Question

A small bucket of mass,$M$ kg, is attached to a long inextensible cord of length $L$ m. 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 upto a height $h$. The height $h$ in meter is:
A.${{\left( \dfrac{M}{M+m} \right)}^{2}}L$
B.$\left( \dfrac{M}{M+m} \right)L$
C.${{\left( \dfrac{M+m}{M} \right)}^{2}}L$
D.$\left( \dfrac{M+m}{M} \right)L$

Answer 1

Hint: in this question we are going to find the correct option. Here gravitational force or conservative force is applied means non-conservatives force not applied means mechanical energy is constant. Use formula ${{v}_{i}}+{{k}_{i}}={{v}_{f}}+{{k}_{f}}$ where ${{v}_{i}}$ and ${{v}_{f}}$ is initial and final potential energy and ${{k}_{i}}$ and ${{k}_{f}}$ is initial and final kinetic energy respectively to solve this problem.

Complete answer:
Let us first find out the speed of the bucket when it touches the water, that is, at the lowest point. We use law of conservation of energy, we have
${{v}_{i}}+{{k}_{i}}={{v}_{f}}+{{k}_{f}}$
$\Rightarrow MgL+\dfrac{1}{2}M{{(0)}^{2}}=0+\dfrac{1}{2}M{{v}^{2}}$
${{v}^{'}}$ $\begin{align}
  & \Rightarrow MgL=\dfrac{1}{2}M{{v}^{2}} \\
 & \Rightarrow v=\sqrt{2gL} \\
\end{align}$
It is the lowest position of velocity of $M$ mass.
Suppose ${{v}^{'}}$ is the velocity of the bucket plus water system, as the bucket scoops up the water.
Now we apply the law of conservation of linear momentum . we have
$\begin{align}
  & M\sqrt{2gL}=(M+m){{v}^{'}} \\
 & \Rightarrow {{v}^{'}}=\dfrac{M\sqrt{2gL}}{M+m} \\
\end{align}$
Now, we again use the law of conservation of energy. We have
${{v}_{i}}+{{k}_{i}}={{v}_{f}}+{{k}_{f}}$
$\begin{align}
  & \Rightarrow 0+\dfrac{1}{2}(M+m){{v}^{'}}^{2}=(M+m)gh+\dfrac{1}{2}(M+m){{(0)}^{2}} \\
 & \Rightarrow \dfrac{1}{2}(M+m){{v}^{'}}^{2}=(M+m)gh \\
 & \Rightarrow \dfrac{{{v}^{'}}^{2}}{2}=gh \\
 & \Rightarrow {{v}^{'}}^{2}=2gh \\
 & \Rightarrow h=\dfrac{{{v}^{'}}^{2}}{2g}........(2) \\
\end{align}$
Now apply the value of ${{v}^{'}}$ in equation (2), we get
$\begin{align}
  & h=\dfrac{{{(\dfrac{M\sqrt{2gL}}{M+m})}^{2}}}{2g} \\
 & h=\dfrac{{{M}^{2}}2gL}{{{(M+m)}^{2}}2g} \\
 & h=\dfrac{{{M}^{2}}}{{{(M+m)}^{2}}}L \\
 & h={{(\dfrac{M}{M+m})}^{2}}L \\
\end{align}$
It is the height of the swing above the lowest position.

Hence, option A. ${{\left( \dfrac{M}{M+m} \right)}^{2}}L$ is the correct answer.

Note:
An alternative approach to the question is: first use the law of conservative energy and find the velocity of lowest position, then again apply law of conservation when the bucket scoops up the water, and find velocity of mass. Then equating first and second, you will get the answer.