Answer:
0.3 m/s
Explanation:
Since there's no external force acting on the 2-cart-spring system, the momentum is conserved. And since the total momentum of the 2 carts before the release is 0 (they were at rest), the momentum of cart 1 equals to the momentum of cart 2 post-release.
[tex]m_1v_1 = m_2v_2[/tex]
[tex]5*0.12 = 2v_2[/tex]
[tex]v_2 = \frac{0.6}{2} = 0.3 m/s[/tex]