Question

At one instant, the center of mass of a system of two particles is located on the \(x\)-axis at \(x\) = 2.0 m and has a velocity of (5.0 m/s)\(\hat{\imath}\). One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the \(x\)-axis at \(x\) = 8.0 m. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

Short answer

(a) The mass is 0.30 kg. (b) The total momentum is 2.0 kg m/s. (c) The velocity is approximately 6.67 m/s.

Step-by-step solution

Understanding the Center of Mass Location

The formula for the center of mass for two particles on the x-axis is given by \[x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\]where one particle is at the origin \(x_1 = 0\) \(\text{m}\), and its mass is \(m_1\), which we need to find. The other particle's position is \(x_2 = 8.0\) \(\text{m}\) with mass \(m_2 = 0.10\) \(\text{kg}\). The center of mass is located at \(x_{cm} = 2.0\) \(\text{m}\). Substituting into the equation gives: \[2.0 = \frac{m_1 \times 0 + 0.10 \times 8}{m_1 + 0.10}\]Simplifying this equation will help us find \(m_1\).

Solving for the Mass of the Particle at the Origin

Substitute the known values into the center of mass equation and solve for \(m_1\): \[2.0 = \frac{0 + 0.8}{m_1 + 0.10}\]Multiply both sides by \(m_1 + 0.10\): \[2.0(m_1 + 0.10) = 0.8\]Expanding gives: \[2.0m_1 + 0.20 = 0.8\]Subtract 0.20 from both sides: \[2.0m_1 = 0.6\]Divide by 2.0 to isolate \(m_1\): \[m_1 = 0.30 \text{ kg}\]

Calculating the Total Momentum of the System

The total momentum \(P_{total}\) of the system is calculated using the formula \[P_{total} = (m_1 + m_2) \times v_{cm}\]where \(v_{cm} = 5.0 \text{ m/s}\). Substitute \(m_1 = 0.30 \text{ kg}\) and \(m_2 = 0.10 \text{ kg}\): \[P_{total} = (0.30 + 0.10) \times 5.0\]\[P_{total} = 2.0 \text{ kg m/s}\]

Finding the Velocity of the Particle at the Origin

Using conservation of momentum, we know that the momentum of the complete system is equal to the momentum of the individual components: \[m_1 v_1 + m_2 v_2 = P_{total}\]where \(v_2 = 0\) because the second particle is at rest. Rearranging, we get: \[m_1 v_1 = P_{total}\]Substitute \(P_{total} = 2.0 \text{ kg m/s}\) and \(m_1 = 0.30 \text{ kg}\): \[0.30 v_1 = 2.0\]Solve for \(v_1\): \[v_1 = \frac{2.0}{0.30} \approx 6.67 \text{ m/s}\]

Key concepts

Momentum Conservation

In physics, momentum conservation is a fundamental concept that follows the law of conservation of momentum. It states that within a closed system, where no external forces are acting, the total momentum before any interaction will be equal to the total momentum after the interaction. This law is very useful in solving problems involving collisions or interactions between particles.

For example, consider a system of two particles where one particle is initially at rest. When a force is applied to one particle, the total momentum of the system will be the same before and after any movement. This principle helps calculate unknown variables like velocity or mass when another variable is known.

• The formula crucial to this conservation is:\[m_1 v_1 + m_2 v_2 = P_{total}\]
• Any change in the system must account for both particles, meaning if one gains momentum, the other must change to keep the system's total momentum constant.

This law is fundamental to understanding how particle interactions affect motion.

Particle Systems

A particle system is a model used in physics to study interactions among multiple particles. In problems like our exercise, we consider a system consisting of two particles. These particles can have different properties like mass and velocity, and they interact in a way that can be analyzed using principles like momentum conservation.

When analyzing a particle system, attention is often focused on concepts like the center of mass, which offers a simplified view of the system's combined motion. The center of mass acts as an "average" position that depends on the masses and positions of all particles within the system.

• For two particles, the center of mass \(x_{cm}\) is calculated as:\[x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\]
• This equation allows determination of the system's central position and, when further developed, offers insights into other aspects like velocities of the particles.

By focusing on particle systems, one can understand more complex interactions using simplified equations and models.

Velocity Calculation

Velocity calculation is an essential aspect of analyzing the dynamics of any particle system. In multiple particle systems, the velocity of each particle may be influenced by its mass and position relative to the center of mass and to other particles.

To determine the velocity of a specific particle, especially in a system where momentum is conserved, apply the conservation of momentum principle.

• For example, the velocity of a particle at the origin, \(v_1\), can be found using the formula:\[m_1 v_1 = P_{total}\]
• This formula rearranges to find \(v_1\) once all other variables are known.

In our system with a total momentum of 2.0 kg m/s, calculating the velocity involves rearranging the equation to solve for the desired velocity. Being comfortable with these calculations is key to understanding how the movement of each part of a system contributes to overall dynamics.