Question

A string with one end fixed on a rigid well passing over a fixed frictionless pulley at a distance of $2m$ from the wall, has a point mass $M=2kg$ attached to it at a distance of $1m$, from the wall. A mass $m=0.5kg$ attached at the free end is held at rest so that the string is horizontal between the wall and the pulley, and vertical beyond the pulley. If the speed with which the mass $M$ will hit the wall when mass $m$ released is $335\times {10}^{-k}$, then $k$ is

Answer 1

First we will use energy conservation$,$

Loss in gravitational potential energy $=$ gain in kinetic energy of the two blocks$,$

$\begin{array}{c}MgH-mgh=\frac{1}{2}M{v_1}^2+\frac{1}{2}m{v_2}^2\\ 2\times 10\times 1-0.5\times 10\times \left(\sqrt{5}-1\right)=\frac{1}{2}\times 2{v_1}^2+\frac{1}{2}\times 0.5\times {v_2}^2\\ 20-6.2={v_1}^2+0.25{v_2}^2\\ alsobyconstriantrelationweknowthat\\ v_1\cos \theta =v_2\\ whereweknowthat\\ \cos \theta =\frac{2}{\sqrt{5}}\\ Now,wehave\\ 13.8={v_1}^2+0.25\times 0.8{v_1}^2\\ {v_1}^2=\frac{13.8}{1.2}\\ v_1=3.4m/s\\ So,thespeedoftheblockMwillbe3.4m/s\end{array}$