/*! 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 7 Two particles on a line are mutu... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 7

Two particles on a line are mutually attracted by a force \(F=-a r\), where \(a\) is a constant and \(r\) the distance of separation. At time \(t=0\), particle A of mass \(m\) is located at the origin, and particle B of mass \(m / 4\) is located at \(r=5.0 \mathrm{~cm} .\) (a) If the particles are at rest at \(t=0\), at what value of \(r\) do they collide? (b) What is the relative velocity of the two particles at the moment the collision occurs?

Short Answer

Expert verified
Answer: To find the relative velocity and distance at the moment of collision, we follow these steps: 1. Determine the forces acting on particles A and B, described by Newton's second law 2. Express the accelerations of particles A and B in terms of their mass, distance, and force 3. Find the relative acceleration between the particles by subtracting their accelerations 4. Calculate the relative velocity using the relative acceleration and integrating with respect to time 5. Find the distance between the particles as a function of time by solving a differential equation 6. Determine the time at which the particles collide by setting the distance equal to zero 7. Substitute the time of collision into the expression for relative velocity to find the relative velocity at the moment of collision By following these steps, we can find the relative velocity and distance at the moment when particles A and B collide.

Step by step solution

01

Determine the forces acting on the particles

First, we express the forces acting on the particles A and B according to Newton's second law (\(F = m \cdot a\)). For particle A, \(F_A = m a_A\) and for Particle B, \(F_B=\frac{m}{4}a_B\). The force of attraction between the particles is given by \(F = -ar\). Therefore, we can write the forces acting on A and B as follows: \(F_A = -ar\) \(F_B = ar\)
02

Determine the accelerations for particles

Now using the force equations, we can express the accelerations \(a_A\) and \(a_B\) in terms of \(r\) and \(m\): \(a_A = -\frac{ar}{m}\) \(a_B = \frac{4ar}{m}\)
03

Combine the accelerations to find the relative acceleration

To find the relative acceleration between the particles, we can subtract the accelerations calculated in Step 2: \(a_{rel} = a_B - a_A = \frac{4ar}{m} - (-\frac{ar}{m}) = \frac{5ar}{m}\)
04

Use the relative acceleration to find the relative velocity

Since the particles are initially at rest, we can use the following equation to find the relative velocity \(v_{rel}\) by integrating the relative acceleration with respect to time: \(v_{rel} = \int a_{rel} dt = \int \frac{5ar}{m} dt = \frac{5a}{m} \int r dt\) Next, we must find the function \(r(t)\), which represents the distance between the particles as a function of time.
05

Calculate r(t) as a function of time

To find \(r(t)\), we can use the constraint that the sum of the displacements of both particles (from their initial positions) is always equal to \(r\): \(x_A + x_B = r\) We know that \(x_A=\int v_A dt\) and \(x_B=\int v_B dt\). So, the equation becomes: \(\int v_A dt + \int v_B dt = r\) Now, we can differentiate this equation with respect to time to find \(r(t)\): \(\frac{d}{dt}(\int v_A dt) + \frac{d}{dt}(\int v_B dt) = \frac{dr}{dt}\) \(v_A+v_B=\frac{dr}{dt}\) Substituting the expressions of \(v_A\) and \(v_B\) in terms of \(a_A\) and \(a_B\), we get: \(-\frac{ar}{m}x_A + \frac{4ar}{m}x_B = \frac{dr}{dt}\) Solving this differential equation will give us \(r(t)\) as a function of time.
06

Find when particles collide

The particles collide when their distance equals zero: \(r(t) = 0\) By solving the equation for \(r(t)\) for when it is equal to zero, we find the time at which the particles collide.
07

Calculate the relative velocity at the moment of collision

Now, substitute the time of collision back into the expression for relative velocity \(v_{rel}\). \(v_{rel}(t) = \frac{5a}{m} \int r(t) dt\) Evaluate the integral at the time of collision and this will give us the relative velocity between the two particles at the moment the collision occurs.

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Key Concepts

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

Force of Attraction
When two particles exert a force on each other, this force is often referred to as a force of attraction. In our specific scenario, the particles are attracted by a force given by the expression \( F = -ar \). The negative sign indicates that the force is attractive, pulling the particles towards one another. Here, \( a \) is a constant coefficient and \( r \) represents the distance between the particles. This relationship reflects Newton's law of universal gravitation, where attractive forces decrease with distance but remain a crucial part of particle interactions.

To better understand:
  • The force's magnitude increases as the distance decreases.
  • The nature of the force ensures both particles move towards each other.
  • Understanding these forces helps us predict how particles behave over time.
A clear grasp of attraction forces is essential for analyzing particle dynamics and motion under various conditions.
Relative Velocity
Relative velocity describes how quickly two objects are moving towards or away from each other. In this exercise, we have two particles moving due to their mutual attraction. At the point of collision, their relative velocity plays a key role.

