A ball of mass m moving with a horizontal velocity v strikes the bob of the pendulum at rest. Mass of the bob of the pendulum is also m. During this collision the ball sticks with the bob. The height to which the combined mass rises (g=acceleration due to gravity)

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$\frac{v^{2}}{4 g}$

$\frac{v^{2}}{2 g}$

$\frac{v^{2}}{8 g}$

$\frac{v^{2}}{g}$

answer is B.

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Detailed Solution

By linear momentum is conserved both before and after the collision,

$mv = (m + m) v' \Rightarrow v' = \frac{v}{2}$
At the maximum height, all of the initial kinetic energy is transformed into gravitational potential energy.

Thus,

$\frac{1}{2} (2 m) v^{' 2} = 2 m \times gh \Rightarrow h = \frac{v^{2}}{8 g}$

Hence the correct answer is $\frac{v^{2}}{8 g} .$

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