In Fig. 9-64, block 2
(mass 1.0 kg) is at rest on a
frictionless surface and touch-
ing the end of an unstretched
spring of spring constant 200
N/m. The other end of the spring is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v1 = 4.0 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?
2. Relevant equations
m1v1+m2v2=MV where M is the mass of the system and V is the velocity of the system
Hooke’s Law: F = kx
3. The attempt at a solution
I was able to find the final velocity of the system:
V = (m1v1)/(m1+m2) = 2.974
How do I get from there to the distance that the spring is compressed?