Question

A heavy particle hanging from a string of length I is projected horizontally with speed $\sqrt{gl}$ . The speed of the particle at the point where the tension in the string equals weight of the particle

• A. $\sqrt{2gl}$
• B. $\sqrt{3gl}$
• C. $\sqrt{gl/2}$
• D. $\sqrt{gl/3}$

Answer 1

The correct option is D $\sqrt{gl/3}$

$\begin{array}{c}mgl\left(1-\cos \theta \right)=\frac{1}{2}m{v}^2-\frac{1}{2}mv{}^{\prime 2}\\ \Rightarrow 2gl\left(1-\cos \theta \right)=gl-v{}^{\prime 2}\\ \Rightarrow 2gl-2gl\cos \theta =gl-v{}^{\prime 2}\\ v{}^{\prime 2}=gl-2gl+2gl\cos \theta \\ =2g\cos \theta -gl\\ =gl\left(2\cos \theta -1\right).........\left(i\right)\\ T-mg\cos \theta -\frac{mv{}^{\prime 2}}{l}\\ mg\left(1-\cos \theta \right)=\frac{m}{l}\left[gl\left(2\cos \theta -1\right)\right]\\ \Rightarrow 1-\cos \theta =2\cos \theta -1\\ \Rightarrow 2=3\cos \theta \\ \therefore \cos \theta \frac{2}{3}\\ \therefore v{}^{\prime 2}=gl\left(\frac{4}{3}-1\right)\\ =\frac{gl}{3}\\ v^{\prime }=\sqrt{\frac{gl}{3}}\end{array}$

$\therefore$ Option $D$ is correct .