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(D) Let the masses of the two balls be $m_1 = m_2 = m$.Let the initial velocities be $u_1 = 6 \, m/s$ and $u_2 = -6 \, m/s$.The coefficient of restitu

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$2$

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(D) Let the masses of the two balls be $m_1 = m_2 = m$.Let the initial velocities be $u_1 = 6 \, m/s$ and $u_2 = -6 \, m/s$.The coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$,where $v_1$ and $v_2$ are the final velocities.Since the masses are equal and the collision is head-on,the balls will exchange velocities if $e=1$. For a general $e$,the final velocities are given by $v_1 = \frac{(m_1 - em_2)u_1 + m_2(1+e)u_2}{m_1 + m_2}$ and $v_2 = \frac{(m_2 - em_1)u_2 + m_1(1+e)u_1}{m_1 + m_2}$.Substituting $m_1 = m_2 = m$,$u_1 = 6$,and $u_2 = -6$:$v_1 = \frac{(m - em)(6) + m(1+e)(-6)}{2m} = \frac{6 - 6e - 6 - 6e}{2} = -6e$.$v_2 = \frac{(m - em)(-6) + m(1+e)(6)}{2m} = \frac{-6 + 6e + 6 + 6e}{2} = 6e$.Given $e = 1/3$,the speed of each ball is $|v| = 6 	imes (1/3) = 2 \, m/s$.

Two balls of equal mass undergo head-on collision while each was moving with speed $6 \, m/s$ in opposite directions. If the coefficient of restitution is $e = 1/3$,the speed of each ball after impact will be ............ $m/s$.

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