A ball of mass m moving with speed v collides | vedclass

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(B) In an elastic collision,both linear momentum and kinetic energy are conserved.Let the mass of both balls be $m$.Initial kinetic energy $K_i = \fra

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{3} v$

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(B) In an elastic collision,both linear momentum and kinetic energy are conserved.Let the mass of both balls be $m$.Initial kinetic energy $K_i = \frac{1}{2} m v^2 + 0 = \frac{1}{2} m v^2$.Final kinetic energy $K_f = \frac{1}{2} m (v/3)^2 + \frac{1}{2} m v_2'^2$,where $v_2'$ is the speed of the second ball.Since the collision is elastic,$K_i = K_f$.$\frac{1}{2} m v^2 = \frac{1}{2} m (v^2/9) + \frac{1}{2} m v_2'^2$.Dividing by $\frac{1}{2} m$,we get $v^2 = \frac{v^2}{9} + v_2'^2$.$v_2'^2 = v^2 - \frac{v^2}{9} = \frac{8v^2}{9}$.Therefore,$v_2' = \sqrt{\frac{8}{9}} v = \frac{2\sqrt{2}}{3} v$.

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