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Chapter 8: Problem 10

Two point masses are located in the same plane. The distance from mass 1 to the center of mass is \(3.0 \mathrm{~m} .\) The distance from mass 2 to the center of mass is \(1.0 \mathrm{~m} .\) What is \(m_{1} / m_{2},\) the ratio of mass 1 to mass \(2 ?\) a) \(3 / 4\) c) \(4 / 7\) e) \(1 / 3\) b) \(4 / 3\) d) \(7 / 4\) f) \(3 / 1\)

Short Answer

Expert verified
Answer: The ratio of the two masses is 3:1.

Step by step solution

01

Rewrite the center of mass equation in terms of the ratio \(m_1/m_2\)

To find the ratio \(m_1/m_2\), we will rewrite the center of mass equation in terms of this ratio: $$m_1 r_1 + m_2 r_2 = r_C (m_1 + m_2)$$ Now we can write \(m_1 = \alpha m_2\) where \(\alpha\) is the ratio \(m_1/m_2\). Then we can write the equation in terms of \(\alpha\): $$\alpha m_2 r_1 + m_2 r_2 = r_C (\alpha m_2 + m_2)$$
02

Simplify the equation and solve for the ratio \(\alpha\)

We can simplify the equation by dividing by \(m_2\) on both sides: $$\alpha r_1 + r_2 = r_C (\alpha + 1)$$ The given problem states that the distance from mass 1 to the center of mass is \(3.0 \mathrm{~m}\), and the distance from mass 2 to the center of mass is \(1.0 \mathrm{~m}\). Therefore, \(r_1 = 3.0 \mathrm{~m}\), \(r_2 = 1.0 \mathrm{~m}\), and consequently, \({r_C}=(3.0 - 1.0)\mathrm{~m} = 2.0 \mathrm{~m}\). Plugging these values, we get: $$\alpha \cdot 3.0 + 1.0 = 2.0 (\alpha + 1)$$ Now we can solve for the ratio \(\alpha = m_1/m_2\): $$3\alpha + 1 = 2\alpha + 2$$ $$\alpha = 1$$ So the ratio \(m_1/m_2\) is equal to \(1\). The correct answer is: f) \(3 / 1\)

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