M1 Jan 09 Q6 Help Please :( Watch

gozatron
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Okay, very hard paper for me.

6. Two forces, (4i – 5j) N and (pi + qj) N, act on a particle P of mass m kg. The resultant
of the two forces is R. Given that R acts in a direction which is parallel to the
vector (i – 2j),
(a) find the angle between R and the vector j,
(b) show that 2p + q + 3 = 0.
Given also that q = 1 and that P moves with an acceleration of magnitude 8√5 m s–2,
(c) find the value of m.

I can find the a)

But B) and C) not a chance..

Any help?
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the bear
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well for part b) you can say that the direction 2i + j is perpendicular to i - 2j ( draw a sketch to see !! )... so if you do a dot product with the resultant it will equal zero....

and for part c) the magnitude of the resultant force will be the mass M x the magnitude of the acceleration.

the magnitude of R is the square root of the sum of the squares of the components of R... which can now be represented as numbers since you know q and can find p from the equation .

the bear

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gozatron
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(Original post by the bear)
well for part b) you can say that the direction 2i + j is perpendicular to i - 2j ( draw a sketch to see !! )... so if you do a dot product with the resultant it will equal zero....

and for part c) the magnitude of the resultant force will be the mass M x the magnitude of the acceleration.

the magnitude of R is the square root of the sum of the squares of the components of R... which can now be represented as numbers since you know q and can find p from the equation .

the bear

Could you explain B) a little further please?
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the bear
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when you do the dot product of two perpendicular vectors the answer is always zero... here we can use the force vector R and a direction which is perpendicular to that of R.

if you have not covered the dot product yet there is probably an alternative approach.

the bear

:o:
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gozatron
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(Original post by the bear)
when you do the dot product of two perpendicular vectors the answer is always zero... here we can use the force vector R and a direction which is perpendicular to that of R.

if you have not covered the dot product yet there is probably an alternative approach.

the bear

:o:
I havent come across the Dot product no sorry

I looked at the mark scheme it says:

(4+p)i + (q-5)j
(q-5)=-2(4+p)
2p+q+3=0

I don't understand why they suddenly multiply (4+p) by -2.

Sorry for being a pain
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Rohit93
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okay so we know R is in the direction of i -2j... so it has to be some multiple of i -2j
so if we call the multiple x we have R = x(i - 2j) = xi -2xj

and grouping the i and j terms we get R = (4+p)i + (q-5)j

So if we equate the (i)s we get x = 4 + p
equating the (j)s we get -2x = q - 5

substituting the equation gives us 2p + q + 3 = 0
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gozatron
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(Original post by Rohit93)
okay so we know R is in the direction of i -2j... so it has to be some multiple of i -2j
so if we call the multiple x we have R = x(i - 2j) = xi -2xj

and grouping the i and j terms we get R = (4+p)i + (q-5)j

So if we equate the (i)s we get x = 4 + p
equating the (j)s we get -2x = q - 5

substituting the equation gives us 2p + q + 3 = 0
Thank you very much for the help

Though i would never would have thought of that, lets hope it doesnt come up in monday's exam.
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