Problem 24 - Thomas Calculus

Question

Beaming electrons An electron in a TV tube is beamed hori- zontally at a speed of \(5 \times 10^{6} \mathrm{m} / \mathrm{sec}\) toward the face of the tube 40 \(\mathrm{cm}\) away, About how far will the electron drop before it hits?

Step-by-Step Solution

Answer

The electron drops approximately \(3.14 \times 10^{-15} \ \mathrm{m}\).

Step 1: Identify Known Values

First, identify the known values given in the problem: the horizontal speed of the electron is \(v = 5 \times 10^{6} \ \mathrm{m/s}\), and the distance to the face of the tube is \(d = 0.40 \ \mathrm{m}\). The acceleration due to gravity is \(g = 9.81 \ \mathrm{m/s^2}\).

Step 2: Calculate Time of Travel

Using the formula for constant speed, \(\text{time} = \frac{\text{distance}}{\text{speed}}\), calculate the time it takes for the electron to reach the face of the tube.\[ t = \frac{0.40 \ \mathrm{m}}{5 \times 10^{6} \ \mathrm{m/s}} \approx 8 \times 10^{-8} \ \mathrm{s}\]

Step 3: Determine Vertical Displacement

The vertical displacement, which is the distance the electron falls, can be calculated using the formula for free-fall motion: \(s = \frac{1}{2}gt^2\). Substituting in the known values, we get: \[ s = \frac{1}{2} \times 9.81 \ \mathrm{m/s^2} \times (8 \times 10^{-8} \ \mathrm{s})^2 \approx 3.14 \times 10^{-15} \ \mathrm{m}\]

Step 4: Conclusion

The electron drops minimally due to its high horizontal speed and the short distance it travels before hitting the face of the tube, leading to an almost negligible vertical displacement.

Key Concepts

Horizontal Motion

Horizontal motion in physics refers to the movement of an object in a straight line across a horizontal plane, without any vertical influence initially. In the given exercise, the electron is moving horizontally towards the screen at a high speed of \(5 \times 10^6\ \text{m/s}\). This constant horizontal velocity indicates that there are no horizontal forces acting on the electron, allowing it to maintain its speed across the distance.
- Horizontal motion remains consistent unless an external force is applied.- The lack of horizontal forces results in a straightforward calculation of time using speed and distance.
In the exercise, the simplicity of horizontal motion lets us ignore any horizontal forces, thus focusing entirely on calculating how far the electron travels over time. Understanding horizontal motion is crucial because it allows us to predict the electron's path without complicated calculations, simplifying our understanding of the overall problem.

Vertical Displacement

Vertical displacement refers to the change in height or position of an object in the vertical direction as it moves. As the electron travels horizontally, gravity acts on it, causing a slight downward motion. This is parallel to the concept of free fall.
The formula for vertical displacement in free fall is given by:\[ s = \frac{1}{2}gt^2 \]where:

• \( s \) is the vertical displacement.
• \( g \) is the acceleration due to gravity, \(9.81 \ \text{m/s}^2\).
• \( t \) is time.

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In our exercise:- Calculating the time it takes for the electron to reach its destination allows us to apply the formula to find its vertical displacement.- Although calculated to be approximately \(3.14 \times 10^{-15}\ \text{m}\), which is almost negligible, it underscores the role of gravity acting on a horizontally traveling object.
Thus, vertical displacement illustrates that even in high-speed horizontal motion, external forces like gravity can indeed cause a small vertical movement.

Constant Speed

Constant speed is a fundamental concept where an object's speed remains unchanged along its path. In such motion, distance covered is directly proportional to time taken, simplifying calculations remarkably.
Constant speed formulas are straightforward: the overall distance travelled (\( d \)) can be easily calculated by multiplying speed (\( v \)) by time (\( t \)), or vice versa in the case of calculating time using distance and speed:\[ t = \frac{d}{v} \]With the exercise focusing on an electron moving at a constant speed of \(5 \times 10^6\ \text{m/s}\), the task of determining the time taken to cover the horizontal distance is simplified:

• The time (\( t \)) is computed as \( 8 \times 10^{-8}\ \text{s}\).

Constant speed allows conditions to predict future positions using simple arithmetic without focusing on force dynamics.