(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

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is:

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$A$ $2 \, kg$ stone at the end of a string $1 \, m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \, m/s$. The tension in the string will be $52 \, N$ when the stone is:

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$A$ bob attached to a string is held horizontal and released. The tension $T$ and vertical distance $h$ from the point of suspension can be represented by which of the following graphs?

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$A$ stone is tied to a string of length $L$ and 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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$A$ stone of mass $40$ g is tied to a light inextensible string of length $50$ cm and whirled in a vertical circle at the rate of $30$ revolutions per minute. The tension in the string when it is at the lowest point of the circle is (Take gravitational acceleration, $g = 10 \text{ m/s}^2$ and $\pi^2 = 10$) (in $N$)

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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

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is:

$A$ $2 \, kg$ stone at the end of a string $1 \, m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \, m/s$. The tension in the string will be $52 \, N$ when the stone is:

$A$ bob attached to a string is held horizontal and released. The tension $T$ and vertical distance $h$ from the point of suspension can be represented by which of the following graphs?

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