According to the image, the cue ball after the collision has a speed of 0.8 m/s with a direction of 20°. Let's use the rectangular components formulas to find each component.
[tex]\begin{gathered} x=v\cdot\cos \theta \\ y=v\cdot\sin \theta \end{gathered}[/tex]Where v = 0.8 m/s and theta = 20°.
[tex]\begin{gathered} x=0.8\cdot\cos 20(m/s)\approx0.75(m/s) \\ y=0.8\cdot\sin 20(m/s)\approx0.27(m/s) \end{gathered}[/tex]Now, we use the components to write the velocity of the cue ball, which is a vector
[tex]\vec{v}=(0.75i+0.27j)(m/s)[/tex]On the other hand, we can observe that the eight ball is still before the collision. To find its speed after the collision, we use the law of conservation of momentum.
[tex]p_{i1}+p_{i2}=p_{f1}+p_{f2}[/tex]Where p = mv. Let's use all the given magnitudes.
[tex]0.6\operatorname{kg}\cdot2m/s+0.6\operatorname{kg}\cdot0m/s=0.6\operatorname{kg}\cdot0.8m/s+0.6\operatorname{kg}\cdot v_{f2}[/tex]Then, we solve for the speed 2 (eight ball speed):
[tex]\begin{gathered} v_{f2}=\frac{1.2kg\cdot m/s-0.48kg\cdot m/s}{0.6\operatorname{kg}} \\ v_{f2}=\frac{0.72\operatorname{kg}\cdot m/s}{0.6\operatorname{kg}} \\ v_{f2}=1.2m/s \end{gathered}[/tex]The final speed of the eight ball is 1.2 m/s. But we have to find the angle of the eight ball after the collision in order to find its components. To find the angle, we just have to subtract 90°-20° = 70°.
Now, we can find the rectangular components.
[tex]\begin{gathered} x=v\cdot\cos \theta \\ y=v\cdot\sin \theta \end{gathered}[/tex]Where v = 1.2 m/s and theta = 70°.
[tex]\begin{gathered} x=1.2\cdot\cos 70(m/s)\approx0.41m/s \\ y=1.2\cdot\sin 70(m/s)\approx1.13m/s \end{gathered}[/tex]The x-component of the velocity of the eight ball is 0.41i (m/s) and the y-component of the velocity of the eight ball is 1.13j (m/s).
The eight ball travels at a 70° angle after the collision.
The magnitude of the eight ball's velocity after the collision is 1.2 m/s.
