Question

B4 solution
Q.B-4. Two billiard balls are rolling on a flat table. One has velocity components $v_X$ = 1 m/s, $v_Y$ = $\sqrt{3}$ m/s and the other has components $v_X$ = 2 m/s and $v_Y$ = 2 m/s. If both the balls start moving from the same point, the angle between their path is -
(A) 60$°$
(B) 45$°$
(C) 22.5$°$
(D) 15$°$

Answer 1

Dear Student,


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