Question

A smooth ring of mass m which is free to slide along a rigid horizontal wire, is connected to a particle of mass M by a string of length $l$ . The system is released from rest when the string is horizontal. At an instant when string becomes vertical. Speed of particle

• A. $0$
• B. $\sqrt{2gl\left(\frac{m+M}{m}\right)}$
• C. $m\sqrt{\frac{2gl}{M(m+M)}}$
• D. $\sqrt{\frac{2mgl}{M+m}}$

Answer 1

The correct option is D $\sqrt{\frac{2mgl}{M+m}}$

$\begin{array}{c}mv=MV......\left(1\right)\\ \Rightarrow U=\frac{MV}{m}\\ Mgl=\frac{1}{2}m{u}^2+\frac{1}{2}M{V}^2\\ \Rightarrow 2Mgl=m.\frac{M^2{V}^2}{m^2}+M.V^2\\ \Rightarrow 2gl=\frac{M{V}^2}{m}+V^2\\ \Rightarrow 2gl=\frac{\left(M+m\right)}{m}{V}^2\\ \Rightarrow V=\sqrt{\frac{2mgl}{M+m}}\\ Hence,\\ optionDiscorrectanswer.\end{array}$