SHM and Conservation of Energy/Momentum

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badatphysics2
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Homework Statement


One end of a spring is attached to a wall to a block of mass X= 2kg (on a frictionless horizontal table). Another mass M of 150g moving at a speed of 7m/s collides (inelastic). This takes 0.4s to compress the spring to its max compression.
I have to find the max force of contact between X and M after the collision. And then the energy of the system when the force between the masses is half this max value.

Homework Equations


0.5kx^2
0.5mv^2 = 11.025 m/s (for mass M)
m1v1=m2v2
p=m*v
T = 1.6s (since it takes 0.4s for 1/4 of the period)

The Attempt at a Solution


I know energy will be conserved, so the energy from the spring, and from the mass X is needed. However, since the spring constant k, and the amount it's compressed by isn't given, I'm not sure how to do 0.5kx^2. I'm confused with what the 'max force of contact' between the two masses is. Is that considered impulse? I think I'd have to find the velocity of the two combined during the collision, and then using conservation to find the velocity after it?
But I'm not sure...how to relate them.
 
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badatphysics2 said:
0.5mv^2 = 11.025 m/s (for mass M)
How can an energy be equal to a velocity?
badatphysics2 said:
m1v1=m2v2
What are m1, v1, m2, v2?
badatphysics2 said:
I know energy will be conserved
Where? Everywhere?

Start with the correct time order, i. e. the inelastic collision first. That will make the second part with the spring easier.