Answer:
[tex]\Delta K = 2164.053\,J[/tex]
Explanation:
Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:
[tex](32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v[/tex]
The speed of the more massive piece is:
[tex]v = 31.202\,\frac{m}{s}[/tex]
The kinetic energy added to the system is:
[tex]\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}[/tex]
[tex]\Delta K = 2164.053\,J[/tex]
