Question

A small bucket of mass $Mkg$ is attached to a long inextensible massless cord of length $L$. 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
$\frac{x}{8}{\left(\frac{M}{M+m}\right)}^{2}L$. Find $x$.

Answer 1

Applying conservation of energy

$\frac{1}{2}M{v}^2=MgL$

$\Rightarrow v=\sqrt{2gL}...\left(1\right)$

{$v$ is speed of the bucket when it touches the water i.e. at lowest point.}
using conservation of linear momentum
$(M+m)v^{\prime }=Mv$
{ $v^{\prime }$ is speed of (bucket + water) system}

Put value of $v$ from equation (1)
$\Rightarrow v^{\prime }=\frac{M\sqrt{2gL}}{M+m}...\left(2\right)$

Maximum height reached is $\left(h\right)=\frac{v^{\prime 2}}{2g}$

$\Rightarrow h=\frac{v^{\prime 2}}{2g}$

put value of $v^{\prime }$ from equation (2)

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

compare it with given expression for $h$ , we get
$x=8$