Two identical balls are suspended side by sid | vedclass

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(D) Let the mass of each ball be $m$. When the first ball is released from a vertical height $h$,its velocity $v$ just before the collision is given b

Vedclass · Accepted Answer

$h/4$

Vedclass · Answer

(D) Let the mass of each ball be $m$. When the first ball is released from a vertical height $h$,its velocity $v$ just before the collision is given by the conservation of energy: $v = \sqrt{2gh}$.Since the collision is perfectly inelastic,the two balls stick together and move with a common velocity $V$ after the collision. Applying the law of conservation of linear momentum:$m \cdot v + m \cdot 0 = (m + m) \cdot V$$m \sqrt{2gh} = 2m \cdot V$$V = \frac{\sqrt{2gh}}{2} = \sqrt{\frac{2gh}{4}}$Let the combined system rise to a new vertical height $h'$. Using the conservation of energy for the combined system:$\frac{1}{2} (2m) V^2 = (2m) g h'$$V^2 = 2g h'$Substituting $V^2 = \frac{2gh}{4}$:$\frac{2gh}{4} = 2g h'$$h' = \frac{h}{4}$

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