Question

A small bucket of mass, $Mkg$, 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 $mkg$ of water and swings upto 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. $\left(\frac{M+m}{M}\right)L$

Answer 1

The correct option is A ${\left(\frac{M}{M+m}\right)}^{2}L$

Lets, first find out the speed of the bucket when it touches the water, i.e., at the lowest point. We use law of conservation of energy, we have
$MgL=\frac{1}{2}M{v}^2\Rightarrow v=\sqrt{2gL}$
Now, we apply the law of conservation of linear momentum. We have
$M\sqrt{2gL}=(M+m)v^{{}^{\prime }}\Rightarrow v^{{}^{\prime }}=\frac{M\sqrt{2gL}}{(M+m)}$ .......... $\left(I\right)$
$v^{{}^{\prime }}$ is the velocity of bucket plus water system, as the bucket scoops up the water.
Now, we again apply the law of conservation of energy. We have
$\frac{1}{2}(M+m){v^{{}^{\prime }}}^2=(M+m)gh\Rightarrow v^{{}^{\prime }}=\sqrt{2gh}$ .......... $\left(II\right)$
equating both $\left(I\right)$ and $\left(II\right)$, we have
$\frac{M\sqrt{2gL}}{(M+m)}=\sqrt{2gh}\Rightarrow h={\left(\frac{M}{M+m}\right)}^{2}L$