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Chapter 7: Problem 72

For a system of three particles moving along a line, an observer in a laboratory measures the following masses and velocities. What is the velocity of the \(\mathrm{CM}\) of the system? $$\begin{array}{cc}\hline \text { Mass }(\mathrm{kg}) & v_{x}(\mathrm{m} / \mathrm{s}) \\\\\hline 3.0 & +290 \\\5.0 & -120 \\\2.0 & +52 \\\\\hline\end{array}$$

Short Answer

Expert verified
Answer: The velocity of the center of mass for the system of three particles is 37.4 m/s.

Step by step solution

01

Calculate the total mass of the system

We need to calculate the total mass of the system: \(M = \sum_{i=1}^{3} m_i\). Using the given values, we have: $$M = 3.0\,\text{kg} + 5.0\,\text{kg} + 2.0\,\text{kg} = 10.0\,\text{kg}$$
02

Calculate the total momentum of the system

Now, we need to calculate the total momentum of the system: \(p_{total} = \sum_{i=1}^{3} m_iv_i\). Using the given values, we have: $$p_{total} = (3.0\,\text{kg}\cdot 290\,\text{m/s}) + (5.0\,\text{kg}\cdot -120\,\text{m/s}) + (2.0\,\text{kg}\cdot 52\,\text{m/s}) = 870\,\text{kg·m/s} - 600\,\text{kg·m/s} + 104\,\text{kg·m/s} = 374\,\text{kg·m/s}$$
03

Calculate the velocity of the center of mass

Finally, we can use the formula \(v_{CM} = \frac{p_{total}}{M}\) to calculate the velocity of the center of mass. $$v_{CM} = \frac{374\,\text{kg·m/s}}{10.0\,\text{kg}} = 37.4\,\text{m/s}$$ The velocity of the center of mass of the system is \(37.4\,\text{m/s}\).

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Most popular questions from this chapter

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