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(B) Let the initial velocity of the particle with mass $2m$ be $u$. The particle with mass $m$ is at rest.By the law of conservation of linear momentu

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$1/3$

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(B) Let the initial velocity of the particle with mass $2m$ be $u$. The particle with mass $m$ is at rest.By the law of conservation of linear momentum, the final velocity $v$ of the combined mass $(2m + m = 3m)$ is:$(2m)u + (m)(0) = (3m)v$$v = \frac{2mu}{3m} = \frac{2u}{3}$Initial kinetic energy $K_i = \frac{1}{2}(2m)u^2 = mu^2$.Final kinetic energy $K_f = \frac{1}{2}(3m)v^2 = \frac{1}{2}(3m)\left(\frac{2u}{3}\right)^2 = \frac{3m}{2} \cdot \frac{4u^2}{9} = \frac{2}{3}mu^2$.The loss in kinetic energy is $\Delta K = K_i - K_f = mu^2 - \frac{2}{3}mu^2 = \frac{1}{3}mu^2$.The fraction of initial kinetic energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{1}{3}mu^2}{mu^2} = \frac{1}{3}$.

In a one-dimensional collision, a particle of mass $2m$ collides with a particle of mass $m$ at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

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