Explanation:
The given data is as follows.
Original kinetic energy = [tex]\frac{1}{2}mv^{2}[/tex] = 16 J
Energy imparted due to explosion = 16 J
Total energy = 32 J
Therefore, expression for total energy will be as follows.
Total K.E = [tex]\frac{1}{2}m_{1}v^{2}_{1} + \frac{1}{2}m_{2}v^{2}_{2}[/tex]
32 J = [tex]2v^{2}_{1} + 2v^{2}_{2}[/tex]
16 J = [tex]v^{2}_{1} + v^{2}_{2}[/tex]
Now, according to the conservation of linear momentum the expression is as follows.
mv = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex]
[tex]8 \times 2 = 4v_{1} + 4v_{2}[/tex]
16 = [tex](v^{2}_{1} + v^{2}_{2})^{2}[/tex]
that is, [tex]v_{1}v_{2}[/tex] = 0
Therefore, it means one of the chunk stops at rest and the velocity of another chunk is 4 m/s in the same direction.
Therefore, we can conclude that the larger velocity is 4 m/s and the smaller velocity is 0 m/s.