Here鈥檚 how it unfolds:
  • Initially, both particles are at rest. Forces then cause acceleration.
  • This eventually leads to velocity as they converge.
  • The relative velocity \( v_{rel} \) at collision is derived by integrating relative acceleration over time.
To find \( v_{rel} \), we use:\[ v_{rel} = \int a_{rel} \, dt = \int \frac{5ar}{m} \, dt \]Understanding relative velocity allows us to appreciate how their movements relate to one another in the context of collision dynamics.
Particle Dynamics
Particle dynamics involves studying how forces affect particle motion. In our context, the particles' paths, velocities, and accelerations are all intertwined.

Let's break down the dynamics:
  • Newton鈥檚 second law helps explain the accelerations \( a_A = -\frac{ar}{m} \) and \( a_B = \frac{4ar}{m} \).
  • Their relative acceleration is \( a_{rel} = \frac{5ar}{m} \).
  • By integrating acceleration, we can determine velocities and positions over time.
Combining these elements:- We can predict when and where the particles will collide.- Particle dynamics provides a framework for understanding complex motion patterns in a straightforward way.

Real-world applications of these principles are abundant, from understanding atomic interactions to celestial body movements.

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Most popular questions from this chapter

Round fruits like oranges and mandarins are typically stacked in alternating rows, as shown in figure \(2.9 .\) Suppose you have a crate with a square base that is exactly five oranges wide. You stack 25 oranges in the crate, then put another 16 on top in the holes, and then add a second layer of 25 oranges, held in place by the sides of the crate. Find the total force on the sides of the crate in this configuration. Assume all oranges are spheres with a diameter of \(8.0 \mathrm{~cm}\) and a mass of \(250 \mathrm{~g}\).

When you cook rice, some of the dry grains always stick to the measuring cup. A common way to get them out is to turn the measuring cup upside-down and hit the bottom (now on top) with your hand so that the grains come off \([32]\). (a) Explain why static friction is irrelevant here. (b) Explain why gravity is negligible. (c) Explain why hitting the cup works, and why its success depends on hitting the cup hard enough.

Objects with densities less than that of water float, and even objects that have higher densities are 'lighter' in the water. The force that's responsible for this is known as the buoyancy force, which is equal but opposite to the gravitational force on the displaced water: \(F_{\text {buoyancy }}=\rho_{\mathrm{w}} g V_{\mathrm{w}}\), where \(\rho_{\mathrm{w}}\) is the water's density and \(V_{\mathrm{w}}\) the displaced volume. In parts (a) and (b), we consider a block of wood with density \(\rho<\rho_{\mathrm{w}}\) which is floating in water. (a) Which fraction of the block of wood is submerged when floating? (b) You push down the block somewhat more by hand, then let go. The block then oscillates on the surface of the water. Explain why, and calculate the frequency of the oscillation. (c) You take out the piece of wood, and now float a piece of ice in a bucket of water. On top of the ice, you place a small stone. When everything has stopped moving, you mark the water level. Then you wait till the ice has melted, and the stone has dropped to the bottom of the bucket. What has happened to the water level? Explain your answer (you can do so either qualitatively through an argument or quantitatively through a calculation). (d) Rubber ducks also float, but, despite the fact that they have a flat bottom, they usually do not stay upright in water. Explain why. (e) You drop a \(5.0 \mathrm{~kg}\) ball with a radius of \(10 \mathrm{~cm}\) and a drag coefficient \(c_{d}\) of \(0.20\) in water (viscosity \(1.002 \mathrm{mPa} \cdot \mathrm{s})\). This ball has a density higher than that of water, so it sinks. After a while, it reaches a constant velocity, known as its terminal velocity. What is the value of this terminal velocity? (f) When the ball in (e) has reached terminal velocity, what is the value of its Reynolds number (see problem 1.3)?

Two blocks, of mass \(m\) and \(2 m\), are connected by a massless string and slide down an inclined plane at angle \(\theta\). The coefficient of kinetic friction between the lighter block and the plane is \(\mu\), and that between the heavier block and the plane is \(2 \mu\). The lighter block leads. (a) Find the magnitude of the acceleration of the blocks. (b) Find the tension in the taut string.

A uniform rod with a length of \(4.25 \mathrm{~m}\) and a mass of \(47.0 \mathrm{~kg}\) is attached to a wall with a hinge at one end. The rod is held in horizontal position by a wire attached to its other end. The wire makes an angle of \(30.0^{\circ}\) with the horizontal, and is bolted to the wall directly above the hinge. If the wire can support a maximum tension of \(1250 \mathrm{~N}\) before breaking, how far from the wall can a \(75.0 \mathrm{~kg}\) person sit without breaking the wire?
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