2000 HSC 5b(vi) (1 Viewer)

[tempco]

The descriptive part of the answer is clear enough, but what about the

v^2 = gl(1-2/3)
v = root(gl/3)

part of it... where'd they get the first expression from?

Thanks!

 

[Jase]

from v) , when T=0. V^2 = 3gl
then take iii) V^2 = v^2 + 2gl(1-cos@) and equate the V^2
so 3gl = v^2 + 2gl(1 - cos@)

since you found @.. cos@ = -1/3 , it should end up as v^2 = gl(3 - 2 -2/3)

i don't think you get a mark for that, so dont worry about it.

 

[tempco]

Oh ok thanks for that.