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Chapter 6: Problem 103

A constant force \(\vec{F}\) acts on an electron for 3.0 s, changing its velocity from \(6.2 \times 10^{6} \mathrm{~m} / \mathrm{s} \hat{\imath}-5.8 \times 10^{5} \mathrm{~m} / \mathrm{s} \hat{\jmath}\) to \(-3.7 \times\) \(10^{6} \mathrm{~m} / \mathrm{s} \hat{\imath}-15.8 \times 10^{6} \mathrm{~m} / \mathrm{s} \hat{\jmath}\). Find \(\vec{F}\)

Short Answer

Expert verified
The force is \(-3.0 \times 10^{-24} \hat{\imath} - 4.6 \times 10^{-24} \hat{\jmath}\) N.

Step by step solution

01

Identify the Given Values

We know that the force \(\vec{F}\) acts on an electron for 3.0 s, changing its velocity. The initial velocity \(\vec{v}_i = 6.2 \times 10^{6} \hat{\imath} - 5.8 \times 10^{5} \hat{\jmath}\) m/s, and the final velocity \(\vec{v}_f = -3.7 \times 10^{6} \hat{\imath} - 15.8 \times 10^{6} \hat{\jmath}\) m/s. The mass of the electron is \(9.11 \times 10^{-31}\) kg.
02

Use the Formula for Force

The force can be calculated using the formula \(\vec{F} = m \frac{\Delta \vec{v}}{\Delta t}\), where \(\Delta \vec{v}\) is the change in velocity and \(\Delta t\) is the time period. We need to find \(\Delta \vec{v} = \vec{v}_f - \vec{v}_i\).
03

Calculate the Change in Velocity

Calculate \(\Delta \vec{v}\) by subtracting the initial velocity from the final velocity:\[ \Delta \vec{v} = \left(-3.7 \times 10^{6} \hat{\imath} - 15.8 \times 10^{6} \hat{\jmath}\right) - \left(6.2 \times 10^{6} \hat{\imath} - 5.8 \times 10^{5} \hat{\jmath}\right) \]Simplifying, we get:\[ \Delta \vec{v} = (-9.9 \times 10^{6}) \hat{\imath} - 15.2 \times 10^{6} \hat{\jmath} \]
04

Substitute Values into the Force Equation

Insert \(\Delta \vec{v}\), \(\Delta t = 3.0\,s\), and \(m = 9.11 \times 10^{-31}\,kg\) into the force equation:\[ \vec{F} = 9.11 \times 10^{-31} \frac{(-9.9 \times 10^{6}) \hat{\imath} - 15.2 \times 10^{6} \hat{\jmath}}{3.0} \]
05

Compute the Force

Calculate each component of the force separately:\[ F_{\hat{\imath}} = 9.11 \times 10^{-31} \times \frac{-9.9 \times 10^{6}}{3.0} = -3.0 \times 10^{-24} \mathrm{~N} \]\[ F_{\hat{\jmath}} = 9.11 \times 10^{-31} \times \frac{-15.2 \times 10^{6}}{3.0} = -4.6 \times 10^{-24} \mathrm{~N} \]
06

Express the Force as a Vector

Combine the components to express the force vector:\[ \vec{F} = (-3.0 \times 10^{-24} \hat{\imath} - 4.6 \times 10^{-24} \hat{\jmath})\,N \]

