(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

Difficult

$A$ 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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$A$ train of mass $3 \times 10^6 \ kg$ is accelerated by an engine such that its velocity increases from $5 \ m/s$ to $25 \ m/s$ in $5 \ minutes$. The power of the engine is ........ $MW$.

$A$ bullet of mass $4\,g$ is fired horizontally with a speed of $300\,m/s$ into a $0.8\,kg$ block of wood at rest on a table. If the coefficient of friction between the block and the table is $0.3$,how far will the block slide approximately (in $,m$)?

$A$ particle of mass $M$ moving along the $X$-axis with speed $V_0$ collides with and sticks to another particle of mass $m$ moving along the $Y$-axis with speed $V_0$. What will be the velocity of the combined mass after the collision?

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