Mathematics: Post your doubts here!

[fatima_jawed]

Uhh we made 2 simultaneous equations for the distance of 100 and 248 considering s-ut+1/2at^2
and then solved them
so a was 3 and u was 19 i think

 

[fatima_jawed]

have u checked your answers with your batch tho I hope we are correct

Everyone in my batch got different weird answers that's why I'm here checking
They made lots of silly errors when we were discussing tho so i'm just making sure here

 

[ThePacifics]

Uhh we made 2 simultaneous equations for the distance of 100 and 248 considering s-ut+1/2at^2
and then solved them
so a was 3 and u was 19 i think

the thing is i didnt make a simultaneous equation. i just used the average speed distance time method to prove my 2.5 m/s^2 was approx equal to 3. do u think i'll get some ecf. i then proceeded to use v^2 = u^2 -2as to find u
what distance did u get?

 

[fatima_jawed]

You can use average speed distance time method for constant velocity i think
I'm not sure if you'll get an ECF
But you'll get your marks for finding u
and the distance CD?
312 m i think
Don't clearly remember

 

[ThePacifics]

You can use average speed distance time method for constant velocity i think
I'm not sure if you'll get an ECF
But you'll get your marks for finding u
and the distance CD?
312 m i think
Don't clearly remember

yes something involving 231 or 312 i dont remember either

 

[ThePacifics]

any guesses on grade threshold for paper 42 anyone.

 

[Thought blocker]

any guesses on grade threshold for paper 42 anyone.

44 near

 

[DudeWithoutName]

some people got the value 5.20 for mew instead of 0.577. What they did was reverse the direction of P such that now it acts down the plane. So the two simultaneous equataions become P(min)+20gsin(60)=20gcos(60)mew and P(max)=20gcos(60)mew+20gsin(60) and then P(max)=2P(min).. Did the question specify the direction of the force P or no?

 

[armeena]

some people got the value 5.20 for mew instead of 0.577. What they did was reverse the direction of P such that now it acts down the plane. So the two simultaneous equataions become P(min)+20gsin(60)=20gcos(60)mew and P(max)=20gcos(60)mew+20gsin(60) and then P(max)=2P(min).. Did the question specify the direction of the force P or no?

no, but mew can't be more than 1 :/

You can use average speed distance time method for constant velocity i think
I'm not sure if you'll get an ECF
But you'll get your marks for finding u
and the distance CD?
312 m i think
Don't clearly remember

ok, my u was also 19 and the method of finding s was the same. and the distance was 312 too. Last answer was 2.1. What power did you get for the second question?

 

[DudeWithoutName]

no, but mew can't be more than 1 :/



ok, my u was also 19 and the method of finding s was the same. and the distance was 312 too. Last answer was 2.1. What power did you get for the second question?

no mew can be more than 1 actually Edit: I got 2.06*10^6W

 

[Mts.99]

some people got the value 5.20 for mew instead of 0.577. What they did was reverse the direction of P such that now it acts down the plane. So the two simultaneous equataions become P(min)+20gsin(60)=20gcos(60)mew and P(max)=20gcos(60)mew+20gsin(60) and then P(max)=2P(min).. Did the question specify the direction of the force P or no?

No they didnt specify the direction, but i think they said P was applied to give equilibrium, so if mgsinx and friction were acting, and if friction was insufficient to resist mgsinx, P was applied to help F overcome mgsinx.

 

[ThePacifics]

someone tell the correct answer for power in question 2 i want to confirm too

 

[abbas haider]

someone tell the correct answer for power in question 2 i want to confirm too

206,000 W

 

[muaz_ali]

206,000 W

2060,000 my dude.

 

[hellodjfos;s'ff]

My answer for Q2 was some big number with 2 at the start but several mates of mine got like 36 something. As for the coefficient of friction, both methods to find coefficient of friction appear to be correct and logical but the only reason you might not choose the method that got the answer 5 is that, 5 is too big to be a coefficient of friction,

 

[muaz_ali]

My answer for Q2 was some big number with 2 at the start but several mates of mine got like 36 something. As for the coefficient of friction, both methods to find coefficient of friction appear to be correct and logical but the only reason you might not choose the method that got the answer 5 is that, 5 is too big to be a coefficient of friction,

r u from lgs jt?

 

[hellodjfos;s'ff]

Btw if someone gets hold of the M1 paper variant 12, please do post it here. Need to check my answers with my teacher

 

[hellodjfos;s'ff]

r u from lgs jt?

How could you possibly know

 

[muaz_ali]

How could you possibly know

theetay udhar hi peeda hotay hen, knew it dude.

 

[Tauqir_hamed]

guys for s2, do they cut marks if we havent written the reason why we have used normal distribution as an approximation to possion ? ( i think i forgot to write lambda > 15 , so normal ). but the answer is correct