To solve this problem we require the concept of conservation of momentum, applied in the particular case of collisions.
The general equation says that,
[tex]M_1V_1 + M_2V_2 = (M_1+M_2) * V_f,[/tex]
Where,
[tex]M_2 = 4 M_1[/tex]
The crash occurs at an intersection so we must separate the two speeds by their respective vector: x, y.
In the case of what happened on the X axis, we have that the body M_2 has a speed = 0, this because it is not the direction in which it travels, therefore
[tex]M_1* 13 = (M_1+M_2) * V_{fx} \\M_1*13 = (M_1+4M_1)*V_{fx}\\M_1*13=5M_1*V_{fx}\\Vx = \frac{13}{5}m/s[/tex]
The same analysis must be given for the particular case in the Y direction, where the mass body M_1 does not act with its velocity here, therefore:
[tex]M_2* 13 = (M_1+M_2) * V_{fy},[/tex]
[tex]4*M_1* 13 = 5Ma * V_{fy} ,[/tex]
[tex]V_{fy} = \frac{52}{5}m/s[/tex] ,
We have the two components of a velocity vector given by
[tex]V_f = \frac{13}{5}\hat{i} + \frac{52}{5}\hat{j}[/tex]
Get the magnitude,
[tex]V_f = \sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}[/tex]
[tex]V_f = 10.72 m/s[/tex]
With a direction given by
[tex]Tan^{-1} \frac{4}{1} = 75.96 \°[/tex]
