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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 \(x=2.0 \mathrm{m}\) and has a velocity of \((5.0 \mathrm{m} / \mathrm{s}) \hat{\imath} .\) One of the particles is at the origin. The other particle has a mass of 0.10 \(\mathrm{kg}\) and is at rest on the \(x\) -axis at \(x=8.0 \mathrm{m}\) . (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) Mass at the origin is 0.30 kg. (b) Total momentum is 2.0 kg*m/s. (c) Velocity at the origin is 6.67 m/s.

Step by step solution

01

Setup the Center of Mass Equation

The center of mass (COM) of two particles along the x-axis is given by the formula: \( x_{COM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \). Here, \( x_{COM} = 2.0 \text{ m}\), \( m_2 = 0.10 \text{ kg} \), and \( x_2 = 8.0 \text{ m} \). \( x_1 \) is the position of the particle at the origin, so \( x_1 = 0 \text{ m} \). Substitute the values into the equation to solve for \( m_1 \).
02

Solve for the Mass of the Particle at the Origin (m鈧�)

Substitute the known values into the center of mass equation: \[ 2.0 = \frac{m_1 \times 0 + 0.10 \times 8.0}{m_1 + 0.10} \]Simplify to get:\[ 2.0(m_1 + 0.10) = 0.80 \]\[ 2.0m_1 + 0.20 = 0.80 \]\[ 2.0m_1 = 0.60 \]\[ m_1 = 0.30 \text{ kg} \]
03

Calculate the Total Momentum

The total momentum (\( p_{total} \)) of the system is given by the formula: \( p_{total} = (m_1 v_1 + m_2 v_2) \), where \( v_1 \) is the velocity of the particle at the origin, and \( v_2 = 0 \) as the 0.10 kg particle is at rest. We know \( v_{COM} = 5.0 \text{ m/s} \), and since momentum is mass times velocity, use the total system mass (m鈧� + m鈧�) to find \( p_{total} = m_{total} v_{COM} = (0.30 + 0.10) \times 5.0 = 2.0 \text{ kg m/s} \).
04

Find the Velocity of the Particle at the Origin

Using the equation \( p_{total} = m_1 v_1 + m_2 v_2 \), substitute the known momentum \( 2.0 \text{ kg m/s} = 0.30 v_1 + 0 \text{ m/s} \). Solve for \( v_1 \):\[ 0.30 v_1 = 2.0 \text{ kg m/s} \]\[ v_1 = \frac{2.0}{0.30} = 6.67 \text{ m/s} \]

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, crucial for understanding motion within a system. In simple terms, momentum
is the product of an object's mass and its velocity. Mathematically, it's represented as:
  • \( p = mv \)
where \( p \) stands for momentum, \( m \) for mass, and \( v \) for velocity.
Understanding momentum becomes particularly important in systems comprising multiple objects, such as the two particles in this exercise.
Here, to discover the system's total momentum, you need to consider the contributions from each particle. In this context:
  • The total momentum \( p_{total} \) can be calculated by summing the momenta of all particles in the system.
  • For the exercise, given that one particle is at rest, its momentum is zero, simplifying the calculation to \( p_{total} = m_1 v_1 + m_2 v_2 \).
By calculating this, you gain insight into the motion dynamics of the overall system.
Particle Velocity
Velocity is the rate at which an object changes its position. It's a vector quantity, meaning it has both magnitude and direction.
In solving physics problems involving systems, understanding individual particle velocities is crucial; it aids in finding kinetic properties like total momentum.
In this scenario:
  • The given velocity of the center of mass (COM) significantly influences calculations.
  • One particle's position and velocity are known (rest at 8.0 m and velocity 0 m/s), allowing focus on the unknowns concerning the other particle.
To deduce the velocity of the particle at the origin:
  1. Utilize the total momentum equation.
  2. Since the system's momentum is influenced majorly by this particle, plug the values into \( p_{total} = m_1 v_1 + m_2 v_2 \) and solve for \( v_1 \).
  3. Embed the known values to find its pace, which happens to be 6.67 m/s through calculation.
With this understanding, you decipher how each particle's velocity affects the overall system鈥檚 dynamic movement.
Mass Calculation
Calculating mass accurately within a multi-particle system is vital for assessing properties like center of mass and momentum.
It involves understanding how the mass of each particle contributes to the system's overall dynamics.
When solving the task of finding the unknown mass at the origin:
  • The center of mass equation \( x_{COM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \) aids in relating an unknown mass or position to known quantities.
  • Given the other particle's mass and positions were known, the equation was rearranged to solve for the missing mass, \( m_1 \).
The solution steps indicated isolating \( m_1 \) by eliminating other variables through substitution:
  • Replace each known value carefully.
  • Resolve the equation to retrieve \( m_1 = 0.30 \, \text{kg} \).
This demonstrated how mathematical rearrangement alone can illuminate part of an unknown variable when dealing with center of mass in equation-based exercises.

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Most popular questions from this chapter

Energy Sharing. An object with mass \(m,\) initially at rest, explodes into two fragments, one with mass \(m_{A}\) and the other with mass \(m_{B},\) where \(m_{A}+m_{B}=m .\) (a) If energy \(Q\) is released in the explosion, how much kinetic energy does each fragment have immediately after the explosion? (b) What percentage of the total energy released does each fragment get when one fragment has four times the mass of the other?

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass \(7.50 \mathrm{kg},\) is sliding to the left at \(5.00 \mathrm{m} / \mathrm{s},\) while the other, of mass \(5.75 \mathrm{kg},\) is slipping to the right at 6.00 \(\mathrm{m} / \mathrm{s} .\) They hold fast to each other after they collide. (a) Find the magnitude and direction of the velocity of these free-spirited otters right after they collide. (b) How much mechanical energy dissipates during this play?

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

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass \(80.0 \mathrm{kg},\) is given a push and slides eastward. Abigail, with mass \(50.0 \mathrm{kg},\) is sent sliding northward. They collide, and after the collision Sam is moving at \(37.0^{\circ}\) north of east with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) and Abigail is moving at \(23.0^{\circ}\) south of east with a speed of 9.00 \(\mathrm{m} / \mathrm{s}\) (a) What was the speed of each person before the collision? (b) By how much did the total kinetic energy of the two people decrease during the collision?

A 15.0 -kg fish swimming at 1.10 \(\mathrm{m} / \mathrm{s}\) suddenly gobbles up a 4.50 -kg fish that is initially stationary. Neglect any drag effects of the water. (a) Find the speed of the large fish just after it eats the small one. (b) How much mechanical energy was dissipated during this meal?
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