# Homework Help: Two masses connected by a spring

1. Apr 30, 2012

### grusini

1. The problem statement, all variables and given/known data
Two masses $m_1=2kg$, $m_2=2m_1$ are placed on the opposite edges of a spring of constant $k_1=3N/m$, compressed of a length $x_1=1.73cm$. The system is located on a smooth plane. At the right end of the plane there is a second spring of constant $k_2=12N/m$. Once the first spring is at rest the masses are free to move (they are not fixed to the spring). So the situation is as follows:

_________________M1 spring1 M2___________spring2 ||

a) Find the velocity $v_1,v_2$ of masses $m_1,m_2$ respectively, when the first spring is at rest.

b) Find the maximum compression $\Delta x$ of the second spring.

2. Relevant equations
Conservation of energy, conservation of momentum, potential energy of a spring

3. The attempt at a solution

a) The velocities must satisfy the system:
$$\begin{cases}\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}k_1x_1^2 \\ m_1v_1+m_2v_2=0\end{cases}$$
$\Rightarrow v_2=\sqrt{\frac{m_1k_1x_1^2}{m_2^2+m_1m_2}}=$
$\sqrt{\frac{2\cdot 3\cdot (0.0173)^2}{4^2+4\cdot 2}}=0.008 m/s$.
Then $v_1=-\frac{m_2}{m_1}v_2=-2v_2=-0.016m/s$.

b) The value $\Delta x$ must satisfy:
$\frac{1}{2}m_2v_2^2=\frac{1}{2}k_2(\Delta x)^2 \Rightarrow \Delta x=\sqrt{\frac{m_2v_2^2}{k_2}}=\sqrt{\frac{4\cdot (0.008)^2}{12}}=0.0046m$

Values are wrong on my textbook..

2. Apr 30, 2012

### Alucinor

What do you mean the values are wrong in your textbook? Do you mean your textbook gives you a solution that is different from your own? What is the given solution?

Your work seems to make sense to me. Perhaps you shouldn't be rounding the value you get for v_2 when you work through the whole problem, ie pretend 0.00865 m/s is entirely significant until the end (also, you should probably round that to 0.009 if you're going to round)

3. Apr 30, 2012

### grusini

I talked to my professor and he agrees that the values found in my textbooks are wrong (what I meant before was that my result differs from that of the textbook). Anyway, thank you Alucinator for the tip about the rounding!