Speed of centre of mass

Consider the resultant momentum of the system. Since the two momenta are perpendicular, the magnitude of the resultant momentum is as follows, where $m_s$ and $v_s$ are the mass and speed of the system.

$\begin{aligned} m_sv_s & = \sqrt{m_1 v_1+m_2v_2} \\ 7v_s & = \sqrt{(3\times 15)^2 + (4 \times 10) ^2} \\ \implies v_s & \approx \boxed{8.601} \end{aligned}$

Let masses A and B both start from the origin such that at time $t$ their respective positions are at: $A(0,-15t), B(-10t,0).$ The center-of-mass for this dynamic system is computed per:

$\bar{x} = \frac{m_{A}x_{A} + m_{B}x_{B}}{m_{A} + m_{B}} = \frac{(3)(0) + (4)(-10t)}{3+4} = -\frac{40t}{7}.$

$\bar{y} = \frac{m_{A}y_{A} + m_{B}y_{B}}{m_{A} + m_{B}} = \frac{(3)(-15t) + (4)(0)}{3+4} = -\frac{45t}{7}.$

The resultant speed of the center-of-mass finally calculates to: $\sqrt{((\bar{x}'(t))^2 + (\bar{y}'(t))^2} = \sqrt{(-40/7)^2 + (-45/7)^2} = \boxed{\frac{\sqrt{3625}}{7}}$ m/s.