To solve the problem we use conservation of momentum. The momentum is given by:
[tex]p=mv[/tex]
before the block exploded the momentum was zero, then we have the conservation expression:
[tex]m_1v_1+m_2v_2=0[/tex]
In this case mass one is 8 kg, velocity one is 5 m/s (positive since is moving to the right) and mass two is 12 kg; plugging these values and solving for velocity one we have:
[tex]\begin{gathered} 12v_2+(8)(5)=0 \\ 12v_2=-40 \\ v_2=-\frac{40}{12} \\ v_2=-3.33 \end{gathered}[/tex]
Therefore, the velocity of the second piece is -3.33 m/s which means that its magnitude is 3.33 m/s and it is moving to the left.
