(A) $1$. When the bucket is released from the horizontal position,its kinetic energy at the lowest point equals its potential energy: $\frac{1}{2} M v

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

(A) $1$. When the bucket is released from the horizontal position,its kinetic energy at the lowest point equals its potential energy: $\frac{1}{2} M v_0^2 = MgL$,which gives $v_0 = \sqrt{2gL}$.$2$. When the bucket scoops up water,using the law of Conservation of Linear Momentum,the velocity $v$ immediately after scooping is: $M v_0 = (M + m) v$.$3$. Therefore,$v = \frac{M}{M + m} \sqrt{2gL}$.$4$. Now,by the law of Conservation of Energy,for the maximum height $h$ reached by the bucket: $\frac{1}{2} (M + m) v^2 = (M + m) gh$.$5$. $h = \frac{v^2}{2g} = \frac{1}{2g} \left( \frac{M}{M + m} \right)^2 (2gL) = \left( \frac{M}{M + m} \right)^2 L$.

Class 11 Physics Work, Energy, Power and Collision Mix Examples-Work, Energy, Power and Collision

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$A$ small bucket of mass $M \, kg$ is attached to a non-extensible string of length $L \, m$. The bucket is released from rest when the string is in a horizontal position. At its lowest point,the bucket scoops up $m \, kg$ of water and rises 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
$\left( {\frac{{M + m}}{M}} \right)L$

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$A$ small bucket of mass $M \, kg$ is attached to a non-extensible string of length $L \, m$. The bucket is released from rest when the string is in a horizontal position. At its lowest point,the bucket scoops up $m \, kg$ of water and rises to a height $h$. The height $h$ (in meters) is:

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$A$ ball is dropped vertically downwards from a height $h$ above the ground. It hits the ground inelastically and bounces up vertically. Neglecting subsequent motion and air resistance,which of the following graphs represents the variation between speed $(v)$ and height $(h)$ correctly?

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