(A) For the pendulum-bob to complete a vertical circle,the minimum velocity $v$ required at the lowest point is $\sqrt{5 g \ell}$.By the principle of

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$\frac{2 M}{m} \sqrt{5 g \ell}$

(A) For the pendulum-bob to complete a vertical circle,the minimum velocity $v$ required at the lowest point is $\sqrt{5 g \ell}$.By the principle of conservation of linear momentum during the collision:$m u + M(0) = m(u/2) + M v$Substituting $v = \sqrt{5 g \ell}$:$m u = m(u/2) + M \sqrt{5 g \ell}$$m u - m u/2 = M \sqrt{5 g \ell}$$m u / 2 = M \sqrt{5 g \ell}$$u = \frac{2 M}{m} \sqrt{5 g \ell}$

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

Difficult

$A$ bullet of mass $m$ strikes a pendulum-bob of mass $M$ with velocity $u$. It passes through it and emerges out with a velocity $u/2$ from the bob. The length of the pendulum is $\ell$. What should be the minimum value of $u$,if the pendulum-bob will swing through a complete circle?

A
$\frac{2 M}{m} \sqrt{5 g \ell}$
B
$\frac{M}{2 m} \sqrt{5 g \ell}$
C
$\frac{2 M}{m} \frac{1}{\sqrt{5 g \ell}}$
D
$\frac{M}{2 m} \frac{1}{\sqrt{5 g \ell}}$

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Work, Energy, Power and Collision

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Vertical Circular Motion

$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)

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$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

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$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

DifficultJEE MAIN 2024

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$A$ bullet of mass $m$ strikes a pendulum-bob of mass $M$ with velocity $u$. It passes through it and emerges out with a velocity $u/2$ from the bob. The length of the pendulum is $\ell$. What should be the minimum value of $u$,if the pendulum-bob will swing through a complete circle?

Similar Questions

$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)

$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

$A$ stone of mass $1 \ kg$ is tied to a string $2 \ m$ long and is rotated at a constant speed of $40 \ ms^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 \ ms^{-2}$].

$A$ particle tied to one end of a string is being rotated in a vertical circle with constant frequency. The tension in the string at points $A, B, C$ and $D$ are $T_1, T_2, T_3$ and $T_4$ respectively. Then:

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

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