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(D) According to the law of conservation of linear momentum for a perfectly inelastic collision:$m_1\vec{u}_1 + m_2\vec{u}_2 = (m_1 + m_2)\vec{V}$Give

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$4.6$

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(D) According to the law of conservation of linear momentum for a perfectly inelastic collision:$m_1\vec{u}_1 + m_2\vec{u}_2 = (m_1 + m_2)\vec{V}$Given:$m_1 = 1 \ g, \vec{u}_1 = 3\hat{i} - 2\hat{j} \ m/s$$m_2 = 2 \ g, \vec{u}_2 = 4\hat{j} - 6\hat{k} \ m/s$Substituting the values:$\vec{V} = \frac{m_1\vec{u}_1 + m_2\vec{u}_2}{m_1 + m_2} = \frac{1(3\hat{i} - 2\hat{j}) + 2(4\hat{j} - 6\hat{k})}{1 + 2}$$\vec{V} = \frac{3\hat{i} - 2\hat{j} + 8\hat{j} - 12\hat{k}}{3} = \frac{3\hat{i} + 6\hat{j} - 12\hat{k}}{3}$$\vec{V} = 1\hat{i} + 2\hat{j} - 4\hat{k} \ m/s$The magnitude of the combined velocity is:$|\vec{V}| = \sqrt{(1)^2 + (2)^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \approx 4.58 \ m/s \approx 4.6 \ m/s$.

$A$ body of mass $1 \ g$ moving with a velocity of $3\hat{i} - 2\hat{j} \ m/s$ collides with a body of mass $2 \ g$ moving with a velocity of $4\hat{j} - 6\hat{k} \ m/s$. If the two bodies stick together after the collision,what is the magnitude of their combined velocity in $m/s$?

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