need help in M1

[T.Q]

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 need help in part (b) with explanation plz

[astarmathsandphysics]

When i get home

[vanibharutham]

You need to realise that the resultant vector is parallel to i – 2j

This means that the resultant vector will have a magnitude of a multiple of Root 5 ( by finding the magnitude of i - 2j)

Anyways,

so to get a resultant force, you add your vectors

4i – 5j + pi + qj = Resultant

Because resultant is parallel to i - 2j, you know that the ratio of i's to j's are equal

therefore

(4+p)/(q-5) = 1/-2

Cross multiply

-2(4-p) = q - 5

-8 - 2p = q - 5
2p + q + 3 = 0

 

[XFlufferzX]

Thannnnnnnnnksss vanibharutham !!
What a coincidence! 
I was going to ask my teacher tomorrow about dat problem

[T.Q]

You need to realise that the resultant vector is parallel to i – 2j

This means that the resultant vector will have a magnitude of a multiple of Root 5 ( by finding the magnitude of i - 2j)

Anyways,

so to get a resultant force, you add your vectors

4i – 5j + pi + qj = Resultant

Because resultant is parallel to i - 2j, you know that the ratio of i's to j's are equal

therefore

(4+p)/(q-5) = 1/-2

Cross multiply

-2(4-p) = q - 5

-8 - 2p = q - 5
2p + q + 3 = 0

 

thank you very much