/*! 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 116 Two particles \(P\) and \(Q\) ar... [FREE SOLUTION] | 91Ó°ÊÓ

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Two particles \(P\) and \(Q\) are released from rest \(1.0 \mathrm{~m}\) apart. \(P\) has a mass of \(0.10 \mathrm{~kg}\), and \(Q\) a mass of \(0.30 \mathrm{~kg} . P\) and \(Q\) attract each other with a constant force of \(1.0 \times 10^{-2} \mathrm{~N}\). No external forces act on the system. (a) What is the speed of the center of mass of \(P\) and \(Q\) when the separation is \(0.50 \mathrm{~m}\) ? (b) At what distance from \(P\) 's original position do the particles collide?

Short Answer

Expert verified
(a) Speed of the center of mass is 0 \, m/s. (b) They collide 0.25 \, m from P's original position.

Step by step solution

01

Initial Setup

First, identify the initial conditions of the problem. The particles are initially at rest, so the initial speed of the center of mass is zero.
02

Understanding the Center of Mass

The system's center of mass does not move since there are no external forces. Therefore, the speed of the center of mass remains zero, even when they are 0.5 m apart.
03

Apply Newton's Second Law

For each particle, Newton's second law states that the force equals the mass times acceleration: \( F = m \cdot a \). Using this, compute accelerations of each particle: \[ a_P = \frac{F}{m_P} = \frac{1.0 \times 10^{-2}}{0.10} \mathrm{\ m/s^2} \]\[ a_Q = \frac{F}{m_Q} = \frac{1.0 \times 10^{-2}}{0.30} \mathrm{\ m/s^2} \]
04

Set Up the Motion Equation

Since the particles start from rest and move towards each other, use the equation of motion to find when they collide. The sum of displacements from each side will equal 1 m:\[ x_P + x_Q = 1 \mathrm{\ m} \]
05

Use the Equations of Motion

Consider the formula: \( x = \frac{1}{2} a t^2 \). For each particle, substitute the acceleration and solve: \[ x_P = \frac{1}{2} a_P t^2 \] and \[ x_Q = \frac{1}{2} a_Q t^2 \], satisfying \( x_P + x_Q = 1 \). Substitute the given accelerations to solve for \( t \).
06

Solve for Time of Collision

From the motion equations:\( \frac{1}{2} \cdot \left( \frac{1.0 \times 10^{-2}}{0.10} \right) t^2 + \frac{1}{2} \cdot \left( \frac{1.0 \times 10^{-2}}{0.30} \right) t^2 = 1 \).Solve for \( t \) to find the time when they collide.
07

Calculate Collision Point

Use \( x_P = \frac{1}{2} a_P t^2 \) and calculate the distance \( x_P \) from the original position of \( P \) when they collide, given the time \( t \) calculated from earlier steps.

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Key Concepts

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

Center of Mass
The center of mass is an important concept in physics that helps us understand how objects move. Imagine it as the "average" position of all the mass in a system. For two particles like in our exercise, the center of mass will determine how the entire system behaves when forces act on it. Since no external forces are acting on the system of particles, the center of mass doesn't move, meaning its velocity remains zero throughout.
This might seem surprising, especially since each individual particle is moving, but it makes sense when considering the system as a whole. The particles are always getting closer, but they pull the same on each other, and this ensures that the center of mass remains at its initial position.
  • Think of the center of mass as a balance point: if you try to measure it, it will not shift unless an outside force acts on it.
  • Because both particles start from rest, and there's no external push or pull, the center of mass naturally stays still.
Understanding this helps us predict and calculate the results of collisions and interactions in larger systems as well.
Equations of Motion
The equations of motion are key tools in physics. They describe how objects move under various forces and conditions. In the case of our two particles, understanding motion gives us insight into how fast they collide and how far they move in that time. We specifically used a form of this equation: \[ x = \frac{1}{2} a t^2 \]This equation tells us how far an object will move over a period of time, depending on its acceleration and starting condition.
Each particle's acceleration was calculated using Newton's second law, showing how their separate forces influenced their motion in this problem. This provides a foundation to measure when and where they will meet, using time as a variable to track their paths.
  • The formula involves understanding initial conditions, such as starting from rest.
  • By calculating each particle's movement, we can piece together the entire motion of the system.
This systematic approach is crucial for predicting outcomes in engineering and many scientific fields.
Collision Dynamics
Collision dynamics involves studying how two or more objects interact when they come into contact. It's all about understanding the changes in motion and forces as they collide. In our problem, the two particles eventually collide after being attracted to each other by a constant force. This dynamics study tells us not only when they will collide but also at what position relative to their starting points.
In such problems, knowing the time of collision helps us determine the precise point of impact. By applying the equations of motion for each particle, we figure out where they meet. This critical analysis is grounded in physics principles like momentum conservation—though not directly listed in our problem, the constant force and resulting accelerations imply this.
  • The collision point is calculated based on both particles' paths measured from their original positions.
  • This type of analysis helps in designing safe transportation systems or understanding planetary movements.
Recognizing these interactions helps us manage and predict complex behaviors, from simple systems to larger, more intricate scenarios.

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Most popular questions from this chapter

A \(75 \mathrm{~kg}\) man rides on a \(39 \mathrm{~kg}\) cart moving at a velocity of \(2.3 \mathrm{~m} / \mathrm{s}\) He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?

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An object, with mass \(m\) and speed \(v\) relative to an observer. explodes into two pieces, one three times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?

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