Maximum spring compression when 2 kg block collides with 5 kg block

[Jngo22]

Homework Statement

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?

Homework Equations

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

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

 

[rl.bhat]

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.