AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
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
- A \(\frac{200}{3} \mathrm{~J}\)
- B \(\frac{500}{3} \mathrm{~J}\)
- C \(\frac{400}{3} \mathrm{~J}\)
- D \(\frac{800}{3} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(\frac{400}{3} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(v_{\mathrm{CM}}=\frac{m_1 \frac{d r_1}{d t}+m_2 \frac{d r_2}{d t}}{m_1+m_2}\) \(=\frac{4 \times 5 \hat{\mathbf{i}}+2 \times 10 \hat{\mathbf{i}}}{4+2}\) \(v_{\mathrm{CM}}=\frac{40 \hat{\mathbf{i}}}{6}=\frac{20}{3} \hat{\mathbf{i}}\) The kinetic energy \(K=\frac{1}{2} m v^2\)…
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