M3 restitution

Hello,

4.a on this paper

http://filestore.aqa.org.uk/subjects/AQA-MM03-QP-JUN13.PDF


It is unclear to me how they get the equation for e:

http://filestore.aqa.org.uk/subjects/AQA-MM03-W-MS-JUN13.PDF

by the specification it is V(1) - V(2) / U(1) - U(2) = -e

thus e = V(2) - V(1) / U(2) - U(1)

But in this question the denominator doesn't seem to have changed?

pls help

I do't see why the denominator would change?

Vb - Va = -e(Ub - Ua)
Vb - Va = e(-Ub - - Ua)
Vb - Va = e(Ua - Ub)

Vb - Va / Ua - Ub = e, which is what they got on the mark scheme?

why is Vb first on LHS and not Va? that implies it matters what V1 and V2 is

That's just the way I've been told to learn it - Vb is first on both the LHS and RHS. But, it shouldn't matter which way you have them, as the negative should cancel out if you're getting the right form of the equation.

I think the problem you had originally was when you multiplied V(1) - V(2) / U(1) - U(2) = -e by -1. When you did this, you multiplied both numerator and denominator by -1, to get e = V(2) - V(1) / U(2) - U(1) when you should have just multiplied the numerator, so that the denominator does not change, and so you should have ended up with e= V2 - V1 / U1 - U2

oh, balls.

Thank you very much for clearing that up.

No worries. You should see the amount of times I've messed up my cos/sin in M3 let alone sign errors