/*! 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 54 A system consists of two particl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A system consists of two particles. 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 \(\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\mathrm{i}}\) , (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
(a) 1.25 kg, (b) \(1.5t\, \mathrm{m/s^2} \hat{\mathrm{i}}\), (c) 5.625 N \(\hat{\mathrm{i}}\) at \(t=3\, \mathrm{s}\).

Step by step solution

01

Understand the given data

We have two particles in the system. At time \(t=0\), one is at the origin \((0,0)\) and the other is on the \(y\)-axis at \(y=6.0\, \mathrm{m}\) with a mass of \(0.50\, \mathrm{kg}\). The center of mass of the system is along the \(y\)-axis at \(y=2.4\, \mathrm{m}\). The velocity of the center of mass \(\mathrm{v_{CM}}\) is given by \(\left(0.75 \mathrm{m/s}^{3}\right)t^{2} \hat{\mathrm{i}}\).
02

Calculate the other mass

The center of mass \(\mathrm{y_{CM}}\) formula for two particles is \(\frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \mathrm{y_{CM}}\). Given that \(y_1 = 0\), \(y_2 = 6.0\, \mathrm{m}\), \(y_{CM} = 2.4\, \mathrm{m}\), and \(m_2 = 0.50\, \mathrm{kg}\), we solve for \(m_1\):\[\begin{align*}\frac{m_1 \cdot 0 + 0.50 \cdot 6.0}{m_1 + 0.50} &= 2.4 \3.0 &= 2.4(m_1 + 0.50) \3.0 = 2.4m_1 + 1.2 \1.8 = 2.4m_1 \m_1 &= \frac{1.8}{2.4} = 0.75\, \mathrm{kg}\end{align*}\]
03

Find the total mass of the system

The total mass of the system \(M\) is the sum of the two masses: \[ M = m_1 + m_2 = 0.75\, \mathrm{kg} + 0.50\, \mathrm{kg} = 1.25\, \mathrm{kg} \]
04

Express the acceleration of the center of mass

The velocity of the center of mass \(\mathrm{v_{CM}}\) is given by \( \left(0.75 \mathrm{m/s}^{3}\right)t^{2} \hat{\mathrm{i}}\). Acceleration is the derivative of velocity with respect to time.\[ a_{\mathrm{CM}} = \frac{d}{dt} \left(0.75 t^2 \right) = 2 \cdot 0.75 \cdot t = 1.5t \mathrm{\, m/s^2} \hat{\mathrm{i}} \]
05

Calculate the net external force acting on the system at \(t = 3.0\, \mathrm{s}\)

Using Newton's second law, the net external force \(\mathbf{F}_{\text{net}}\) acting on the system is given by:\[\mathbf{F}_{\text{net}} = M \cdot \mathbf{a}_{\mathrm{CM}}\]Substituting the known values:\[\mathbf{F}_{\text{net}} = 1.25\, \mathrm{kg} \cdot 1.5 \times 3.0\, \mathrm{m/s^2} = 5.625\, \mathrm{N} \hat{\mathrm{i}} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Particle Systems
A particle system is a collection of particles considered as a whole within a physics problem. Each particle in the system is defined by its own properties, such as mass and position. In the problem discussed, there are two particles in the system. The first particle is at the origin of the coordinate system, and the second particle is at a specific point along the y-axis. By analyzing the properties of each particle, we can determine various characteristics of the system, like the center of mass, which indicates how the mass is distributed in space.
  • Particle 1: Positioned at the origin (0,0), unknown mass.
  • Particle 2: Located at (0, 6.0 m) on the y-axis, with a mass of 0.50 kg.
The concept of particle systems becomes crucial when analyzing the overall motion or properties of the system, such as finding the center of mass or using these properties to solve for unknowns, like the missing mass in this scenario.
Newton's Second Law
Newton's Second Law is pivotal in understanding motion and forces in physics. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, expressed by the equation:\[ F = ma \]In our scenario, this law helps us determine the net external force acting on the particle system. Once the acceleration of the center of mass is known, and the total mass of the system is calculated, we can apply Newton's Second Law to find the net force:
  • Formula: \( F_{\text{net}} = M \cdot a_{\mathrm{CM}} \)
  • At \( t = 3.0 \) s: Given the acceleration and mass, this law provides the necessary force (in Newtons) acting on the system.
By applying Newton's Second Law to the whole system, we connect the acceleration of the center of mass with the forces involved, culminating in a precise understanding of the forces at work.
External Forces
External forces are influences that come from outside a given system and affect its motion. For the particle system under discussion, external forces are the key reason for any change in velocity or acceleration. Understanding these forces is critical for predicting and analyzing how the system will behave over time.In mechanics, the equation from Newton's Second Law \( F = ma \) shows how these external forces relate to acceleration. By knowing the acceleration of the center of mass at any time, we can infer the sum of all external forces acting on the system. At time \( t = 3.0 \) seconds, the net external force was calculated based on the rate of change of velocity of the center of mass.
  • Magnitude: The external force magnitude tells how strong the external influence is.
  • Analysis: Evaluating external forces aids in understanding the dynamics of complex particle systems.
Kinematics
Kinematics is the branch of physics that studies motion without considering the forces that cause such motion. It focuses on parameters like displacement, velocity, and acceleration. The current problem uses kinematics to analyze the motion of the center of mass of the particle system.In this problem, the velocity of the center of mass is given as a function of time, indicating how it moves as the system evolves. From velocity, we can derive the acceleration systematically using kinematic equations:\[ a_{\mathrm{CM}} = \frac{d}{dt}v_{\mathrm{CM}} = \frac{d}{dt} \left(0.75t^2\right) = 1.5t \mathrm{\, m/s^2} \]The calculated acceleration is then utilized in conjunction with the masses to explore further physical properties like forces.
  • Velocity Function: Described as \( \left(0.75 \mathrm{m/s}^{3}\right)t^{2} \hat{\mathrm{i}} \)
  • Acceleration Derivation: Derive acceleration by taking the time derivative of velocity.
This structured approach allows us to comprehend how each particle's motion contributes to the system's physical behavior, an essential component in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(0.150-\mathrm{kg}\) frame, when suspended from a coil spring, stretches the spring 0.050 \(\mathrm{m}\) . A \(0.200-\mathrm{kg}\) lump of putty is dropped from rest onto the frame from a height of 30.0 \(\mathrm{cm}\) (Fig. 8.42\()\) . Find the maximum distance the frame moves downward from its initial position.

A \(1500-\mathrm{kg}\) blue convertible is traveling south, and a \(2000-\mathrm{kg}\) red SUV is traveling west. If the total momentum of the system consisting of the two cars is 8000 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) directed at \(60.0^{\circ}\) west of south, what is the speed of each vehicle?

Force of a Golf Swing. A \(0.0450-\mathrm{kg}\) golf ball initially at rest is given a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) when a club strikes. If the club and ball are in contact for 2.00 \(\mathrm{ms}\) , what average force acts on the ball? Is the effect of the ball's weight during the time of contact significant? Why or why not?

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}(\text { Fig } .8 .46)\) . You can ignore friction between the cart and the floor. A 15.0 \(\mathrm{kg}\) package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

A 68.5 -kg astronaut is doing a repair in space on the orbiting space station. She throws a \(2.25-\mathrm{kg}\) tool away from her at 3.20 \(\mathrm{m} / \mathrm{s}\) relative to the space station. With what speed and in what direction will she begin to move?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.