Two point masses `m_1` and `m_2` are connected by a spring of natural length `l_0`. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity `v_0` along positive x-axis. When the system reached the origin, the string breaks `(t=0)`. The position of the point mass `m_1` is given by `x_1=v_0t-A(1-cos omegat)` where A and `omega` are constants. Find the position of the second block as a function of time. Also, find the relation between A and `l_0`.

Explanation

To solve the problem step by step, we will start with the information provided and derive the position of the second mass as a function of time, as well as the relationship between \( A \) and \( l_0 \).

### Step 1: Understanding the System
We have two point masses \( m_1 \) and \( m_2 \) connected by a spring of natural length \( l_0 \). Initially, the spring is compressed, and the two masses touch each other. The system moves with a velocity \( v_0 \) until the string holding them together breaks at \( t = 0 \).

### Step 2: Position of Mass \( m_1 \)
The position of mass \( m_1 \) is given by:
\[ x_1 = v_0 t - A(1 - \cos(\omega t)) \]

### Step 3: Finding the Velocity of Mass \( m_1 \)
To find the velocity of mass \( m_1 \), we differentiate the position with respect to time:
\[ v_1 = \frac{dx_1}{dt} = v_0 - A \omega \sin(\omega t) \]

### Step 4: Conservation of Momentum
Since no external forces are acting on the system in the x-direction, the momentum must be conserved. Initially, the total momentum is:
\[ P_{\text{initial}} = (m_1 + m_2)v_0 \]
After the string breaks, the momentum is:
\[ P_{\text{final}} = m_1 v_1 + m_2 v_2 \]
Setting these equal gives us:
\[ (m_1 + m_2)v_0 = m_1(v_0 - A \omega \sin(\omega t)) + m_2 v_2 \]

### Step 5: Solving for \( v_2 \)
Rearranging the equation for \( v_2 \):
\[ m_2 v_2 = (m_1 + m_2)v_0 - m_1(v_0 - A \omega \sin(\omega t)) \]
\[ v_2 = v_0 + \frac{m_1}{m_2} A \omega \sin(\omega t) \]

### Step 6: Finding the Position of Mass \( m_2 \)
To find the position \( x_2 \) of mass \( m_2 \), we integrate \( v_2 \):
\[ x_2 = \int v_2 \, dt = \int \left( v_0 + \frac{m_1}{m_2} A \omega \sin(\omega t) \right) dt \]
\[ x_2 = v_0 t - \frac{m_1}{m_2} A \cos(\omega t) + C \]
Assuming the constant of integration \( C = 0 \) at \( t = 0 \):
\[ x_2 = v_0 t - \frac{m_1}{m_2} A (1 - \cos(\omega t)) \]

### Step 7: Relation Between \( A \) and \( l_0 \)
For the spring to return to its natural length \( l_0 \), the distance between the two masses must equal \( l_0 \):
\[ x_2 - x_1 = l_0 \]
Substituting the expressions for \( x_1 \) and \( x_2 \):
\[ \left(v_0 t - \frac{m_1}{m_2} A (1 - \cos(\omega t))\right) - \left(v_0 t - A(1 - \cos(\omega t))\right) = l_0 \]
Simplifying gives:
\[ \left(- \frac{m_1}{m_2} A (1 - \cos(\omega t)) + A(1 - \cos(\omega t))\right) = l_0 \]
Factoring out \( (1 - \cos(\omega t)) \):
\[ A(1 - \cos(\omega t))\left(1 - \frac{m_1}{m_2}\right) = l_0 \]
Thus, we find the relationship between \( A \) and \( l_0 \):
\[ A(1 - \frac{m_1}{m_2}) = l_0 \]

### Final Answers
1. The position of mass \( m_2 \) as a function of time is:
\[ x_2 = v_0 t - \frac{m_1}{m_2} A (1 - \cos(\omega t)) \]
2. The relation between \( A \) and \( l_0 \) is:
\[ A(1 - \frac{m_1}{m_2}) = l_0 \]