Shortcut formula for conservation of kinetic energy

See attached:

The course I am doing is a pre-uni course, so the question seems a little complex to me. I am also given little space to work with.
My approach involves a lot of expanding. I didn’t learn a more direct approach in this module but when it comes to maths, there’s always a quicker way.
Can anyone assist?

You usually do restitution with a coefficient of 1
https://en.wikipedia.org/wiki/Coefficient_of_restitution
so
|relative velocity before collision| = |relative velocity after collision|
So that with momentum gives two linear simultaneous equations to get v1 and v2.
The wiki link shows that its equivalent to conservation of ke when e=1.

My brain is having a hard time understanding any of that wiki page lol.
I decided to do the math and I found that the final
Velocity for 5kg block was 3ms or 5.67ms so, its increased by e if this has any relevance but the 10kg block as final v as 5ms or -11/3 but I know it continues on the same direction.
According to and if my maths is correct the velocity remains unchanged after collision.
Edit: V1 is 5ms or 3.67 not negative.

Agree with the 17/3 and 11/3. Momentum gives
65 = 10v1 + 5v2
restitution (KE) gives
5-3 = v2 - v1
and solving gives the above.
Id have to see your working for the "or" parts.

I saw in a video something about the restitution part but I wasn’t sure if it was derived from the conservation of kinetic energy formula or from something else, but the person used like differentiation symbol like U1+U2’=V1+V2’ …
I shall attach my working later this evening as I’m in the process of travelling to my parents place.

Like the wiki page, conservation of KE is combined with conservation of momentum to give
|relative velocity before collision| = |relative velocity after collision|
for an elastic collision, otherwise their ratio is e.
The magnitude of the relative velocity before the collision is 2m/s as theyre moving in the same direction and have a difference of 2. So 5-3. Assuming this direction is positive, then v2 must be moving faster than v1 after the collision (as the 10 kg mass cant pass through the 5 kg mass) so v2-v1 is the relative velocity after the collision and its positive so
2 = v2 - v1
Thats it.

This is the working.
So, KE=1/2mv² and P=MV….
Conservation of momentum
m₁u₁+m₂u₂=m₁v₁+m₂v₂
Conservation of KE
½m₁u²₁+½m₂u²₂=…..
When you say combine, how are they combined? I can see the difference between two two formulas are. If I take out the ½u₁+½u₂= oh, so cancel the ½.
It gives us the u₁-u₂=v₁-v₂ . This is only true in elastic collision and the same direction? Or direction is irrelevant?
So, 5-3=v₁-v₂
2=(11/3)-(17/3)
2=-2… erm? Is that right?

What youve done is correct, but too long using the KE directly. The post impact v=5,3 solutions are rejected as they are the preimpact velocities and will trivially satisfy both the momentum and KE equations.
For the proof about restitution being conservation of KE (with momentum) the wiki link goes through it, but say you have masses m and M and velocities u, U and v, V, the momentum gives
mu + MU = mv + MV
and rearranging
m(u-v) = M(V-U)
KE gives
1/2 mu^2 + 1/2 MU^2 = 1/2 mv^2 + 1/2 MV^2
so multiplying by 2 and rearranging
m(u^2-v^2) = M(V^2-U^2)
and difference of two squares
m(u-v)(u+v) = M(V-U)(U+V)
Divide by the momentum equation (note you divide out the u=v, U=V "solution")
u+v = U+V
or
u-U = V-v
so the relative velocity before, u-U, equals the relative velocity after, V-v, and a bit of fiddling to define a positive restitution coefficient if KE is not conserved.
You have two linear simultaneous equations rather than a linear and quadratic. It sounds like they expect you to know it.

I see. Thank you for explaining.
That’s the thing though. In the material that I read provided by them. The only example provided where a heavier block collides with a smaller block, the only thing stated, was that momentum is transferred to the small block but depends on velocity and mass of the bigger block.
I was then given the conservation of momentum formula. I had to find the conservation of kinetic energy formula separately.
I did look at it and think this is a long process and there must be a shorter route.
I don’t know if they will like the fact that I have shown the process as is or whether they’d want me to simplify it.
If did show the restitution concept, that would probably be just as long going through those steps. Unless, I just pull it out of that hat and find a reference page to provide with it to show how it’s found.

At a level, youd be expected to use restitution rather than (conservation of) KE as its equivalent, it results in a simpler method and the trivial solution (no collision) is removed.

Okay. So, if I have understood correctly. Restitution is only valid in elastic collision due to kinetic energy and momentum being conserved. I read something about KE= e (between 0 and 1) depending on the material. The closer to one means the more elastic the collision. In this case e should equal 1. Is this related to what I am trying to find out?

like on the wiki page, e=0 represents perfectly inelastic (max loss of KE when objects coalesce), e=1 represents perfectly elastic (conservation of KE) and for other values represents the square root of the final/initial KE ratio. So its equivalent.

I’ve revised my solution with using what I’ve now learnt from our conversation. Simplified the whole process a lot!!
I’ve attached a reference, if they need to look where I’ve got it from. Wiki page.
Is what I’ve put sufficient enough?

Looks about right, remembering units. The only thing to be careful about (make sure you understand/justify) is making sure that the velocity differences (relative velocities) are positive. Its trivial for the pre impact ones so the relative velocity is 5-3 = 2. For the post impact, v2 (positive direction) must be moving faster so the relative velocity v2-v1 will be positive.

Ah, yes, the units!!
So, the pre-velocity must be positive.
is that explained by why I've seen this formula include the magnitude lines e=|u1-u2/v2-v1| in order to ensure that the pre and after are positive?