Problem 115 - Fundamentals of Physics

Question

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.

Step-by-Step Solution

Answer

(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 1: 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 \(2.00 \times 10^{-9} \text{ s}.\)

Step 2: 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 \(2.00 \text{ cm}.\)

Step 3: 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 \(v_{x0}\): \[v_{x} = 1.00 \times 10^9 \text{ cm/s}.\]

Step 4: 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 \(2.00 \times 10^8 \text{ cm/s}.\)

Key Concepts

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.