Mathematics: Post your doubts here!

[Math Buddy]

 

[Math Buddy]

View attachment 48061

(i)

 

[Abdel Moniem]

http://teachers.cie.org.uk/docs/dynamic/72493.pdf
question 5 help plz.....

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
Work done by driving force=gain in P.E + gain in K.E+Work done against resistance
1800x=1000x+550x^2+700x
100x=55ov^2
kv^2=x
55ok=100
k=5.5
In A.B there is no increase in height so Gain in P.E equals zero
Driving Force-Resistance=mass*acceleration
1800 – 700 = 1100a
a=1m/s^2

 

[Sehrish Ahmad]

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
Work done by driving force=gain in P.E + gain in K.E+Work done against resistance
1800x=1000x+550x^2+700x
100x=55ov^2
kv^2=x
55ok=100
k=5.5
In A.B there is no increase in height so Gain in P.E equals zero
Driving Force-Resistance=mass*acceleration
1800 – 700 = 1100a
a=1m/s^2

Thanks Alot Brother...
btw best of luck for tom

 

[RoOkaYya G]

View attachment 48061

i)
using newtons 2nd law of motion
at A :
F=ma
Tension- weight=ma
T - 2.5 = 0.25a ---------equation 1

at B :
F=ma
weight-tension =ma (in this we do weight - tension coz the motion of the ball is in direction of weight so weight is the driving force)
7.5 - T = 0.75a --------equation 2

solve equation 1 and 2 simultaneously
u will obtain a=5 ms^-2<-------------

using kinematics formula s=ut = 0.5 * a * t^2

u=0 (strts from rest)
t=0.6

s= 0 + 0.5 *5 * (0.6) ^2
s= 0.9 m

 

[bludlynsolja]

hey folks! can someone help me understand this thing straight..
at times t0 find work done against resistance, we subtract the pe from ke. at times, we add them, at times it vice versa, and at times theres a driving force.
now which equation is to be used and in which situation?
coz theres so many versions of equations that it actually gets me baffled and dazzled as to which one to use and in which situation.
have 3 hours to my exam, if anyone could explain, i would really appreciate.
and good luck to all!

 

[Math Buddy]

i)
using newtons 2nd law of motion
at A :
F=ma
Tension- weight=ma
T - 2.5 = 0.25a ---------equation 1

at B :
F=ma
weight-tension =ma (in this we do weight - tension coz the motion of the ball is in direction of weight so weight is the driving force)
7.5 - T = 0.75a --------equation 2

solve equation 1 and 2 simultaneously
u will obtain a=5 ms^-2<-------------

using kinematics formula s=ut = 0.5 * a * t^2

u=0 (strts from rest)
t=0.6

s= 0 + 0.5 *5 * (0.6) ^2
s= 0.9 m

Thank Bro

 

[bludlynsolja]

what times ur exam bruv?
and u got any explanation about the work energy principal like i asked above.
esp when to knw if u have to subtract gpe and ke and when to add?

 

[bludlynsolja]

thx a lot for the help.
but also, there was one instance when there was a loss in both ke and pe as it was descending an undulated slope, so in that case the mark scheme took the change in energy as the loss in ke plus loss in pe and didnt subtract.
so does that mean that if theres a loss in both, u add both the loses to get the change?
and if so, then does it also mean that if theres also a gain in both then u add both of them to get the change?
thx.

 

[bludlynsolja]

oh oritee i see!
thx a lot bro!! u too!!
the key time is 12 right so the exam starts at 10.45 innit if its the morning session?

 

[f_m_r]

Can someone please help me with this, particularly the ii) with full working please
Winter 2013 33
Thanks in advance

 

[Haya Ahmed]

Questions realted to Trignometry A2 ..
Q40 the underlined part
Q44 (ii) b
Q45
Q46

Answers :

ZaqZainab
Thought blocker
RoOkaYya G

 

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[RoOkaYya G]

Thank Bro

sis

 

[Math Buddy]

sis

Oh Sis sorry

 

[RoOkaYya G]

Oh Sis sorry

i thought my gender was specified on my profile summary hihi
np ^_^

 

[cheemaboyz]

Can anyone please help me out with the (ii) and (iii) part .
Thankyou in advance

 

[cheemaboyz]

Can anyone plz help in part (iii)
thanks

 

[RoOkaYya G]

View attachment 48108
Can anyone plz help in part (iii)
thanks

AB is perpendicular to OAB. hence the normal to AB lies on OAB as a direction.
find n hat using cross product taking directions 1. the normal to line (OP) 2. direction vector of plane OAB ( n hat of plane which is the normal to the plane OAB n here it is AB direction)

in short do cross product for OP and direction of AB

replace this in formula r.n hat = a. n hat
where a can be any point on plane. use any u want.

 

[RoOkaYya G]

AB is perpendicular to OAB. hence the normal to AB lies on OAB as a direction.
find n hat using cross product taking directions 1. the normal (OP) 2. direction vector of plane OAB

replace this in formula r.n hat = a. n hat
where a can be any point on plane. use any u want.

View attachment 48106
Can anyone please help me out with the (ii) and (iii) part .
Thankyou in advance

ii)
use scalar product for direction vector of line n plane

cos theta= (direction vector of line dot direction vector of plane) / (magnitude of direction vector of line dot magnitude of direction vector of plane)
angle theta = cos inverse [(direction vector of line dot direction vector of plane) / (magnitude of direction vector of line dot magnitude of direction vector of plane) ]

iii)
find n hat using direction vectors of line and plane
replace in formula r.n hat =a.n hat
use poitn A to replace as a