Momentum Question

http://i-want-to-study-engineering.org/q/momentum/# <-- The question, I managed to reach u = root((2 - root2)g) but I don't know where to go from there to get the answer of 4/3 root((2 - root2)g). The solutions didn't help me out.

http://i-want-to-study-engineering.org/q/momentum/# <-- The question, I managed to reach u = root((2 - root2)g) but I don't know where to go from there to get the answer of 4/3 root((2 - root2)g). The solutions didn't help me out.

Would I have to form a simultaneous equation of some sort to solve them? My physics teacher said he didn't know either.

yes it does involve simultaneous equations - what level of study are you at? This question would be a pretty hard AS question

I'm doing A-levels, currently in year 13. I am trying to get into Cambridge to study engineering.

Yeah, no matter what I try, I can't seem to solve it, I reach up until the point of making a momentum and kinetic energy equation. If I say the velocity of ball A before the collision is u, velocity after is v, and the velocity of ball B is w, I get 2 equations of 2u = w - 2v and u^2 = v^2 + (w^2)/2

those equations are right.
Now you want an equation for the kinetic energy the first ball gained before it's collision.
This is caluclated using the length and equations mgh=1/2 m u^2

I can't tell exactly if this question is high end A-level or university based. The question seems so confusing and it's "mathematics and physics problems at the level of university engineering admissions interviews"

This is defo a university admissions level question - I only said hard AS because mechanics is AS and the kind of stuff they ask in the A-Level exams isn't far off this and can be harder tbh.

essentially you have three equations like in the summary(have you derived all three?)
What might help is leaving u and u and subbing in it's value at the end once you have values for v/w in terms if u..

I subbed in w=2u+2v into u^2=v^2+w^2/2 then completed the square to get v=-u or v=-u/3

If v=-u then w=0 and that's not really possible if you think about it for the smaller ball to stay still and the larger ball to just rebound with the same speed in the opposite direction.

Subbing v=-u/3 into w=2u+2v gives w=4u/3 which is the rigth answer if you sub in u

So if I substitute w = 2u + 2v into u^2 = v^2 + (w^2)/2, I would get u^2 = 2u^2 + 2uv + 2v^2 + v^2 which equals 0 = u^2 + 2uv + 3v^2, then I complete the square to get 0 = (u+v)^2 + 2v^2 ? Sorry about this, I seem so dam slow at figuring out where the v=-u and v= -u/3 comes from, but I understand that once you get those values you substitute them in.

I got 4uv then completed square to get

(V+2u/3)^2 = u^2/9

Dam I see what I did wrong with expanding (2u+2v)^2, I am so stupid...

no problem the question is pretty hard tbh theres alot to it. Are you preparing for Oxbridge/Imperial?

Beginning to prepare for Cambridge, I doubt I will have a chance to be honest, but it's worth a try.

defo worth a try, I was in the same position last year and wouldn't have been able to do this at all im pretty sure. Im reapplying too, make sure to use isaac physics btw and this is a good link for some physics if you're interrested:
http://www.physics.ox.ac.uk/olympiad/downloads/uyp/upgrade_your_physics.pdf

Thinking of reapplying to Cambridge too - why did you pick it over Oxford may I ask?

I live in Cambridge