M1 mk2

2 forces (4i-2j)N and (2i+qj)N act on a particle P of mass 1.5kg. The resultant of these 2 forces is parallel to the vector (2i+j)
fin the value of q

so far i've done this

4i-2j+2I+qj= 6i+(q-2)j

i know that if a vector is parallel to another, it must be a scalar multiple of each other

thus 2i+j multiplied by 3 = 6i+3j

q-2=3
q=5

is this right?
@Zacken

Reply 1

Zacken

22

Original post by Steelmeat

is this right?
@Zacken

Yes.

Reply 2

Steelmeat

OP

2

Original post by Zacken

Yes.

The second part of this question is
at time t=0 P is moving with velocity (-2i+4j)

ms^{-1}

Find the speed of P at time t=2 seconds

so initial position vector is (-2i+4j) and i'm not sure what to do from there

Edit: wait a second f=ma i have the mass and i have a resultant force (6i+3j) so i can get acceleration, so i can do this question?
F=ma
(6i+3j)=1.5a
a= (4i+2j)???

so v=(-2i+4j)+2(4i+2j)
so the speed of P after 2 secs would be
(6i+8j)???

(edited 7 years ago)

Reply 3

Steelmeat

OP

2

@notnek is that right?? ^^

Reply 4

Notnek

21

Original post by Steelmeat

The second part of this question is
at time t=0 P is moving with velocity (-2i+4j)

ms^{-1}

Find the speed of P at time t=2 seconds

so initial position vector is (-2i+4j) and i'm not sure what to do from there

Edit: wait a second f=ma i have the mass and i have a resultant force (6i+3j) so i can get acceleration, so i can do this question?
F=ma
(6i+3j)=1.5a
a= (4i+2j)???

so v=(-2i+4j)+2(4i+2j)
so the speed of P after 2 secs would be
(6i+8j)???

It's nearly right but notice the question asks for speed and not velocity...