# Homework Help: Spring compression with masses in motion

1. Mar 27, 2006

### scavok

https://chip.physics.purdue.edu/protected/Halliday6Mimg/h10p33.jpg
A block of mass m1 = 2.5 kg slides along a frictionless table with a speed of 12 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 6.2 kg moving at 3.7 m/s. A massless spring with spring constant k = 1100 N/m is attached to the near side of m2, as shown in figure above. When the blocks collide, what is the maximum compression of the spring?

There's just too much going on here and I don't know where I'm going on wrong.

m1v1 + m2v2 = (m1+m2)vf
vf=velocity of both masses at the point where maximum compression is reached.

Is this correct?

If so, then the change in kinetic energy of mass m1 equals the work done on mass m1 by the spring:
.5m1vf2-.5m1v12=.5kx2

But this gets me a negative value, which it should since it is losing velocity, but I can't take the square root of a negative value which I need to do when solving for x. This makes me think I'm doing something wrong.

2. Mar 27, 2006

### durt

Apply conservation of energy: state that the sum of initial kinetic energies of the blocks equals their sums at the point of compression plus the energy stored in the spring.

3. Mar 28, 2006

### topsquark

If you are comfortable with it, a way to simplify the problem would be use center of mass coordinates, do the spring problem, then convert back to "lab" coordinates.

-Dan