A body of mass $m_1=4 \mathrm{~kg}$ moves at $5 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$ and another body…

A body of mass $m_1=4 \mathrm{~kg}$ moves at $5 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$ and another body of mass $m_2=2 \mathrm{~kg}$ moves at $10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$. The kinetic energy of centre of mass is
  1. $\frac{200}{3} \mathrm{~J}$
  2. $\frac{500}{3} \mathrm{~J}$
  3. $\frac{400}{3} \mathrm{~J}$
  4. $\frac{800}{3} \mathrm{~J}$

Solution

$v_{\mathrm{CM}}=\frac{m_1 \frac{d r_1}{d t}+m_2 \frac{d r_2}{d t}}{m_1+m_2}$ $=\frac{4 \times 5 \hat{\mathbf{i}}+2 \times 10 \hat{\mathbf{i}}}{4+2}$ $v_{\mathrm{CM}}=\frac{40 \hat{\mathbf{i}}}{6}=\frac{20}{3} \hat{\mathbf{i}}$ The kinetic energy $K=\frac{1}{2} m v^2$ $=\frac{1}{2} \times(4+2) \times \frac{20 \times 20}{3 \times 3}$ $K=\frac{400}{3} \mathrm{~J}$ $K=\frac{400}{3} \mathrm{~J}$

Asked in: AP EAMCET 2010

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