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

$\begin {array} {1 1} (1)\;\large\frac{v^2}{4g} & \quad (2)\;\large\frac{v^2}{8g} \\ (3)\;\large\frac{v^2}{g} & \quad (4)\;\large\frac{v^2}{2g} \end {array}$

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1 Answer

(2) $ \large\frac{v^2}{8g}$
answered Nov 7, 2013 by pady_1
 

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