Since the collision is elastic, the total kinetic energy of the system will remain the same after the collision.
For the conservation of momentum, we have the following equation:
[tex]\begin{gathered} m_1v_{o1}+m_2v_{o2}=m_1v_{f1}+m_2v_{f2} \\ 20\cdot4+10\cdot0=20\cdot v_{f1}+10\cdot v_{f2} \\ 20v_{f1}+10v_{f2}=80 \\ 2v_{f1}+v_{f2}=8 \end{gathered}[/tex]Now, from the kinetic energy conservation, we have:
[tex]\begin{gathered} \frac{m_1v^2_{o1}}{2}+\frac{m^{}_2v^2_{o2}}{2}=\frac{m_1v^2_{f1}}{2}+\frac{m_2v^2_{f2}}{2} \\ m_1v^2_{o1}+m_2v^2_{o2}=m_1v^2_{f1}+m_2v^2_{f2} \\ 20\cdot4^2+10\cdot0=20\cdot v^2_{f1}+10\cdot v^2_{f2} \\ 20v^2_{f1}+10v^2_{f2}=320 \\ 2v^2_{f1}+v^2_{f2}=32 \end{gathered}[/tex]From the first equation, we have vf2 = 8 - 2vf1. Using this value on the second equation, we have:
[tex]\begin{gathered} 2v^2_{f1}+(8-2v^{}_{f1})^2=32 \\ 2v^2_{f1}+64-32v^{}_{f1}+4v^2_{f1}=32 \\ 6v^2_{f1}-32v^{}_{f1}+32=0 \end{gathered}[/tex]Solving this quadratic equation, we have vf1 = 1.33 or vf1 = 4.
vf1 = 4 represents the initial situation, before the collision, so we have vf1 = 1.33 m/s.
Calculating vf2, we have:
[tex]\begin{gathered} v_{f2}=8-2\cdot1.33 \\ v_{f2}=8-2.66 \\ v_{f2}=5.33 \end{gathered}[/tex]Rounding to two significant figures, we have vf1 = 1.3 m/s and vf2 = 5.3 m/s.