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Chapter 5: Problem 37

Two particles of masses \(4 \mathrm{~kg}\) and \(8 \mathrm{~kg}\) are separated by a distance of \(6 \mathrm{~m}\). If they are moving towards each other under the influence of a mutual force of attraction, then the two particles will meet each other at a distance of (A) \(6 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (B) \(2 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (C) \(4 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (D) \(8 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass

Short Answer

Expert verified
The two particles will meet at a distance of \(2 \mathrm{~m}\) from the \(8 \mathrm{~kg}\) mass.

Step by step solution

01

Calculate the initial center of mass

To calculate the initial center of mass, use the formula: \[x_{cm} = \frac{\sum m_ix_i}{\sum m_i}\] where \(m_i\) are the masses and \(x_i\) are their positions. Let the positions of the 4 kg and 8 kg masses be x=0 m and x=6 m, respectively. Then, \[x_{cm} = \frac{4(0) + 8(6)}{4+8}\]
02

Compute the initial center of mass position

Substitute the given values into the formula and solve for \(x_{cm}\): \[x_{cm} = \frac{0 + 48}{12}\] \[x_{cm} = 4\mathrm{~m}\] The initial center of mass is located at 4 meters from the 4 kg mass (and 2 meters from the 8 kg mass).
03

Conclude the final position

Since the center of mass remains constant due to the mutual force of attraction, the two particles will meet at the location of the center of mass: 4 meters from the 4 kg mass (and 2 meters from the 8 kg mass). Therefore, the correct answer is: (B) \(2 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass.

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