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take lcmView attachment 58034
last part
Did you try partial fractions?View attachment 58034
last part
Mass = 1100
Fd = 1800
Fr = 700N
Distance = x
Speed = v
i) K.E = 1/2(1100)v^2 = 550v^2
P.E = (1100)(10)(160) = 1760000 Now we have to find it in terms of x which is distance. So it will become 1760000/1760 = 1000x
kv^2 = x
1800x = 700x+ 550v^2+ 1000x
^This is the WD equation. I just substituted everything in terms of x and v
Solve it further you will get
100x = 550v^2
x = 550/100v^2
x = 5.5v^2
Hence k = 5.5 Ans.
ii) 5.5v^2 = 1760
v^2 = 320
1800(x-1760) = 700(x-1760)+ 550(v^2-320) [we subtracted 320 because it was our initial velocity and so to find gain in K.E we already have v^2(initial velocity)
1800x-3168000 = 700x - 1232000+ 550v^2 - 176000
1800x-700-3168000 = -1408000+ 550v^2
1100x-3168000+ 1408000 = 550v^2
1100x+ 1760000 = 550v^2
(1100x+ 176000)/550 = v^2
v^2 = 2x - 3200 (shown)
take lcm
dh/dt = (100-h^2)/20h^2
(20h^2/(100-h^2)) dh = dt
Use result from part 2.
-20 + 2000/(10-h)(10+h) dh = dt
use partial fractions
-20 + 100/(10-h) + 100/(10+h) dh = dt
then integrate
Thaks a lotDid you try partial fractions?
Both of those things are in the examiner report.Thaks a lot
Yes i did but my C is zero just wanted to confirm because in ms there's no correct answer given.
Btw in 2 nd part can we just say assume h=1 anwers are 2/100 and 2/99 so since the answers are close enough hence we can say that they are eqivalent
Further Pure Mathematics by Brian and Mark Gaultercan u plz suggest a good book for further math
thnxFurther Pure Mathematics by Brian and Mark Gaulter
A Concise Course in Advanced Level Statistics by Crawshaw and Chambers
Mechanics for A Level by Bostock and Chandler
They don't cover all the topics, but you can find the majority of them in these books.
consider points A and C, u have...https://drive.google.com/file/d/0B6rlbmTCHUFZRm0tN2RkZEJVNW8/view?usp=sharing
Q8..! Mechanics..? need help..!
Thanx bro..!consider points A and C, u have...
time = 6s
displacement = 60m
initial velocity = 4 m/s
use s = ut + 1/2 at^2, and u get a = 2 m/s^2
now that u have acceleration,
consider points A and B,
time = 3s
initial velocity = 4 m/s
acc = 2
use again s = ut + 1/2 at^2
s = 21 m
http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Thankshttp://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Use these notes for these type of questions. You will find what u need in the last heading that's all about maximum and minimum values
Can u please post the solution. I already refered to this site.. nahi aa raha mekohttp://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Use these notes for these type of questions. You will find what u need in the last heading that's all about maximum and minimum values
Sure. Wait upCan u please post the solution. I already refered to this site.. nahi aa raha meko
IDK no solution with me thats why m confused.Sure. Wait up
Whats the answer ? I'll post if I get it correct
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