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# M1 mk2

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1. 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
2. (Original post by Steelmeat)
is this right?
Zacken
Yes.
3. (Original post by Zacken)
Yes.
The second part of this question is
at time t=0 P is moving with velocity (-2i+4j)
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)???
4. notnek is that right?? ^^
5. (Original post by Steelmeat)
The second part of this question is
at time t=0 P is moving with velocity (-2i+4j)
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...
6. (Original post by notnek)
It's nearly right but notice the question asks for speed and not velocity...
i see well i tripped up there
so speed is right?
7. (Original post by Steelmeat)
i see well i tripped up there
so speed is right?
Correct
8. (Original post by notnek)
Correct
sorry to bother you again but

can i say that |v|=?
9. (Original post by Steelmeat)
sorry to bother you again but

can i say that |v|=?
Yes that's fine.
10. (Original post by notnek)
Yes that's fine.
thanks

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