What Happens to Block Velocities and Spring Compression in a Collision?

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AI Thread Summary
In a collision between two blocks, M1 (13.6 kg) and M2 (4.4 kg), with initial velocities of 4.7 m/s and -4.7 m/s respectively, the problem focuses on finding the speed of M1 when both blocks have equal velocities and the compression of the spring attached to M2. The conservation of momentum and energy principles are relevant for solving this problem. The user initially struggled with applying these principles but ultimately realized the need for an additional equation to solve for the two unknowns. After some thought, they managed to figure out the solution independently. The discussion highlights the importance of understanding conservation laws in collision scenarios.
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Homework Statement
A block of mass M1 = 13.6 kg and initial velocity v0 = 4.7 m/s collides with a block of mass M2 = 4.4 kg and initial velocity of -4.7 m/s. Attached to M2 is a spring with a force constant k = 10000 N/m. At one point the velocities of block M1 and block M2 are equal. What is the speed of block M1 at that point?
What is the compression of the spring when the velocities of the blocks are equal?

Relevant equations
conservation of momentum
conservation of energy

The attempt at a solution

I figure the moment when both masses have equal velocity is when they are presses together with M1 pressing against the string. At that point both masses will act as one and have the same mechanical energy, so using conservation of energy, I get:
0.5*(M1+M2)*V^2 + 0.5*K*X^2 = 0.5*M1*V1(ini)^2 + 0.5*M2*V2(ini)^2
I get the 2 variables that I'm trying to find but I'm missing another equation.
I tried using conservation of momentum, but I'm not sure if I can, since I'm looking for the moment right after the collision and I'm not sure final velocities to put there..

Any help would be good. I would prefer to figure it out by myself, but a little push in the right direction would be great :)
 
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Just figured it out and feeling a little idiotic for not doing it sooner, so thanks anyway :)
 
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