Question

Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Answer 0.305 m

My question is how did they solve for that answer? Why did you choose the steps you did?

Answer 1

the initial momentum of the 1.7 kg block = 1.7 x 3.6 = 6.12 kgm/s
the momentum is conserved before and after the collision.
after the collision, both the masses move together with same velocity. let this velocity be v
applying conservation of momentum,
6.12 = (1.7+1.1)v
v = 2.186 m/s
the kinetic energy of the two block system = 1/2(M+m)v² = 6.69 J

when the blocks come to a stop, this kinetic energy is totally converted to the potential energy of the spring which is equal to 1/2 kx² , where x is the compression.

6.69 = (1/2 )144 x²=72x²

x = 0.305 m