JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    A particle has two velocities of equal magnitude inclined to each other at an angle \[\theta \]. If one of them is halved, the angle between the- other and the original resultant velocity is bisected by the new resultant. Then, 0 is     AIEEE  Solved  Paper-2006

    A) \[120{}^\circ \]                                 

    B) \[45{}^\circ \]                   

    C)        \[60{}^\circ \]                   

    D)        \[90{}^\circ \]

    Correct Answer: A

    Solution :

    Let the magnitude of two velocities be\[u\]. We know that, \[R=\sqrt{{{u}^{2}}+{{u}^{2}}+2{{u}^{2}}\cos \theta }\] \[R=\sqrt{2u}\sqrt{1+\cos \theta }\] \[R=2u\,\,\cos \frac{\theta }{2}\] We know that by angle bisector theorem. \[\frac{AB}{AC}=\frac{BD}{DC}\Rightarrow \frac{u}{2u\cos \frac{\theta }{2}}=1\] \[\Rightarrow \]\[\cos \frac{\theta }{2}=\frac{1}{2}=\cos \frac{\pi }{3}\] \[\Rightarrow \]\[\theta ={{120}^{o}}\]

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