Two bodies of mass 1 kg and 3 kg have position vectors î + 2ĵ + k̂ and −3î − 2ĵ + k̂ respectively. The centre of mass of this system has a position vector:

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Accepted Answer

−2î − ĵ + k̂. The position vector of the center of mass is given by $ \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} $. Substituting $ m_1 = 1 $, $ \vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k} $, $ m_2 = 3 $, and $ \vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k} $, we get $ \vec{R}_{cm} = \frac{(1)(\hat{i} + 2\hat{j} + \hat{k}) + (3)(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k} $.

Two bodies of mass 1 kg and 3 kg have position vectors î + 2ĵ + k̂ and −3î − 2ĵ + k̂ respectively. The centre of mass of this system has a position vector:

Solution

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