Can somebody help me with Q9 of the 2015 paper?
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Thanks in advance.
P and Q are fixed in place. Since P and R are in the same direction, the force between them pulls R to the left. But since Q and R are in opposite directions, the force between them pushed R to the right. So pretty much find both forces and add them:
Force : F ∝ I
1I
2L / d
Since I and L are constant; F = 1 / d
Now you're given the force between Q and R, thats F, so:
F = 1 / 10 to the right between Q and R
d is 20mm between P and R so force between P and R = 1 / 20
Since F = 1/10 from before and youre tryna get an answer in terms of F, you can sub F as 1/10:
F between P and R = 1/10(2) = F/2
So F
QR to the right (positive) plus F
PR to the left (negative)
= F + - F/2 = F/2
Positive means to the right, so its B
(There is probably an easier way for this but this is what makes most sense to me
)