Mathematics: Post your doubts here!

[daredevil]

Suchal Riaz yeah u can do it now.. i thought maybe u had chemistry paper so i said that u cud do it after that...


i'll be waiting... thanks a lot for ur help man!
this is the one concept in the pulleys topic that is not going that well for me... :/

 

[usama321]

Why is the formula s=1/2 at2 when u is not zero? What gave you taken u as then?

I said take displacement from the highest point of the signal. U is zero there

 

[Suchal Riaz]

Suchal Riaz yeah u can do it now.. i thought maybe u had chemistry paper so i said that u cud do it after that...


i'll be waiting... thanks a lot for ur help man!
this is the one concept in the pulleys topic that is not going that well for me... :/

daredevil
my best attempt at explaining. please ask me if you have any queries. make sure you understand everything.

 

[Suchal Riaz]

daredevil here is a little video which will play without proxy.
http://screencast.com/t/XwDUt8PMJ3L

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_43.pdf
What's the purpose of the coefficient of friction in question 1?

 

[daredevil]

daredevil
my best attempt at explaining. please ask me if you have any queries. make sure you understand everything.

oohh thanksss a lottt man!! that was a real good explanation ...!! :`)

i'll taag u if i have some more questions... that cool?

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_43.pdf
What's the purpose of the coefficient of friction in question 1?

F=ma
Forward force - backward force = ma

-> backward force = resistance = Reaction (coefficient of friction) and reaction=weight

WSin14 - 0.02W = ma

-> W=mg

mgSin14 - 0.02mg = ma
take m common and cancel from both sides:

10(Sin14) - 0.2 = a
a=2.219 ms^-2

2as = v^2 - u^2
2(2.219)(50) = v^2 - (8)^2
v = 16.9 ms^-1

 

[Snowysangel]

F=ma
Forward force - backward force = ma

-> backward force = resistance = Reaction (coefficient of friction) and reaction=weight

WSin14 - 0.02W = ma

-> W=mg

mgSin14 - 0.02mg = ma
take m common and cancel from both sides:

10(Sin14) - 0.2 = a
a=2.219 ms^-2

2as = v^2 - u^2
2(2.219)(50) = v^2 - (8)^2
v = 16.9 ms^-1

No according to the ms, 'w' is the vertical component of acceleration due to free fall...can that happen...can friction = u x g(cos a)? Why cause g(cos a) isn't the normal reaction :S

 

[Snowysangel]

No according to the ms, 'w' is the vertical component of acceleration due to free fall...can that happen...can friction = u x g(cos a)? Why cause g(cos a) isn't the normal reaction :S

Oh wait never mind I get it

 

[Asad Moosvi]

Hey guys, may someone please help me with an m1 question? Q7 part (ii) and (iii) from the following paper: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf


I would really appreciate it if someone could help me out as soon as possible as I still have to do more questions. Thank you in advance.

 

[Jelleh Belleh]

Hey guys, may someone please help me with an m1 question? Q7 part (ii) and (iii) from the following paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf


I would really appreciate it if someone could help me out as soon as possible as I still have to do more questions. Thank you in advance.

View attachment 41931

 

[Asad Moosvi]

Could you please explain how you solved it?

 

[Asad Moosvi]

I don't understand it...

 

[Browny]

I have seen equations like F<uR and so on without the traditional F=uR equations. Can anyone please explain to me the conditions for all inequality equations involving uR?

 

[daredevil]

yeah plz explain it...

 

[usama321]

I have seen equations like F<uR and so on without the traditional F=uR equations. Can anyone please explain to me the conditions for all inequality equations involving uR?

They usually require a bit of logic. E-g sometimes the question says that the block is stationary instead of limiting equlibrium. Thus if the block is not moving, the resistance could be equal to or much greater than the force trying to budge it. Thus we will use >=
Another such condition is usually involved in newton's second law. E-g F - uR would be > 0 if the object is accelerating and so on.
However there could be countless possibilities for such questions. Just try to get the hang of these questions, as they can be quite tricky

 

[usama321]

They usually require a bit of logic. E-g sometimes the question says that the block is in equilibrium instead of limiting equlibrium. Thus if the block is not moving, the resistance could be equal to or much greater than the force trying to budge it. Thus we will use >=
Another such condition is usually involved in newton's second law. E-g F - uR would be > 0 if the object is accelerating and so on.
However there could be countless possibilities for such questions. Just try to get the hang of these questions, as they can be quite tricky

Oops i meant that the block is not moving.... not in equlibrium. I am editing it

 

[daredevil]

i dont get the part about no slipping.... like if they dont slip then they keep sticking together or wat?

 

[Snowysangel]

If they don't slip, friction is greater than the applied force

 

[Snowysangel]

Could someone please explain question 7 to me http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_41.pdf