The magnitude of the final momentum (\(p_2\)) is approximately \(5.273 \times 10^{17} \, \text{kg} \cdot \text{m/s}\).
To determine the final momentum magnitude (\(p_2\)), we can apply the conservation of angular momentum since there are no external torques acting on the system. Angular momentum (\(L\)) is given by the product of the moment of inertia (\(I\)) and angular velocity (\(ω\)):
\[ L = I \cdot ω \]
For a point mass like the rock, \(I = m \cdot r^2\), where \(m\) is the mass and \(r\) is the distance from the axis of rotation. As the rock moves, \(L\) remains constant.
\[ L_1 = L_2 \]
\[ m_1 \cdot r_1^2 \cdot ω_1 = m_2 \cdot r_2^2 \cdot ω_2 \]
The momentum \(p\) is related to angular momentum by \(L = p \cdot r \cdot \sin(α)\), where \(α\) is the angle between \(r\) and \(p\).
\[ p_1 \cdot d_1 \cdot \sin(α) = p_2 \cdot d_2 \cdot \sin(180°) \]
\[ p_1 \cdot d_1 = p_2 \cdot d_2 \]
Now, solving for \(p_2\):
\[ p_2 = \frac{p_1 \cdot d_1}{d_2} \]
Substitute the given values:
\[ p_2 = \frac{(1.52 \times 10^{17} \, \text{kg} \cdot \text{m/s}) \cdot (5.2 \times 10^{13} \, \text{m})}{1.5 \times 10^{13} \, \text{m}} \]
\[ p_2 \approx 5.273 \times 10^{17} \, \text{kg} \cdot \text{m/s} \]
Therefore, the magnitude of the final momentum (\(p_2\)) is approximately \(5.273 \times 10^{17} \, \text{kg} \cdot \text{m/s}\).