Completely Inelastic Collision: A block and a block with a spring

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brendan3eb
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I actually have two problems that I have been struggling with that are very similar, so I believe that I am making the same incorrect assumption for both problems, but I am at that point where I have been re-trying the problems for so long that I do not think I am going to find my mistake.

Homework Statement


A block of mass m1=2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2=5.0 kg moving at 3.0 m/s. A massless spring with spring constant k=1120 N/m is attached to the near side of m2, as shown in Fig. 10-35. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.)


Homework Equations


m1v1i+m2v2i=(m1+m2)v
K2-K1=W
W=(1/2)kx^2

The Attempt at a Solution


m1=2.0 kg
v1i=10 m/s
m2=5.0 kg
v2i=3 m/s
k=1120 N/m
Pi=Mv
m1v1i+m2v2i=(m1+m2)v1
v1=(m1v1i+m2v2i)/(m1+m2)
v1=(2*10+5*3)/(2+5)=5 m/s
K2-K1=W
(1/2)m1(v1)^2-(1/2)m1(v1i)^2=-(1/2)kx^2
(1/2)s cancel out
(2)(5)^2-(2)(10)^2=-(1120)(x)^2
x^2=.134
x=.37 m

It looks as though all the units cancel out right and I do not have to convert any the units.
 
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re-tried it a few more times to no avail
 
Your problem is with computing the energies.

Before the collision, what is the total kinetic energy of the two blocks?
After the collisiion, what is the total kinetic energy of the two blocks?

The difference equals (1/2)kx^2, but both your initial and final KE's are wrong.