# Collisions

First part of the question should be fine however in my attempt of part 2 I just ended up with a bunch of algebra so I stopped there lol, there must be an easier/direct way to solve it that I can't think of
(edited 1 year ago)
There's an easier way. It's essentially a 1 sentence argument - it's hard to give a hint that's not pretty big but:

Spoiler

(edited 1 year ago)
Original post by DFranklin
There's an easier way. It's essentially a 1 sentence argument - it's hard to give a hint that's not pretty big but:

Spoiler

I'm sorry I'm still confused as to what you're hinting at, do mean as to setting the relative velocity of m2 and m3 as 0 since the speed of approach of m1 is going to be u-v and work from there?
I see how does has no effect on the newton's law of restitution equation but doesn't alter the conservation of momentum since m1 and m2 are different masses?
Original post by Student 999
I'm sorry I'm still confused as to what you're hinting at, do mean as to setting the relative velocity of m2 and m3 as 0 since the speed of approach of m1 is going to be u-v and work from there?
I see how does has no effect on the newton's law of restitution equation but doesn't alter the conservation of momentum since m1 and m2 are different masses?

General conservation of momentum for 2 masses looks like: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
Note that subtracting a constant C from each velocity doesn't change whether there's equality:

$m_1 (u_1 - C) + m_2 (u_2 - C) = m_1 (v_1 - C) + m_2(v_2-C)$

$\iff m_1 u_1 - m_1 C + m_2 u_2 - m_2 C = m_1 v_1 - m_1 C + m_2 v_2 - m_2C$

$\iff m_1 u_1 + m_2 u_2 - (m_1 + m_2) C = m_1 v_1 + m_2 v_2 - (m_1 + m_2) C$

$\iff m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

If you've covered inertial frames in physics, you're just shifting to a different inertial frame where m2, m3 are stationary. This doesn't affect the actual physics as longs as you do all your calculations in that inertial frame.