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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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Most popular questions from this chapter

A toy car of mass \(2.0 \mathrm{~kg}\) is stationary, and a child rolls a toy truck of mass 3.5 kg straight toward it with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) a) What is the velocity of the center of mass of the system consisting of the two toys? b) What are the velocities of the truck and the car with respect to the center of mass of the system consisting of the two toys?

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