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:
(1)
Where:
,
,
- Masses of the first, second and third fragments, in kilograms.
- Initial velocity of the object, in meters per second.
,
,
- Velocities of the first, second and third fragments, in meters per second.
If we know that
,
,
,
,
and
, the velocity of the third fragment is:
![(-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)](https://tex.z-dn.net/?f=%28-1.4%2C0%29%20%2B%20%280%2C-1.95%29%20%2B%201.2%5Ccdot%20%5Cvec%20v_%7B3%7D%20%3D%20%280%2C0%29)
![1.2\cdot \vec v_{3} = (1.4,1.95)](https://tex.z-dn.net/?f=1.2%5Ccdot%20%5Cvec%20v_%7B3%7D%20%3D%20%281.4%2C1.95%29)
![\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=%5Cvec%20v_%7B3%7D%20%3D%20%281.167%2C%201.625%29%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
The speed of the third fragment is the magnitude of the result found above:
![v_{3} = 2\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7B3%7D%20%3D%202%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
And the direction of the third fragment is:
![\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)](https://tex.z-dn.net/?f=%5Ctheta_%7B3%7D%20%3D%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B1.625%7D%7B1.167%7D%5Cright%29)
![\theta_{3} \approx 54.316^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7B3%7D%20%5Capprox%2054.316%5E%7B%5Ccirc%7D)
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.