Question

A particle moving with a velocity v changes its direction of motion by an angle theta without change in speed, then how themagnitude of change in velocity is 2vsin(θ/2)

Answer 1

$$\begin{gathered} Letinitialvelocityisinxaxiswhichisv. \\ Nowfinalvelocityismakinganangle\theta withxaxis. \\ componentofthisvelocityinxdirection=v\cos \theta \\ componentofthisvelocityinydirection=v\sin \theta \\ changeinvelocityinxdirection=∆V_x=v\cos \theta -v \\ and \\ ∆V_y=v\sin \theta \\ ∆v=\sqrt{{\left(∆V_x\right)}^2+\left(∆V_y\right)} \\ =\sqrt{v^2{\cos }^2\theta +v^2-2v^2\cos \theta +v^2{\sin }^2\theta } \\ =v\sqrt{2-2\cos \theta } \\ =v\sqrt{2}\sqrt{1-\cos \theta } \\ =v\sqrt{2}\sqrt{1-1+2{\sin }^2\left(\theta /2\right)} \\ =2v\sin \left(\frac{\theta }{2}\right) \end{gathered}$$