy_wonder_95

Original post by Bromers

Does anyone have any tips on when to take moments and when to just equate?

it's best to take moments at the place where most forces are acting, in most cases it's 'A' so that you can ignore the forces acting there

You equate when you compare vertical forces and similarly horizontal forces

Reply 373

Exams v__v

Original post by Boy_wonder_95

I always get confused between which moment is clockwise and anticlockwise!

Reply 374

JenniS

Original post by Boy_wonder_95

Yeh I take moments either when there's an unknown force you don't need to know for the part you're doing, or as you've said, it there's a few at one place

Reply 375

Boy_wonder_95

Original post by Exams v__v

I always get confused between which moment is clockwise and anticlockwise!

Visualize it! Even if you had to use your fingers in the exam (that's what I do :biggrin:

)

Reply 376

.Username.

Can anyone tell me the all of the assumptions I.e. what rod, plank, particle, lamina mean (and any others)

Reply 377

RoyalBlue7

Original post by Exams v__v

I always get confused between which moment is clockwise and anticlockwise!

Well you could determine them like this:
Keep you index finger at the point around which you are taking moments, then without moving it swing you thumb over and in the direction of the arrow of the force. After you have passed the arrow head see how best it can make a circle around that point. That circle will show if it is clockwise or anticlockwise.

You can use either the thumb or the index finger for either. What's seems more comfortable... :wink:

Reply 378

Olive123

Original post by .Username.

Can anyone tell me the all of the assumptions I.e. what rod, plank, particle, lamina mean (and any others)

I know 2 after doing an exam q to do with a rod lol

A rod doesn't bend and it has negligible thickness

If you have the M1 book the first chapter has a few..