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Key Concepts

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

Constant Force
Constant force refers to a force that remains unchanged over time, regardless of the conditions or changes in motion that occur. In physics, a constant force is vital for understanding motion, as described by Newton's second law. **Key Characteristics of Constant Force**
  • The magnitude and direction of the force do not vary with time.
  • It results in uniform acceleration when acting on an object with a constant mass.
  • This kind of force often simplifies complex problems, making them easier to analyze and solve.
In the context of the exercise, the constant force acts on an electron for a duration of three seconds, changing its velocity. This indicates that the force was stable throughout the interaction period, which is essential for using standard kinematic equations to solve the problem. Without constant force, the calculations could become significantly more complex, requiring advanced calculus or computational methods.
Change in Velocity
Change in velocity is a measure of how the speed and direction of an object's movement differ over time. This concept is fundamental for understanding acceleration and is calculated as the difference between the final and initial velocities.**Calculating Change in Velocity**To find the change in velocity (\( \Delta \vec{v} \)), use the formula:\[\Delta \vec{v} = \vec{v}_f - \vec{v}_i\]Where:
  • \( \vec{v}_f \) is the final velocity.
  • \( \vec{v}_i \) is the initial velocity.
By determining the change in velocity, one can calculate acceleration鈥攁 stepping stone to understanding forces acting on an object, such as the force exerted on an electron as in the exercise. In this particular scenario, the change in velocity provides insight into how the electron's path was altered by the exerted force.
Electron Dynamics
Electron dynamics deals with the motion and forces on electrons, the tiny charged particles essential for chemical reactions and electricity. Understanding electron dynamics is crucial for areas such as electronics, electromagnetism, and chemistry.**Key Aspects of Electron Dynamics**
  • Electrons have a very small mass: \(9.11 \times 10^{-31} \) kg, which means even tiny forces can result in significant accelerations.
  • Electrons carry a negative charge, so they are influenced by electric fields and other charged particles.
  • Their behavior can be described using Newtonian mechanics when speeds are not close to the speed of light.
In physical exercises, dynamics of an electron are examined under forces to predict their paths and effects. The exercise provided uses the force acting on an electron, along with its mass and change in velocity to determine the net force vector. This is a practical application of electron dynamics, involving calculations to predict how forces influence the electron's movement over time.

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Most popular questions from this chapter

A \(2.0-\mathrm{kg}\) ball and a \(3.0-\mathrm{kg}\) ball, each moving at \(0.90 \mathrm{~m} / \mathrm{s},\) undergo a head-on collision. The lighter ball rebounds opposite its initial direction, with speed \(0.90 \mathrm{~m} / \mathrm{s}\). (a) Find the post-collision velocity of the heavier ball. (b) How much mechanical energy was lost in this collision? Express your answer in \(\mathrm{J}\) and as a fraction of the system's initial mechanical energy.

Model rocket engines are characterized by the total impulse they deliver, measured in newton-seconds. (a) Show that \(1 \mathrm{~N} \cdot \mathrm{s}\) is the same as \(1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\), the unit we've used for impulse. (b) What speed can a \(7.5-\mathrm{N} \cdot \mathrm{s}\) engine give a rocket whose mass is \(140 \mathrm{~g}\) at the end of the engine firing?

Two skaters stand a short distance apart, facing each other on frictionless ice. Explain what happens when they begin tossing a ball back and forth.

A 2.64 -g bullet moving at \(280 \mathrm{~m} / \mathrm{s}\) enters and stops in an initially stationary \(2.10-\mathrm{kg}\) wooden block on a horizontal frictionless surface. (a) What's the speed of the bullet/block combination? (b) What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision? (c) How much work was done in stopping the bullet? (d) If the bullet penetrated \(5.00 \mathrm{~cm}\) into the wood, what was the average stopping force?

Consider an elastic head-on collision between two masses \(m_{1}\) and \(m_{2} .\) Their velocities before colliding are \(v_{1 x i} \hat{\imath}\) and \(v_{2 x i} \hat{\imath}\) afterward, they're \(v_{1 x f} \hat{\imath}\) and \(v_{2 x f} \hat{\imath}\). (a) Write equations expressing conservation of momentum and conservation of energy. (b) Use your equations to show that \(v_{1 x \mathrm{i}}-v_{2 x \mathrm{i}}=-\left(v_{1 x \mathrm{f}}-v_{2 x \mathrm{f}}\right) \cdot(\mathrm{c}) \mathrm{In}-\) terpret the results of part (b).
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