Completely inelastic collision with object attached to spring

AI Thread Summary
In a completely inelastic collision, block 1 (2.0 kg) collides with block 2 (1.0 kg) at a speed of 4.0 m/s, causing them to stick together. The final velocity of the combined blocks is calculated to be approximately 2.974 m/s. The discussion raises the question of how to determine the distance the spring compresses after the collision. It is suggested to consider conservation of energy principles, as the kinetic energy of the blocks converts into potential energy stored in the spring. The key focus is on applying Hooke's Law to find the compression distance of the spring after the collision.
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Homework Statement



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?


Homework 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

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?
 
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