A body of mass 2 , kg moving with a velocity | vedclass

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(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.$m_1\vec{v}_

Vedclass · Accepted Answer

$(\hat{i} - \frac{8}{7}\hat{j} + \frac{17}{7}\hat{k})$

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(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.$m_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_{final}$Given $m_1 = 2 \, kg$,$\vec{v}_1 = (\hat{i} + \hat{j} + \hat{k}) \, m/s$ and $m_2 = 5 \, kg$,$\vec{v}_2 = (\hat{i} - 2\hat{j} + 3\hat{k}) \, m/s$.Substituting the values:$2(\hat{i} + \hat{j} + \hat{k}) + 5(\hat{i} - 2\hat{j} + 3\hat{k}) = (2 + 5)\vec{v}_{final}$$(2\hat{i} + 2\hat{j} + 2\hat{k}) + (5\hat{i} - 10\hat{j} + 15\hat{k}) = 7\vec{v}_{final}$$(7\hat{i} - 8\hat{j} + 17\hat{k}) = 7\vec{v}_{final}$$\vec{v}_{final} = \frac{7}{7}\hat{i} - \frac{8}{7}\hat{j} + \frac{17}{7}\hat{k} = (\hat{i} - \frac{8}{7}\hat{j} + \frac{17}{7}\hat{k}) \, m/s$.

$A$ body of mass $2 \, kg$ moving with a velocity $(\hat{i} + \hat{j} + \hat{k}) \, m/s$ collides with another body of mass $5 \, kg$ moving with velocity $(\hat{i} - 2\hat{j} + 3\hat{k}) \, m/s$. If they stick together,the velocity (in $m/s$) of the composite body is:

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