Question

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 $\theta$ is.

• A. $120^{°}$
• B. $45^{°}$
• C. $60^{°}$
• D. $90^{°}$

Answer 1

Answer: (A)

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