Answer:
[tex]v_{2'}=8.1\:\mathrm{m/s}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot8.4^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot 4.4\cdot {v_{2'}}^2,\\\\{v_{2'}}=\sqrt{64.31},\\\\{v_{2'}}\approx\fbox{$8.1\:\mathrm{m/s}$}[/tex]..
