Oxford Physics: PAT test discussion

Y equation was made of omega*t also. So, if u saw both the equations in terms of t and not omega*t... It won't make a difference.

See... Let's say the X equation was a(t - sine(t)) sine(t) will take the value of 1/2 ONLY. But.... t can take MANY values: pi/6, pi - pi/6, 2pi + pi/6 are some examples. (note that The sine of all these values is the same: 1/2)

Yes indeed, but was everwhere where i wrote t omega t? Becouse than the solution would be the same

Yes indeed, but was everwhere where i wrote t omega t? Becouse than the solution would be the same

But the solution of root3-2cosx is x is +- 1/6 pi so when you plug this into sine you get +-1/2

for the circles question, if I got one of the radii wrong but the centres of both and the other radius right and correctly drew in all four tangents but was unable to work out their lengths, how many marks do you think that would be?

How much did you get the radii as?

To answer your question, I think very less. Probably 2/9.

2.5 and 3

Oh yeah... Then there are 4 sets of solutions (general solutions).

I got the radii as 5 and 3. I am pretty sure I didn't make a mistake, but everyone here seems to be getting 2.5 and 1.5. Idk.

Okay here is the full and I believe correct solution to the binary system:

For the 2 star binary system, notice that the graviational force on a star is
$\frac{Gm^2}{4R^2}$, this is because the gravitational force between the 2 stars is the inversely proportion to the square of the distance BETWEEN THEM, which in this case is $2R$, notice that the radius of orbit of a star is $R$ so we have the centripetal force is $\frac{mv_2^2}{R}$. So after rearranging we have that ${v_2^2}=\frac{Gm}{4R}$. The distance squared between 2 stars in the 3 star system is 3, so the graviational force is $\frac{Gm^2}{3R^2}$, resolving this force towards the center and recognising that there are 2 stars we get that the centripetal force is $\frac{\sqrt{3}Gm^2}{3R^2}$, we finally get that $v_3^2=\frac{\sqrt{3}Gm}{3R}$. I'll leave it as an exercise for you to find the ratio.

No 2 i think

Exactly the answer I wrote before. V_3 = 2*(1/3)^(1/4)*v_2

2 sets of sine solutions with 1/2 and 2 sets with -1/2. Of course it can be totally 2 sets depending on how u write the general solution (U probably write both the 2pi + theta and pi - theta as a combined general solution. In that case, 2 sets)

No becouse with -1/6 pi comes - 1/2 and 1/6pi vice versa so you get two general