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