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(D) Let the initial velocity of the first ball be $u_1 = u$ and the second ball be $u_2 = 0$. After the collision, the velocity of the first ball is $

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$1.0$

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(D) Let the initial velocity of the first ball be $u_1 = u$ and the second ball be $u_2 = 0$. After the collision, the velocity of the first ball is $v_1 = \frac{u}{3}$.Since the collision is elastic, the coefficient of restitution $e = 1$.The formula for the velocity of the first ball after an elastic collision is $v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2$.Substituting the known values: $\frac{u}{3} = \frac{2 - M}{2 + M} u + 0$.Dividing by $u$: $\frac{1}{3} = \frac{2 - M}{2 + M}$.Cross-multiplying: $2 + M = 3(2 - M) = 6 - 3M$.Rearranging terms: $M + 3M = 6 - 2$, which gives $4M = 4$.Therefore, $M = 1 \,kg$.

$A$ ball of mass $2 \,kg$ collides with another ball of mass $M$ at rest. If the collision is elastic and after the collision, the first ball moves with $\frac{1}{3}$ of its initial velocity in the same direction, then the mass of the second ball is: (in $\,kg$)

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