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(a) If an electron is projected horizontally with a speed of \(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s},\) how far will it fall in traversing \(1.0 \mathrm{~m}\) of horizontal distance? (b) Does the answer increase or decrease if the initial speed is increased?

Short Answer

Expert verified
(a) The electron falls approximately \(5.44 \times 10^{-13} \, \text{m}\). (b) The fall distance decreases with increased initial speed.

Step by step solution

01

Determine Time of Flight

To determine how long the electron is in motion, we use the horizontal motion equation. The horizontal speed is given as \(v = 3.0 \times 10^{6} \, \text{m/s}\) and the horizontal distance \(d = 1.0 \, \text{m}\). The time \(t\) can be found using \(t = \frac{d}{v}\). Substituting the values, we have \(t = \frac{1.0}{3.0 \times 10^{6}}\approx 3.33 \times 10^{-7} \, \text{s}\).
02

Calculate Vertical Displacement

The electron falls under gravity. The formula for vertical displacement \(s\) is given by \(s = \frac{1}{2} g t^{2}\), where \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)). Using the time from Step 1, we have \(s = \frac{1}{2} \times 9.8 \times (3.33 \times 10^{-7})^2\approx 5.44 \times 10^{-13} \, \text{m}\).
03

Analyze the Effect of Speed on Fall Distance

To find out if increasing the horizontal speed affects the fall distance, observe that the vertical displacement \(s\) relies on the time \(t\), which is the inverse of the horizontal speed. As the speed increases, \(t\) decreases, thereby reducing the vertical displacement \(s\). Hence, increasing the initial speed decreases the fall distance.

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Key Concepts

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

Horizontal Motion
When considering projectile motion, it's important to differentiate between horizontal and vertical components. Horizontal motion is defined by a constant velocity, as there is no horizontal acceleration (assuming air resistance is negligible). In our scenario, the electron is projected with a speed of \(3.0 \times 10^{6} \, \text{m/s}\) and needs to cover a horizontal distance of \(1.0\, \text{m}\). To find out how long this takes, we use the formula for constant velocity:
  • \( t = \frac{d}{v} \)
where:
  • \(d\) is the horizontal distance
  • \(v\) is the horizontal velocity
Plugging in the values, we calculate:
  • \( t = \frac{1.0}{3.0 \times 10^{6}} \approx 3.33 \times 10^{-7} \, \text{s}\)
This very short time frame highlights the electron's rapid movement across the horizontal plane.
Vertical Displacement
While the electron moves horizontally, it's also affected by vertical forces, particularly due to gravity. Vertical displacement in projectile motion is calculated using the equation:
  • \( s = \frac{1}{2} g t^2 \)
where:
  • \(s\) is the vertical displacement
  • \(g\) is the acceleration due to gravity, typically \(9.8 \, \text{m/s}^2\)
  • \(t\) is the time of flight, which we've already calculated as \(3.33 \times 10^{-7} \, \text{s}\)
Substituting these values, we compute:
  • \( s = \frac{1}{2} \times 9.8 \times (3.33 \times 10^{-7})^2 \approx 5.44 \times 10^{-13} \, \text{m}\)
This result shows the infinitesimal fall the electron experiences over such a brief span.
Acceleration due to Gravity
A key factor impacting vertical motion is gravity, which consistently acts on objects in projectile motion. In this context, the acceleration due to gravity, denoted as \(g\), is \(9.8 \, \text{m/s}^2\). This force is what causes the electron to fall as it travels horizontally. Importantly, while the horizontal velocity influences the time duration for which gravity acts, gravity itself does not alter horizontal motion.
Increasing the initial horizontal speed decreases the time \(t\), as seen with the inverse relationship in the formula \(t = \frac{d}{v}\). Consequently, with a shorter \(t\), the vertical displacement \(s\) becomes smaller since:
  • \( s = \frac{1}{2} g t^2 \)
This understanding clarifies that while gravity is a vertical force, its effect is indirectly felt in how quickly horizontally moving objects like the electron traverse their path.

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Most popular questions from this chapter

A frightened rabbit moving at \(6.00 \mathrm{~m} / \mathrm{s}\) due east runs onto a large area of level ice of negligible friction. As the rabbit slides across the ice, the force of the wind causes it to have a constant acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2},\) due north. Choose a coordinate system with the origin at the rabbit's initial position on the ice and the positive \(x\) axis directed toward the east. In unit-vector notation, what are the rabbit's (a) velocity and (b) position when it has slid for 3.00 s?

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