(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy

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(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy between $A$ and $B$:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$v^2 = u^2 - 2gL$$v = \sqrt{u^2 - 2gL}$.At position $B$, the velocity is directed vertically upward, so $\vec{v} = v \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u} = v \hat{j} - u \hat{i}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + u^2} = \sqrt{(u^2 - 2gL) + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

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$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

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A
$\sqrt{u^2 - 2gL}$
B
$\sqrt{2gL}$
C
$\sqrt{u^2 - gL}$
D
$\sqrt{2(u^2 - gL)}$

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Work, Energy, Power and Collision

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Vertical Circular Motion

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$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)

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