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Chapter 7: Problem 36

If a particle of mass \(5.0 \mathrm{kg}\) is moving east at $10 \mathrm{m} / \mathrm{s}\( and a particle of mass \)15 \mathrm{kg}\( is moving west at \)10 \mathrm{m} / \mathrm{s},$ what is the velocity of the CM of the pair?

Short Answer

Expert verified
In a two-particle system with masses \(m_1 = 5.0 \ kg\) and \(m_2 = 15 \ kg\), and velocities \(v_1 = 10 \ m/s\) east and \(v_2 = -10 \ m/s\) west, the velocity of the center of mass is \(-5 \ m/s\). This indicates that the center of mass is moving westward at a speed of \(5 \ m/s\).

Step by step solution

01

Note the given values

We are given: - Mass of particle 1, \(m_1 = 5.0 \mathrm{kg}\) - Velocity of particle 1, \(v_1 = 10 \mathrm{m/s}\) (east) - Mass of particle 2, \(m_2 = 15 \mathrm{kg}\) - Velocity of particle 2, \(v_2 = -10 \mathrm{m/s}\) (west)
02

Determine the sum of particle masses

Calculate the total mass of the system by adding the masses of the two particles: $$m_{total} = m_1 + m_2 = 5.0 \mathrm{kg} + 15 \mathrm{kg} = 20 \mathrm{kg}$$
03

Calculate the weighted sum of particle velocities

Next, we'll calculate the weighted sum of the particles' velocities using their masses: $$m_1 v_1 + m_2 v_2 = (5.0 \mathrm{kg})(10 \mathrm{m/s}) + (15 \mathrm{kg})(-10 \mathrm{m/s}) = 50 \mathrm{kg \cdot m/s} - 150 \mathrm{kg \cdot m/s} = -100 \mathrm{kg \cdot m/s}$$
04

Calculate the velocity of the center of mass

Now, we'll use the formula for the velocity of the center of mass and plug in our values for the weighted sum of velocities and the total mass: $$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{-100 \mathrm{kg \cdot m/s}}{20 \mathrm{kg}} = -5 \mathrm{m/s}$$
05

Interpret the result

The velocity of the center of mass of the two particles is \(-5 \mathrm{m/s}\). Since it is negative, the center of mass is moving westward at a speed of \(5 \mathrm{m/s}\).

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Most popular questions from this chapter

For a safe re-entry into the Earth's atmosphere, the pilots of a space capsule must reduce their speed from \(2.6 \times 10^{4} \mathrm{m} / \mathrm{s}\) to \(1.1 \times 10^{4} \mathrm{m} / \mathrm{s} .\) The rocket engine produces a backward force on the capsule of \(1.8 \times 10^{5} \mathrm{N}\) The mass of the capsule is 3800 kg. For how long must they fire their engine? [Hint: Ignore the change in mass of the capsule due to the expulsion of exhaust gases.]
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