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(D) Let the initial velocity of the $1 \ kg$ ball be $u$. After the collision,the first ball moves with velocity $v_1$ along the $Y$-axis,and the seco

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$2$

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(D) Let the initial velocity of the $1 \ kg$ ball be $u$. After the collision,the first ball moves with velocity $v_1$ along the $Y$-axis,and the second ball of mass $m$ moves with velocity $v_2$ at an angle $30^{\circ}$ below the $X$-axis.Applying the law of conservation of linear momentum along the $X$-axis:$1 \cdot u = m v_2 \cos(30^{\circ}) \implies u = m v_2 \frac{\sqrt{3}}{2} \implies v_2 = \frac{2u}{m\sqrt{3}} \quad ... (1)$Applying the law of conservation of linear momentum along the $Y$-axis:$0 = 1 \cdot v_1 - m v_2 \sin(30^{\circ}) \implies v_1 = m v_2 \sin(30^{\circ}) = m v_2 \cdot \frac{1}{2} \implies v_2 = \frac{2v_1}{m} \quad ... (2)$From $(1)$ and $(2)$,$\frac{2u}{m\sqrt{3}} = \frac{2v_1}{m} \implies v_1 = \frac{u}{\sqrt{3}}$.Since the collision is elastic,kinetic energy is conserved:$\frac{1}{2} (1) u^2 = \frac{1}{2} (1) v_1^2 + \frac{1}{2} m v_2^2$$u^2 = v_1^2 + m v_2^2$Substitute $v_1 = \frac{u}{\sqrt{3}}$ and $v_2 = \frac{2v_1}{m} = \frac{2u}{m\sqrt{3}}$:$u^2 = \left(\frac{u}{\sqrt{3}}\right)^2 + m \left(\frac{2u}{m\sqrt{3}}\right)^2$$u^2 = \frac{u^2}{3} + m \cdot \frac{4u^2}{3m^2} = \frac{u^2}{3} + \frac{4u^2}{3m}$$1 = \frac{1}{3} + \frac{4}{3m} \implies \frac{2}{3} = \frac{4}{3m} \implies m = 2 \ kg$.

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