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Chapter 8: Problem 56

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

Expert verified
(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

01

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 \(m_1\).
02

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 \(m_1\): \[m_1 = 0.30 \text{ kg}\]
03

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 \(m_2 = 0.10 \text{ kg}\): \[P_{total} = (0.30 + 0.10) \times 5.0\]\[P_{total} = 2.0 \text{ kg m/s}\]
04

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 \(v_1\): \[v_1 = \frac{2.0}{0.30} \approx 6.67 \text{ m/s}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 kg) is moving toward puck \(B\) (with mass 0.350 kg), which is initially at rest. After the collision, puck A has a velocity of 0.120 m/s to the left, and puck \(B\) has a velocity of 0.650 m/s to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0-ms contact time. The fastest-known served tennis ball was served by "Big Bill" Tilden in 1931, and its speed was measured to be 73 m/s. (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of 55 m/s, what force and what impulse did he exert on the ball, assuming only horizontal motion?

A 20.0-kg projectile is fired at an angle of 60.0\(^\circ\) above the horizontal with a speed of 80.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. Ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

You are called as an expert witness to analyze the following auto accident: Car \(B\), of mass 1900 kg, was stopped at a red light when it was hit from behind by car \(A\), of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 m long. The coefficient of kinetic friction between the tires and the road was 0.65. (a) What was the speed of car \(A\) just before the collision? (b) If the speed limit was 35 mph, was car \(A\) speeding, and if so, by how many miles per hour was it \(exceeding\) the speed limit?

A 1200-kg SUV is moving along a straight highway at 12.0 m/s. Another car, with mass 1800 kg and speed 20.0 m/s, has its center of mass 40.0 m ahead of the center of mass of the SUV (\(\textbf{Fig. E8.54}\)). Find (a) the position of the center of mass of the system consisting of the two cars; (b) the magnitude of the system's total momentum, by using the given data; (c) the speed of the system's center of mass; (d) the system's total momentum, by using the speed of the center of mass. Compare your result with that of part (b).
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