# Spring with masses on either side - force constant and compression remain the same?

1. Jul 15, 2011

### Hobbit300

1. The problem statement, all variables and given/known data

A spring is compressed by 0.100m from its equilibrium position, and two blocks of masses 3kg and 5kg are resting motionless on each side on a frictionless surface. The spring has a force constant of 20 N/m, and a rope holds the block in position against the compressed spring. If the rope is cut, the spring drives the blocks apart - find the resulting speed of each block.

2. Relevant equations

Ee (elastic energy) = 1/2kx^2
Ek(block 1 or 2) = 1/2mv^2

3. The attempt at a solution

I know that the elastic energy of the spring will be equal to the kinetic energy of the blocks as they move and am fine with solving Ee=Ek, and isolating for v.

However, my problem is that I'm not sure whether my Ee for each block is Ee/2 (would the energy be divided evenly between the blocks?), so basically - whether my force constant (k) should be divided by two (20/2)- and if whether the 0.100m compression should also be divided by two, since when the spring is released, I'm thinking the displacement should be equally divided on either side.

As far as the change in force constant, I've read that if a spring is cut in half, its k value is actually doubled - but i'm not sure if this situation can be likened to that in which the spring is cut in half. And I cannot figure out for sure whether compression should be divided in half or not. Please help!

2. Jul 16, 2011

### ehild

Re: Spring with masses on either side - force constant and compression remain the sam

There is no external force acting on the system of blocks and spring, therefore the CM stays in rest. This determines the ratio of the velocities. The masses are not equal, so neither the magnitude of displacements nor the speeds are the same. To cut the spring into half is senseless. The same force acts at both sides of the spring, on both boxes.

ehild

3. Jul 16, 2011

### darkxponent

Re: Spring with masses on either side - force constant and compression remain the sam

there can be two cases::

1)boxes are fixed to spring:

in thjis case the system will do SHM on same position of centre of mass

2) boxes are not fixed to spring::

in this case both boxes will leave in diff directions but the centre of mass remains at same position and net momentum remains conserved

Equation1: (m1*v1) + (m2*v2) =0

equation 2: 1/2 (m1*v1^2) +1/2(m2*v2^2) =1/2(k*x^2)

2 equations and two variables solve it