Mathematics: Post your doubts here!

[RoOkaYya G]

ive

to be honest, i didnt knw! bt thanks for the help...
i might bug u more tomorrow as i have my exam on friday!

exams too on friday
n im leaving today :S
ask Thought blocker and others if u want help
consult chishtyguy for s1

 

[bludlynsolja]

oh my!!
ohrite then...i shall ask them.
good luck with the exam!!

 

[RoOkaYya G]

oh my!!
ohrite then...i shall ask them.
good luck with the exam!!

thanks same to u
uve any more quest for today?
btw where r u from?

 

[bludlynsolja]

thx so much!!
naah not really, if i do, i shall exhaust em in here before retiring to bed...
i am from tanzania and urself?

 

[bludlynsolja]

hmm i see!!
orite then..

 

[RoOkaYya G]

hmm i see!!
orite then..

hmm ok

 

[Abdel Moniem]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Q7
i know that for the first part of the question we apply newton's 2nd law.
sum of forces in the direction of motion-sum of forces in opposite to motion=mass*acceleration
-(0.12)=0.15*a a=-0.8. v=3+(2*-0.8)=1.4 m/s
The block decelerates causing the speed to decrease from 3 m/s till 1.4 m/s. at speed 1.4 m/s the block hits the wall and rebounds. why when the block rebounds it has the same deceleration when it first approached to the wall?

 

[loveyuan]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_33.pdf

question number 9iii
why x is approaches to 10 instead of 20？？？？

 

[RoOkaYya G]

first part!!! how come potential energy is 1000x??? Thought blocker

 

[emilyy]

how to find value of alpha??

 

[Thought blocker]

I cannot see the pics! I said before as well.
Provide the link.

 

[emilyy]

i dnt knw its link , its frm a bk

 

[Thought blocker]

i dnt knw its link , its frm a bk

Then wait for Rookz to answer it.

 

[emilyy]

k

 

[Abdel Moniem]

View attachment 48044

first part!!! how come potential energy is 1000x??? Thought blocker View attachment 48044

Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2
Gain in P.E=mgh*sinѲ=1100*10(x-0)*1/11=1000x

 

[RoOkaYya G]

Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2
Gain in P.E=mgh*sinѲ=1100*10(x-0)*1/11=1000x

thank u loads!!

 

[Abdel Moniem]

thank u loads!!

my pleasure

 

[RoOkaYya G]

how to find value of alpha??

use Newtons second law odf motion or use the formula below if u want easier though
a=-g sin alpha

 

[RoOkaYya G]

my pleasure

actually y dont we take h as 160 itself?
we shld always take sin theta?

 

[emilyy]

but no mass was provided