A bob of mass ' m ' is suspended by a light string of length ' L '. It is imparted a

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5.Work, Energy, Power and Collision

hard

A bob of mass ' $m$ ' is suspended by a light string of length ' $L$ '. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $\frac{(\text { K.E. })_A}{(\text { K.E. })_B}$ is:

$3:2$

$5:1$

$2:5$

$1:5$

(JEE MAIN-2024)

Solution

Apply energy conservation between $A$ & $B$

$ \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) $

$ \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}$

So, $\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$

$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$

Standard 11

Physics

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easy

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hard

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medium

(NEET-2019)

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hard

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hard

A bob of mass ' $m$ ' is suspended by a light string of length ' $L$ '. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $\frac{(\text { K.E. })_A}{(\text { K.E. })_B}$ is:

Solution

Similar Questions

A sphere is suspended by a thread of length $l$. What minimum horizontal velocity has to be imparted the ball for it to reach the height of the suspension

A particle of mass $m$ is tied to string of length $l$ and whirled in vertical circle. The difference of tension and kinetic energy at highest and lowest position of circular path will be

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$A$ particle of mass $m$ is released from $a$ height $H$ on $a$ smooth curved surface which ends into a vertical loop of radius $R$, as shown Choose the correct alternative $(s)$ if $H = 2R$

In the given figure for $u\ =\ u_0/3$ , find the height from the ground at which the block leaves the hemisphere [where $u_0\ =\ \sqrt {\frac{{gr}}{3}} ]$

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