question_answer

A particle is moving on a circular path with a constant speed v. The magnitude of the change in its velocity after it has described an angle of \[60{}^\circ \] is

A)                  \[v\]                                    

B)  \[\sqrt{2v}\]

C)                  \[\sqrt{3v}\]                                     

D)  \[\frac{\sqrt{3}}{2}v\]

Correct Answer: A

Solution :

                Change in velocity                                 \[\Delta v=\sqrt{v_{1}^{2}+v_{2}^{2}+2{{v}_{1}}{{v}_{2}}\cos \theta }\]                 \[=\sqrt{{{v}^{2}}+{{v}^{2}}+2v.v\cos {{120}^{o}}}\]                 \[=\sqrt{2{{v}^{2}}+2{{v}^{2}}\left( -\frac{1}{2} \right)}=\sqrt{{{v}^{2}}}=v\]