Conservation of momentum & energy

1. The problem statement, all variables and given/known data
A block of mass m1 = 2 kg slides along a frictionless table with speed 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3 m/s is a block of mass m2 = 5kg. A massless spring with spring constant k = 1120 N/m is attached to the second block. The spring is compressed a maximum amount x. What is the value of x?

2. Relevant equations
m1v1 + m2v2 = m1vf1 + m2vf2
KEi = KEf + 1/2kx^2

3. The attempt at a solution
I knew that there would be a maximum compression of the spring when the first block has velocity = 0 m/s. I used cons. of momentum :
(2 kg)(10 m/s) + (5 kg)(3 m/s) = (5 kg)(Vf) to find Vf = 7 m/s
then used cons. of energy:
(0.5)(5 kg)(7 m/s)^2 = 1/2(1120)x^2
and got x = 0.4677 m

but the correct answer is 0.250 m
can anyone help me find the correct answer


Homework Helper
After collision both the masses move together until the compression is maximum. At that instant the common velocity is given by
m1v1 + m2v2 = (m1+m2)vf.
Applying the conservation of energy, you can write
m1v1^2 + m2v2^2 = (m1+m2)vf^2 +1/2*k*x^2.
Now proceed to find x.

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