Question

A small bucket of mass M kg is attached to a long inextensible cord of length $Lm$ . The bucket is released from rest when the cord is in a horizontal position. At its lowest position, the bucket scoops up $m$ kg of water and swings up to a height h. The height $h$ in meters is

• A. ${\left(\frac{M}{M+m}\right)}^{2}L$
• B. $\left(\frac{M}{M+m}\right)L$
• C. ${\left(\frac{M+m}{M}\right)}^{2}L$
• D. $L$

Answer 1

Answer: (A)

${\left(\frac{M}{M+m}\right)}^{2}L$

$\begin{array}{c}\text{given-}x\text{A small bucket mass}=M\text{g}\\ \text{* Length of cord}=L\\ \text{Kinetic energy of Bucket at lowest point}\\ kE=MgL=\frac{1}{2}M{v}^2\\ \Rightarrow \text{velocity}v=\sqrt{2gL}\end{array}$

$\begin{array}{c}\text{Then it scoops water of mass}\left(m\right)kg\\ \Rightarrow \text{Final weight}=(M+m)kg\\ \text{Henu,}By\text{momentum conservation}\\ \begin{array}{cc}M& \sqrt{2gL}=(M+m)V^{\prime }\\ & v^{\prime }=\text{upward velocity}\\ \Rightarrow v^{\prime }& =\frac{M\sqrt{2gL}}{M+m}\end{array}\end{array}$$\Rightarrow \text{Maximum height is}=\frac{v^2}{2g}={\left(\frac{M}{M+m}\right)}^2\cdot L$