A sphere of mass 0.1,kg is attached to a cord | vedclass

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(A) $1$. First,calculate the velocity of the sphere just before the collision using the principle of conservation of energy. The potential energy at t

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(A) $1$. First,calculate the velocity of the sphere just before the collision using the principle of conservation of energy. The potential energy at the suspension height is converted into kinetic energy at the bottom: $mgh = \frac{1}{2}mv^2$.$2$. Given $h = 1\,m$ and $g = 10\,m/s^2$,the velocity $v$ is $\sqrt{2gh} = \sqrt{2 	imes 10 	imes 1} = \sqrt{20}\,m/s$.$3$. The collision is perfectly elastic between two bodies of equal mass $(m_1 = m_2 = 0.1\,kg)$. In a one-dimensional elastic collision between two equal masses,the bodies exchange their velocities.$4$. Since the sphere had velocity $v$ and the block was at rest,after the collision,the sphere comes to rest and the block moves with velocity $v$.$5$. The kinetic energy of the block after the collision is $K.E. = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 0.1 	imes (\sqrt{20})^2 = 0.05 	imes 20 = 1\,J$.

$A$ sphere of mass $0.1\,kg$ is attached to a cord of $1\,m$ length. Starting from the height of its point of suspension,this sphere hits a block of the same mass at rest on a frictionless table. If the impact is elastic,then the kinetic energy of the block after the collision is ............. $J$.

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