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(B) Let the velocity of the ball of mass $2m$ be $u_0$ and the system of two balls be at rest. After the elastic collision, let the velocity of the ba

Vedclass · Accepted Answer

$\frac{4}{9}$

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(B) Let the velocity of the ball of mass $2m$ be $u_0$ and the system of two balls be at rest. After the elastic collision, let the velocity of the ball of mass $2m$ be $v_1$ and the velocity of the first ball of mass $m$ be $v_2$.By conservation of linear momentum: $2m u_0 = 2m v_1 + m v_2 \implies 2u_0 = 2v_1 + v_2$.Since the collision is perfectly elastic, the coefficient of restitution $e = 1 = \frac{v_2 - v_1}{u_0} \implies v_2 - v_1 = u_0$.Solving these two equations: $v_2 = u_0 + v_1$. Substituting this into the momentum equation: $2u_0 = 2v_1 + (u_0 + v_1) = 3v_1 + u_0 \implies 3v_1 = u_0 \implies v_1 = \frac{u_0}{3}$.Then $v_2 = u_0 + \frac{u_0}{3} = \frac{4u_0}{3}$.The vibrational energy is the kinetic energy of the two-ball system in the center-of-mass frame. The velocity of the center of mass of the two-ball system is $v_{cm} = \frac{m(v_2) + m(0)}{2m} = \frac{v_2}{2} = \frac{2u_0}{3}$.The vibrational energy $E_v = \frac{1}{2} \mu v_{rel}^2$, where $\mu = \frac{m \cdot m}{m+m} = \frac{m}{2}$ and $v_{rel} = v_2 - 0 = \frac{4u_0}{3}$.$E_v = \frac{1}{2} \left( \frac{m}{2} \right) \left( \frac{4u_0}{3} \right)^2 = \frac{m}{4} \cdot \frac{16u_0^2}{9} = \frac{4mu_0^2}{9}$.The initial kinetic energy of the ball of mass $2m$ is $K_i = \frac{1}{2} (2m) u_0^2 = mu_0^2$.The ratio is $\frac{E_v}{K_i} = \frac{4mu_0^2 / 9}{mu_0^2} = \frac{4}{9}$.

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