Answer:
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
Explanation:
The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:
[tex](m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the first, second and third fragments, in kilograms.
[tex]\vec v_{o}[/tex] - Initial velocity of the object, in meters per second.
[tex]\vec v_{1}[/tex], [tex]\vec v_{2}[/tex], [tex]\vec v_{3}[/tex] - Velocities of the first, second and third fragments, in meters per second.
If we know that [tex]m_{1} = 0.5\,kg[/tex], [tex]m_{2} = 1.3\,kg[/tex], [tex]m_{3} = 1.2\,kg[/tex], [tex]\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right][/tex], the velocity of the third fragment is:
[tex](-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)[/tex]
[tex]1.2\cdot \vec v_{3} = (1.4,1.95)[/tex]
[tex]\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right][/tex]
The speed of the third fragment is the magnitude of the result found above:
[tex]v_{3} = 2\,\frac{m}{s}[/tex]
And the direction of the third fragment is:
[tex]\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)[/tex]
[tex]\theta_{3} \approx 54.316^{\circ}[/tex]
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
