(D) By the law of conservation of linear momentum during the collision:$mv = M V' + m(v/2)$$mv - mv/2 = M V'$$mv/2 = M V'$$V' = \frac{mv}{2M}$For the

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(D) By the law of conservation of linear momentum during the collision:$mv = M V' + m(v/2)$$mv - mv/2 = M V'$$mv/2 = M V'$$V' = \frac{mv}{2M}$For the bob to complete a vertical circle,the minimum velocity $V'$ at the lowest point must be $\sqrt{5g\ell}$.Equating the two expressions for $V'$:$\frac{mv}{2M} = \sqrt{5g\ell}$$v = \frac{2M}{m} \sqrt{5g\ell}$

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

Difficult

$A$ bullet of mass $m$ and velocity $v$ strikes the bob of a pendulum of mass $M$ and emerges out with velocity $v/2$. What is the minimum value of $v$ such that the bob of the pendulum will swing through a complete circle? (Take the length of the pendulum $= \ell$)

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

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Work, Energy, Power and Collision

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Vertical Circular Motion

$A$ stone of mass $m$ is moving in a vertical circle of radius $20 \ cm$. What is the difference in kinetic energy between the lowest and highest points?

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$A$ stone of mass $1\, kg$ tied to a light inextensible string of length $L = \frac{10}{3}\, m$ is whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is $4$ and if $g$ is taken to be $10\, m/s^2$,the speed of the stone at the highest point of the circle is ....... $m/s$.

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$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

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$A$ particle originally at rest at the highest point of a smooth vertical circle (radius $R = 1.8 \ m$) is slightly displaced. It will leave the circle at a vertical distance $h$ from the bottom. The value of $h$ will be ..... $m$.

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$A$ particle is tied to a $20 \ cm$ long string. It performs circular motion in a vertical plane. What is the angular speed of the particle when the tension in the string at the top is zero? (Take $g = 9.8 \ m/s^2$)

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$A$ bullet of mass $m$ and velocity $v$ strikes the bob of a pendulum of mass $M$ and emerges out with velocity $v/2$. What is the minimum value of $v$ such that the bob of the pendulum will swing through a complete circle? (Take the length of the pendulum $= \ell$)

Similar Questions

$A$ stone of mass $m$ is moving in a vertical circle of radius $20 \ cm$. What is the difference in kinetic energy between the lowest and highest points?

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

$A$ particle originally at rest at the highest point of a smooth vertical circle (radius $R = 1.8 \ m$) is slightly displaced. It will leave the circle at a vertical distance $h$ from the bottom. The value of $h$ will be ..... $m$.

$A$ particle is tied to a $20 \ cm$ long string. It performs circular motion in a vertical plane. What is the angular speed of the particle when the tension in the string at the top is zero? (Take $g = 9.8 \ m/s^2$)

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