/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A system consists of two particl... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 58

A system consists of two particles. \(\mathrm{At} t=0\) one particle is at the origin; the other, which has a mass of \(0.50 \mathrm{~kg}\). is on the \(y\) -axis at \(y=6,0 \mathrm{~m} .\) At \(t=0\) the center of mass of the system is on the \(y\) -axis at \(y=2.4 \mathrm{~m} .\) The velocity of the center of mass is given by \(\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2} \hat{i}\) (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time \(t\). (c) Find the net external force acting on the system at \(t=3.0 \mathrm{~s}\)

Short Answer

Expert verified
The total mass of the system is 1.667kg. The acceleration of the center of mass at any time \(t\) is \(1.5t m/s^2\). The net external force acting on the system at \(t = 3.0s\) is \(7.5N\) towards the positive \(x\)-axis.

Step by step solution

01

Calculate Total Mass

The formula for the center of mass for a system of particles is \(y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\), where \(y_{cm}\) represents the y-coordinate of the center of mass, \(m_i\) the mass of the \(i^{th}\) particle and \(y_i\) the y-coordinate of the \(i^{th}\) particle. At \(t=0\) the particle at the origin has y-coordinate \(0\) and the mass of this particle will denote as \(m_1\). The particle on the y-axis at \(y = 6m\) has mass \(m_2 = 0.50kg\). Substituting these values into the formula for center of mass we have \(2.4m = \frac{m_1\cdot0 + 0.50kg\cdot6m}{m_1 + 0.50kg}\). Solving this equation for \(m_1\) yields the total mass of the system.
02

Find the acceleration of the center of mass

The velocity of the center of mass is given by \(v = 0.75t^2\) which is a function of time. The derivative of velocity with respect to time will give the acceleration, denoted by \(a\). Hence by differentiating the equation for velocity, we get \(a = \frac{dv}{dt} = 1.5t\).
03

Calculate Net External Force

The net external force acting on a particle is given by \(F = ma\), where \(a\) is the acceleration and \(m\) is the mass of the particle in question, in this case, we use the total mass of the system, which we have calculated in step 1. Substituting the acceleration calculated in step 2 and the total mass into this equation, we find the net external force at any time. Finally evaluating the equation of net external force at \(t = 3.0s\) gives the solution.

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Key Concepts

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

Total Mass Calculation
Understanding the total mass of a system is a fundamental concept in physics, serving as a base for various other calculations. When dealing with a system of particles, the total mass is simply the sum of the masses of individual particles.

In our textbook exercise, the total mass is found by utilizing the y-coordinate equation for the center of mass, which is \(y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\). This equation essentially states that the center of mass is the weighted average position of all the masses in the system. Here, one particle is at the origin with an unknown mass \(m_1\), and the other particle has a known mass of \(0.50 \text{kg}\) and is located at \(y=6,0 \text{m}\).

Given the center of mass is at \(y=2.4 \text{m}\), we set up the equation \(2.4m = \frac{m_1\cdot0 + 0.50\text{kg}\cdot6m}{m_1 + 0.50\text{kg}}\), which upon solving gives us the value of \(m_1\). The sum of \(m_1\) and \(0.50 \text{kg}\) yields the total mass of the system. Knowing this allows us to delve deeper into understanding the system's dynamics, especially when observing how external forces interact with the mass.
Acceleration of Center of Mass
The acceleration of the center of mass is pivotal in describing how the velocity of the system's mass center changes over time. In this exercise, the velocity of the center of mass is provided as a function of time: \(v = 0.75t^2\).

To find the acceleration at any point in time, we look at the rate of change of velocity, which is the derivative of the velocity function with respect to time. By differentiating the velocity \(v\) with respect to time \(t\), we obtain the acceleration equation \(a = \frac{dv}{dt} = 1.5t\). This simple relation shows that the acceleration of the system's center of mass is directly proportional to time and will help us later in calculating the net external force acting on the system.
Net External Force
The net external force is a concept that emerges from Newton's second law of motion, which states that force is the product of mass and acceleration \(F = ma\). This law implies that the total force acting on an object or system is the cause of its acceleration.

In order to find the net external force acting on the system at a particular time, we need both the total mass calculated previously and the acceleration of the center of mass at that instant. After evaluating the expression for acceleration derived from the velocity鈥擻(a = 1.5t\)鈥攁t \(t = 3.0 \text{s}\), we plug this value into the equation for force, along with the total mass of the system. This calculation deciphers the net external force acting on the system at the specified time, providing insight into the influence of the surroundings on the system's motion.

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Most popular questions from this chapter

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd collisions, thus causing less damage, In one such accident, a \(1750 \mathrm{~kg}\) car traveling to the right at \(1.50 \mathrm{~m} / \mathrm{s}\) collides with a \(1450 \mathrm{~kg}\) car going to the left at \(1.10 \mathrm{~m} / \mathrm{s}\) Measurements show that the heavier car's speed just after the collision was \(0.250 \mathrm{~m} / \mathrm{s}\) in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic eneryy of the two-car system during this collision.

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You (mass 55kg) are riding a frictionless skateboard (mass 5.0 kg) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\). \(\Lambda\) friend standing on a balcuny ahove you drops a \(2.5 \mathrm{~kg}\) sack of flont straight down into your arms. (a) What is your new speed while you hold the suck? (b) since the sack was dropped vertically, how can it affect your horizontal thotion? Explain. (c) Now you try to rid yourscif of the extra weight hy throwing the sack straight up. What will be your speed while the sack is in the air? Explain.

Two carts of cqual mass are on a horizontal, frictionless air track. Initially cart \(A\) is moving toward stationary cart \(B\) with a speed of \(v_{A}\). The carts undergo an inelastic collision, and after the collision the total kinetic energy of the two carts is one-half their initial total kinctic energy before the collision. What is the speed of cach cart after the collision?
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