Question

Two billiard balls are rolling on a flat table one has the velocity components $v_x=1\mathrm{m}.{\mathrm{s}}^{-1},v_y=\sqrt{3}\mathrm{m}.{\mathrm{s}}^{-1}$ and the other has components $v_x^{\text{'}}=2\mathrm{m}.{\mathrm{s}}^{-1}\mathrm{and}{v}_y^{\text{'}}=2\mathrm{m}.{\mathrm{s}}^{-1}$. If both the balls start moving from the same point, what is the angle between their paths?

Answer 1

Dear Student,


$$\begin{gathered} velocityoffirstball \\ V=i+\sqrt{3}j \\ velocityofsecondball \\ V\text{'}=2i+2j \\ tofindtheanglebetweenthemusedotproduct. \\ \cos \theta =\frac{V.V\text{'}}{\left|V\right|\left|V\text{'}\right|}=\frac{2+2\sqrt{3}}{2*2\sqrt{2}}=\frac{1+\sqrt{3}}{2\sqrt{2}} \\ \theta ={\cos }^{-}\frac{1+\sqrt{3}}{2\sqrt{2}}=15degree \\ Regards \end{gathered}$$