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(D) According to the law of conservation of linear momentum,the initial momentum of the rock is zero. Therefore,the vector sum of the momenta of the t

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$5$

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(D) According to the law of conservation of linear momentum,the initial momentum of the rock is zero. Therefore,the vector sum of the momenta of the three parts must be zero.Let the momenta of the three parts be $\vec{p}_1$,$\vec{p}_2$,and $\vec{p}_3$.$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$.The magnitude of the momentum of the first part is $p_1 = m_1 v_1 = 1 \, kg 	imes 12 \, m s^{-1} = 12 \, kg \, m s^{-1}$.The magnitude of the momentum of the second part is $p_2 = m_2 v_2 = 2 \, kg 	imes 8 \, m s^{-1} = 16 \, kg \, m s^{-1}$.Since the two parts move at right angles to each other,the magnitude of the resultant momentum of these two parts is $p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, kg \, m s^{-1}$.The third part must have a momentum equal in magnitude and opposite in direction to this resultant momentum to satisfy the conservation law.Thus,$p_3 = p_{12} = 20 \, kg \, m s^{-1}$.Given the speed of the third part $v_3 = 4 \, m s^{-1}$,its mass $m_3$ is:$m_3 = \frac{p_3}{v_3} = \frac{20 \, kg \, m s^{-1}}{4 \, m s^{-1}} = 5 \, kg$.

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