# Kleppner - blocks and a compressed spring

Homework Statement
A system is composed of two blocks of mass m1 and m2 connected
by a massless spring with spring constant k. The blocks slide on a
frictionless plane. The unstretched length of the spring is L. Initially
m2 is held so that the spring is compressed to L/2 and m1 is forced
against a stop. m2 is released at t = 0.
Find the motion of the center of mass of the system as a function
of time.

I understand that when string is back to full Length L, M2 has speed v=0.5L*sqrt(K/m2).
And that is the moment when M1 loses contact with the stop and start moving with M2.

My question 1.
So once M1 leaves the stop, to keep momentum constant, the center of mass should have speed
m2*v/ (m1+m2)
Am I right? I am asking because I read some other solution which states M1 would move at the same speed v as m2, then momentum is not conserved.

My question 2.
Is energy also conserved? In above, if both M1 and M2 moves in speed m2*v/(m1+m2), or both in speed v, in either case energy is not conserved. What is causing energy to be not conservative here??

Thanks a lot!

haruspex
Homework Helper
Gold Member
the center of mass should have speed m2*v/ (m1+m2)
Yes.
M1 would move at the same speed v as m2
Not at the instant m2 loses contact.
if both M1 and M2 moves in speed m2*v/(m1+m2) ... energy is not conserved
No, the mass centre will be moving at that speed. The individual speeds vary, so the total KE will be more.

Thanks Haruspex!
So m1 and m2 moves in different speeds but center of mass moves constant speed as momentum is conserved.
And I believe total energy (kinetic plus string potential) is also conserved. if m1 and m2 moves in individual speeds then all is not lost.

I am surprised at some blunt mistakes in the Kleppner "solution" I found online.

haruspex