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Mathematics: Post your doubts here!

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

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kIW44v.jpg


QEcISA.jpg


Answers :
afUCYy.jpg


ZaqZainab
Thought blocker
RoOkaYya G
 
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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.
 
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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
 
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