/*! 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 39 An electron of mass \(m\) is dis... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron of mass \(m\) is discharged with an initial horizontal velocity of \(\mathbf{v}_{0}\). If it is subjected to two fields of force for which \(F_{x}=F_{0}\) and \(F_{y}=0.3 F_{0}\), where \(F_{0}\) is constant, determine the equation of the path, and the speed of the electron at any time \(t\). Prob. 13- 39

Short Answer

Expert verified
The velocities at any time \(t\) are \(v_{x} = \mathbf{v}_{0} + (F_{0}/m) * t\) and \(v_{y} = 0.3 * (F_{0}/m) * t\). The speed of the electron at any time is \(v = \sqrt{v_{x}^2 + v_{y}^2}\). The equation of the path is obtained by expressing \(y\) in terms of \(x\) after integrating the velocity components to get displacement.

Step by step solution

01

Determine the acceleration due to forces

Acceleration is obtained by using Newton's second law of motion which states that force \(F\) is the product of mass \(m\) and acceleration \(a\). In this case, we have force components \(F_{x}=F_{0}\) and \(F_{y}=0.3 F_{0}\), implying acceleration components \(a_{x} = F_{x} / m = F_{0} / m\) and \(a_{y} = F_{y} / m = 0.3 F_{0} / m\).
02

Determine the Velocity at any time \(t\)

The velocity at any time can be calculated using the equation \(v = u + at\) where \(u\) is the initial velocity and \(a\) is the acceleration. Here, \(u\) along x-axis is given as \(\mathbf{v}_{0}\) and along y-axis is 0 (as no initial vertical velocity is mentioned). Hence, the velocities are \(v_{x} = \mathbf{v}_{0} + (F_{0}/m) * t\) and \(v_{y} = 0 + 0.3 * (F_{0}/m) * t\). As these are vector quantities, the total velocity at any time \(t\) will be the vector sum, which can be found using Pythagoras theorem, \(v = \sqrt{v_{x}^2 + v_{y}^2}\)
03

Determine the Equation of the Path

The equation of the path can be found by integrating the velocity components to get displacement, as \(x = \int v_{x} dt\) and \(y = \int v_{y} dt\). After integrating, we express \(y\) in terms of \(x\) to get the equation of the path.

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Key Concepts

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

Understanding Electron Motion
The motion of an electron can be fascinating when we consider its interaction with forces. In this scenario, an electron, which is a charged particle, is subjected to specific force fields. These forces cause it to move in a particular path, influenced by its initial conditions and the fields' characteristics. It's vital to first recognize that the electron has a mass denoted by \( m \) and starts with an initial horizontal velocity \( \mathbf{v}_{0} \).

When the electron enters a region with two force components, \( F_{x}=F_{0} \) and \( F_{y}=0.3 F_{0} \), its motion changes. Newton's Second Law is key here, linking the force acting on the electron to its acceleration. Understanding this relationship helps us determine how its velocity and path evolve over time.
Acceleration Components and Their Role
The electron's acceleration is directly influenced by the forces acting upon it. Due to Newton's Second Law, which tells us \( F = ma \), we can find out how the electron accelerates.
  • The horizontal force component \( F_x = F_0 \) gives rise to the horizontal acceleration \( a_x = \frac{F_0}{m} \).
  • The vertical force component \( F_y = 0.3 F_0 \) results in a vertical acceleration \( a_y = \frac{0.3 F_0}{m} \).

This split into different components helps understand how the electron speeds up or changes direction. Separating the force into \( x \) and \( y \) components makes complex motion more manageable, as it allows us to independently analyze and predict the electron's behavior along each axis.
Velocity Calculation at Any Time \( t \)
Calculating the velocity of the electron at any moment involves recognizing how it changes over time. Starting with its known initial velocity, we apply acceleration to discover how fast it's moving later on.
  • For horizontal motion, the velocity is \( v_{x} = \mathbf{v}_{0} + \left( \frac{F_0}{m} \right) t \).
  • Vertically, the velocity becomes \( v_{y} = 0 + 0.3 \left( \frac{F_0}{m} \right) t \).

These expressions allow us to calculate the velocities based on how long the electron has been moving. To find the actual speed of the electron, we combine these components to form the resultant velocity: \( v = \sqrt{v_{x}^2 + v_{y}^2} \). This illustrates how the different forces affect the electron's motion and helps predict its position at various time points.
Equation of Path: Mapping the Journey
To understand the path taken by the electron, we integrate its velocity components over time to find displacement.
  • By integrating the horizontal velocity \( v_{x} \), we find the horizontal position, \( x = \int v_{x} \, dt \).
  • Similarly, integrating \( v_{y} \) gives the vertical position, \( y = \int v_{y} \, dt \).

Once we have these positions, we can describe the electron's path by expressing \( y \) as a function of \( x \). It's a bit like connecting dots to form a curve. This helps visualize how the forces steer the electron across its course. By understanding this mapping, we can predict where the electron will be at any given point along its trajectory. It's a practical application of physics in mapping motion through equations.

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Most popular questions from this chapter

The 4 -kg smooth cylinder is supported by the spring having a stiffness of \(k_{A B}=120 \mathrm{~N} / \mathrm{m}\). Determine the velocity of the cylinder when it moves downward \(s=0.2 \mathrm{~m}\) from its equilibrium position, which is caused by the application of the force \(F=60 \mathrm{~N}\). Prob. 13-34

The \(300-\mathrm{kg}\) bar \(B\), originally at rest, is being towed over a series of small rollers. Determine the force in the cable when \(t=5 \mathrm{~s}\), if the motor \(M\) is drawing in the cable for a short time at a rate of \(v=\left(0.4 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds \((0 \leq t \leq 6 \mathrm{~s})\). How far does the bar move in \(5 \mathrm{~s}\) ? Neglect the mass of the cable, pulley, and the rollers. Prob. 13-38

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's speed is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\), directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3\), determine the speed at which each crate slides off the ramp at \(B\). Assume that no tipping occurs. Take \(\theta=30^{\circ}\).

Rod \(O A\) rotates counterclockwise at a constant angular rate \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s}\). The double collar \(B\) is pinconnected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation \(r=(1.6 \cos \theta) \mathrm{m}\). If both collars have a mass of \(0.5 \mathrm{~kg}\), determine the force which the circular rod exerts on one of the collars and the force that \(O A\) exerts on the other collar at the instant \(\theta=45^{\circ}\). Motion is in the horizontal plane.

A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is \(\mu_{s}=0.4\), determine the minimum speed at which he must travel if he is to ride along the wall when \(\theta=90^{\circ}\). The mass of the motorcycle and rider is \(250 \mathrm{~kg}\), and the radius of curvature to the center of gravity is \(\rho=20 \mathrm{ft}\). Neglect the size of the motorcycle for the calculation. Prob. 13-66

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