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Standard XII

Physics

Throwing and Catching at Same Height

A bullet of m...

Question

A bullet of mass m and velocity v passes through a pendulum of mass M and emerges with velocity $v / 2$ . What is the minimum value of $v$ such that the pendulum will swing through a complete cycle?

\frac{M}{m} \sqrt{2 l g}

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\frac{2 M}{m} \sqrt{2 l g}

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\frac{M}{2 m} \sqrt{5 l g}

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\frac{2 M}{m} \sqrt{5 l g}

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Solution

The correct option is D $\frac{2 M}{m} \sqrt{5 l g}$
$M v_{1} = m (v - \frac{v}{2})$
$M v_{1} = \frac{m v}{2} \Rightarrow v_{1} = \frac{m}{M} (\frac{v}{2})$
To describe be a vertical circle
$v_{1}$ should be $\sqrt{5 g l}$
$∴ \sqrt{5 g l} = \frac{m}{M} (\frac{v}{2}) \Rightarrow v = \frac{2 M}{m} \sqrt{5 g l}$

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A bullet of mass m and velocity v passes through a pendulum of mass M and emerges with velocity v/2. What is the minimum value of v such that the pendulum will swing through a complete cycle?

The correct option is D 2Mm√5lgMv1=m(v−v2)Mv1=mv2⇒v1=mM(v2)To describe be a vertical circlev1 should be √5gl∴√5gl=mM(v2)⇒v=2Mm√5gl

A bullet of mass m and velocity v passes through a pendulum of mass M and emerges with velocity $v / 2$ . What is the minimum value of $v$ such that the pendulum will swing through a complete cycle?