Answer:
Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.
Explanation:
By conservation of the linear momentum:
[tex]m*V_{1o}+m*V_{2o}=m*V_{1f}+m*V_{2f}[/tex] Since both masses are the same, and expressing the equation for each axis x,y:
[tex]V_{1ox}+V_{2ox}=V_{1fx}[/tex] eqX
[tex]0=V_{1fx}+V_{2fx}[/tex] eqY
From eqX: [tex]V_{1fx}=1m/s[/tex]
From eqY: [tex]V_{1fy}=-1.37m/s[/tex]
The module is:
[tex]V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s[/tex]
The angle is:
[tex]\theta=atan(-1.37/1)=-53.87\°[/tex] This is 53.87° due south of east
