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Mathematics: Post your doubts here!

qwertypoiu

qwertypoiu

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ashcull14 said:
View attachment 52128
in part 1 i used v=u+at but the ans turned out to be wrong why??
i used vP= 1.3 +(0.1*20)=3.3m/s
ince vP=vQ so 3.3= vQ
thus, at t=20s 3.3= uQ+(0.016*20) uQ= 3.3-0.32=2.98m/s
Whereas, in markscheme the formula s=ut+1/2at^2 is used which gives the ans 0.1m/s
Can someone explain why cant we use v=u+at here???
Click to expand...
Firstly, equations of motion are to be used only when acceleration is constant. Particle P is moving at a constant speed, but Q is not. Your using v=u+at to find v = 3.3 for P is correct. For Q however, none of the equations of motion can be used, because Q's acceleration is not constant. You will have to integrate the acceleration to find its velocity function. When you integrate, the velocity function will include a +c, the arbitrary constant. By knowing that when t=20, v=3.3 for Q, you're able to substitute this into velocity function, to find the arbitrary constant, and thus able to find v when t=0.
 
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ashcull14

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P13 nov 2014 maths.PNG
to find f '(x) integration= y= -18/x^2 +c ...why do we have to use x=0 to find c why not the x of stationary point that is 3?
 
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ashcull14

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qwertypoiu said:
Firstly, equations of motion are to be used only when acceleration is constant. Particle P is moving at a constant speed, but Q is not. Your using v=u+at to find v = 3.3 for P is correct. For Q however, none of the equations of motion can be used, because Q's acceleration is not constant. You will have to integrate the acceleration to find its velocity function. When you integrate, the velocity function will include a +c, the arbitrary constant. By knowing that when t=20, v=3.3 for Q, you're able to substitute this into velocity function, to find the arbitrary constant, and thus able to find v when t=0.
Click to expand...
how can i figure out from the question that the acceleration is not constant?
 
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Ahmed Aqdam

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ashcull14 said:
View attachment 52134
to find f '(x) integration= y= -18/x^2 +c ...why do we have to use x=0 to find c why not the x of stationary point that is 3?
Click to expand...
x=3 will be used and the equation will be equal to 0 as dy/dx is 0.
ashcull14 said:
how can i figure out from the question that the acceleration is not constant?
Click to expand...
For P the acceleration is constant as the acceleration is just a number but for Q the expression involves time too. This means it varies according to time and hence is not constant.
 
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ashcull14

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Ahmed Aqdam said:
x=3 will be used and the equation will be equal to 0 as dy/dx is 0.

For P the acceleration is constant as the acceleration is just a number but for Q the expression involves time too. This means it varies according to time and hence is not constant.
Click to expand...
thnks :D
 
TheJDOG

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crazytaylorfanXD said:
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_42/

can someone help me solve question 5?
Click to expand...

Hi,
5)
i) they want the KE gain of the whole system, we know that KE=1/2mv^2
We know block A mass= 5 kg and B= 16 kg
So KE of whole system = (1/2)5v^2 + (1/2)16v^2 = 10.5v^2 J

ii) a) we need the loss of PE of the system. We know that PE= mgh and we know they moved a distance of x
We also know that the loss of PE of the system= Loss of b - gain of A
Well, logically B lost PE it moved downwards, so PE Loss of b= mgh= 16(10)(x) J
Also, A gained PE it moved up the slope, use basic trigonometry to find the distance it moved vertically up in terms of x.
So PE gain of A= mgh= 5(10)(xsin30) J
Loss of PE of the system= 16(10)(x) - 5(10)(xsin30)= 160x - 25x= 135x J

b) here we need the work done against friction. We know that f=uR and work = FxD
given u= 1/root3
We have to resolve, take y axis perpendicular to slope. We have R acting upwards along y axis and we have the component of weight of A acting downwards along y axis. Now R= Wcos30=50cos30= 25root3.
So friction = uR= (1/root3)(25root3)= 25 N
So work done against friction = 25x J since it moved a distance of x

iii) we know that Gain in KE= Loss in PE - work done to oppose friction
So 10.5v^2= 135x - 25x
10.5v^2 = 110x
Multiply both sides by 2
21v^2= 220x
 
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cadburrylover

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can someone tell me when continuity correction has to be done for calculation mean/variance for a histogram data in s1?
 
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