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An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

Short Answer

Expert verified
(a) 2.00 x 10^-9 s, (b) 2.00 cm, (c) 1.00 x 10^9 cm/s, (d) 2.00 x 10^8 cm/s.

Step by step solution

01

Determine the Time of Travel

To find the time it takes for the electron to travel the horizontal distance of \(2.00 \text{ cm}\), we can use the formula for constant speed: \( \text{speed} = \frac{\text{distance}}{\text{time}} \). The initial horizontal velocity \(v_{x0}\) is given as \(1.00 \times 10^9 \text{ cm/s}\). Thus,\[\text{time} = \frac{\text{distance}}{v_{x0}} = \frac{2.00 \text{ cm}}{1.00 \times 10^9 \text{ cm/s}} = 2.00 \times 10^{-9} \text{ s}.\] So, the time taken is \(2.00 \times 10^{-9} \text{ s}.\)
02

Calculate the Vertical Distance Traveled

The vertical motion is uniformly accelerated, so we use the kinematic equation: \(s_y = v_{y0}t + \frac{1}{2}a_y t^2\). The initial vertical velocity \(v_{y0}\) is zero (since the electron initially only moves horizontally), and \(a_y = 1.00 \times 10^{17} \text{ cm/s}^2\). Substituting the values, \[s_y = 0 \times 2.00 \times 10^{-9} + \frac{1}{2} \times 1.00 \times 10^{17} \times (2.00 \times 10^{-9})^2\] \[s_y = \frac{1}{2} \times 1.00 \times 10^{17} \times 4.00 \times 10^{-18} = 2.00 \text{ cm}.\] So, the vertical distance traveled is \(2.00 \text{ cm}.\)
03

Determine the Final Horizontal Velocity

The horizontal velocity of the electron remains constant because there is no horizontal acceleration. Therefore, the final horizontal velocity \(v_{x}\) is the same as the initial horizontal velocity \(v_{x0}\): \[v_{x} = 1.00 \times 10^9 \text{ cm/s}.\]
04

Determine the Final Vertical Velocity

Using the formula for final velocity under uniform acceleration, \(v_{yf} = v_{y0} + a_y t\), the initial vertical velocity \(v_{y0} = 0\) and \(a_y = 1.00 \times 10^{17} \text{ cm/s}^2\). So,\[v_{yf} = 0 + 1.00 \times 10^{17} \times 2.00 \times 10^{-9} = 2.00 \times 10^8 \text{ cm/s}.\] Therefore, the final vertical velocity is \(2.00 \times 10^8 \text{ cm/s}.\)

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Key Concepts

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

Uniform Acceleration
In kinematics, uniform acceleration means an object's acceleration remains constant over time. It's a key idea when analyzing motion under the influence of constant forces, such as gravity or in this case, the electric force acting between charged plates.

Uniform acceleration is particularly useful for understanding how an object's velocity and position change. When we deal with uniform acceleration, we often use the kinematic equations:

  • To find the displacement: \( s = v_0 t + \frac{1}{2} a t^2 \)
  • To find the final velocity: \( v = v_0 + a t \)
Here, \( v_0 \) is the initial velocity, \( a \) is the acceleration, \( t \) is the time, and \( s \) is the displacement.

In the given problem, the electron experiences a constant downward acceleration of \(1.00 \times 10^{17} \text{ cm/s}^2\), which allows us to apply these equations to determine how far it travels vertically and how its velocity changes over time.
Horizontal and Vertical Motion
Objects that move along a path affected by forces can have both horizontal and vertical components of motion. Understanding these two components separately helps in solving complex motion problems.

Horizontal motion often involves constant velocity, meaning no horizontal force is acting on the object. This can be derived from the formula: \[\text{Velocity} = \frac{\text{Distance}}{\text{Time}}.\]

Vertical motion usually includes acceleration due to forces like gravity or electric fields. We calculate this using kinematic equations for uniformly accelerated motion.

In the exercise, the electron maintains a constant horizontal velocity (\(1.00 \times 10^9\text{ cm/s}\)) while facing downward acceleration from the charged plates, which affect its vertical movement. By calculating separately, we can find how far it goes horizontally without any change in speed and how much its path curves due to vertical forces.
Velocity Components
Velocity components are essential for breaking down an object's overall movement into manageable parts. We usually split velocity into horizontal and vertical components, especially when analyzing projectiles or objects in motion under multiple forces.

In this exercise, the electron starts with a significant horizontal velocity (\(1.00 \times 10^9\text{ cm/s}\)) and no initial vertical velocity. The lack of horizontal acceleration means the horizontal velocity remains constant.

Vertical velocity, however, changes due to uniform acceleration. By the formula \( v = v_0 + a t \), where the initial vertical velocity \( v_0 \) is zero, we calculate the vertical velocity component by multiplying acceleration by time (\(2.00 \times 10^{-9}\text{ s}\) in this case). The result yields the final vertical velocity of \(2.00 \times 10^8\text{ cm/s}\), showing how differently each component behaves under various forces.

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Most popular questions from this chapter

You are kidnapped by political-science majors (who are upset because you told them political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction of travel are you?

A woman can row a boat at \(6.40 \mathrm{~km} / \mathrm{h}\) in still water. (a) If she is crossing a river where the current is \(3.20 \mathrm{~km} / \mathrm{h}\), in what direction must her boat be headed if she wants to reach a point directly opposite her starting point? (b) If the river is \(6.40 \mathrm{~km}\) wide, how long will she take to cross the river? (c) Suppose that instead of crossing the river she rows \(3.20 \mathrm{~km}\) down the river and then back to her starting point. How long will she take? (d) How long will she take to row \(3.20 \mathrm{~km} u p\) the river and then back to her starting point? (e) In what direction should she head the boat if she wants to cross in the shortest possible time, and what is that time?

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

A radar station detects an airplane approaching directly from the east. At first observation, the airplane is at distance \(d_{1}=360 \mathrm{~m}\) from the station and at angle \(\theta_{1}=40^{\circ}\) above the horizon (Fig. \(4-49\) ). The airplane is tracked through an angular change \(\Delta \theta=123^{\circ}\) in the vertical east-west plane; its distance is then \(d_{2}=790 \mathrm{~m}\). Find the (a) magnitude and (b) direction of the airplane's displacement during this period.

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flagpole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0 \mathrm{~s}\) later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0\) s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration?

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