MEI OCR M2 June 12

Hey guys, I am really stuck on question 3)b)ii) on the June 12 M2 paper.I can find Tab and Tac, but it goes downhill when looking for Tbc. Link is below:

http://www.mei.org.uk/files/papers/m2_june_2012.pdf

Thanks

If you can find Tab, then you should be able to work out Tbr, and hence Tbc.

But how come I can't just look at joint B and consider only the horizontal forces in order to get Tbc?

Tbr and Tbc both have horizontal components. So you'd need further information.

You could do horizontal components, and do vertical components, but it's easier to resolve along Tba, as Tbc has no component in that direction, and you'll get Tbr straight off, and then along Tbc, giving you an equation involving just Tbc and Tbr.

How does Tbr have a horizontal component? Forgive me, I find this really confusing....

I assume where you're refering to horizontal that this corresponds to the horizontal on the diagram, i.e. the direction CA. As such BR is at an angle of 60 degrees to this horizontal and has a horizontal component.

Ohhhh I drew my diagram wrong. So with this, I can just equate Tbr and Tba to get Tbr and go from there?

Yeah I noticed that, thanks Can you help me with 4) vi) please? I don't know why the velocity for P changes.

Not having worked through all the previous parts. I presume that P and Q are travelling at different velocities prior to the final collision, and so when they collide and coalesce conservation of momentum says that their individual velocities must change.

I'd need to see the markscheme if you need more input.

The markscheme is in the link I provided, right at the bottom

Oops! So it is.

So, the velocity of the combined mass (including P) is different to the velocity of P before the collision, because of the collision - not sure how else to explain it.

So how come Vp becomes 10e-3 from 3-10e when e>0.3 ?

Huh!

You asked about part vi). Which part are you looking at?

Whoops! My mistake, iv) was the problem!

Hmmmm. I guess what makes sense, Thank you

You don't sound convinced.

3-10e is negative if P has changed direction (e > 3/10). So, if you want the speed of that velocity you need to take the negative of it, i.e. 10e-3

So as e>0.3, with 10e-3 no mater what what value e has, as long as it satisfies the inequality the speed has to be positive?