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A proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) passes through a metal film of thickness \(0.010 \mathrm{~mm}\) and emerges with a speed of \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). How large an average force opposed its motion through the film?

Short Answer

Expert verified
The average force opposing the proton's motion is \(-1.75 \times 10^{-9} \text{ N}\).

Step by step solution

01

Identify the known variables

The mass of the proton is given by \( m = 1.67 \times 10^{-27} \text{ kg} \). The initial velocity \( v_i \) is \( 5.0 \times 10^6 \text{ m/s} \), and the final velocity \( v_f \) is \( 2.0 \times 10^6 \text{ m/s} \). The thickness of the film is \( 0.010 \text{ mm} = 10^{-5} \text{ m} \).
02

Calculate the change in velocity (Δv)

The change in velocity \( \Delta v \) can be calculated as: \[ \Delta v = v_f - v_i = (2.0 \times 10^6) - (5.0 \times 10^6) = -3.0 \times 10^6 \text{ m/s} \]
03

Use kinetic theory to find acceleration (a)

The acceleration \( a \) is given by: \[ a = \frac{\Delta v}{\Delta t} \]First, find the time \( \Delta t \) using the equation \( \Delta t = \frac{\Delta x}{\bar{v}} \), where \( \Delta x \) is the thickness of the film and \( \bar{v} \) is the average velocity.\[ \bar{v} = \frac{v_i + v_f}{2} = \frac{5.0 \times 10^6 + 2.0 \times 10^6}{2} = 3.5 \times 10^6 \text{ m/s} \]\[ \Delta t = \frac{10^{-5}}{3.5 \times 10^6} = 2.857 \times 10^{-12} \text{ s} \]Now, substitute \( \Delta v \) and \( \Delta t \) to find \( a \):\[ a = \frac{-3.0 \times 10^6}{2.857 \times 10^{-12}} = -1.05 \times 10^{18} \text{ m/s}^2 \]
04

Calculate the average force (F)

Using Newton's second law, the average force \( F \) is calculated as: \[ F = m \times a \]Substitute the values for \( m \) and \( a \):\[ F = (1.67 \times 10^{-27} \text{ kg})(-1.05 \times 10^{18} \text{ m/s}^2) = -1.7535 \times 10^{-9} \text{ N} \]The negative sign indicates the force is opposing the motion.

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Key Concepts

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

Proton Physics
Protons are fundamental particles found in the nucleus of an atom. They have a positive charge, and their mass is incredibly small compared to the objects we interact with daily. A single proton has a mass of approximately \(1.67 \times 10^{-27} \ \mathrm{kg}\). Understanding protons is crucial in both physics and chemistry because they contribute to the atomic number, which defines the element.

In high-energy physics, protons are often accelerated to high speeds in particle accelerators to investigate their properties and interactions with other particles. The study of protons helps us comprehend the forces that hold atomic nuclei together, such as the electromagnetic force and the strong nuclear force.

Protons are also essential in nuclear reactions, such as those occurring in the sun, where hydrogen nuclei (protons) fuse to create helium, releasing vast amounts of energy. Classroom experiments often involve protons passing through materials to study their behavior and the forces they experience, as seen in problems involving metal films or other obstacles.
Acceleration Calculation
Acceleration is a fundamental concept in physics that describes how the velocity of an object changes over time. It can be caused by forces acting on the object, such as gravity, friction, or in the case of a proton moving through a metal film, resistance forces within the material.
  • The basic formula for acceleration, \( a \), is given by \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.
  • For small particles like protons, these changes happen over very short time intervals and distances, often requiring precise measurements.
In the example of a proton slowing down as it passes through a film, the acceleration is negative, indicating a deceleration due to the force opposing its motion.

To find the precise value, one must first calculate the average velocity \( \bar{v} \) used to determine \( \Delta t \) or the time taken for the proton to pass through the film. By knowing the thickness of the film and the average speed of the proton, we can calculate \( \Delta t \). This value is crucial for computing the acceleration and understanding the forces involved.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. It focuses on concepts like velocity, acceleration, and displacement. This topic is crucial for understanding the motion of particles, such as protons.
  • In kinematics, we often begin by identifying initial and final velocities, as well as the distance traveled, which in our example, is the thickness of the metal film a proton passes through.
  • By using these values, we can calculate the change in velocity \( \Delta v \) and subsequently, the acceleration, as these are critical in predicting future motion.
Kinematics also includes the study of average speed, which simplifies the analysis by providing a constant value for speed over a specified distance and time, making it easier to compute time intervals or accelerations.

Understanding kinematics allows physicists to predict the motion of particles under various conditions or constraints, crucial for experiments involving small particles like protons. Such understanding is foundational to higher levels of physics, such as dynamics, which incorporates the forces that cause the motion.

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Most popular questions from this chapter

A 10.0-kg block is launched up a \(30.0^{\circ}\) inclined plane at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). As it slides it loses \(200 \mathrm{~J}\) to friction. How far along the incline will it travel before coming to rest?

A \(0.50\) -kg ball falls past a window that is \(1.50 \mathrm{~m}\) in vertical length. (a) How much did the KE of the ball increase as it fell past the window? \((b)\) If its speed was \(3.0 \mathrm{~m} / \mathrm{s}\) at the top of the window, what was its speed at the bottom?

A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

How much work is done on an object by the force that supports it as the object is lowered at a constant speed through a vertical distance \(h\) ? How much work does the gravitational force on the object do in this same process? The supporting force is \(m g\), where \(m\) is the mass of the object. It is directed upward while the displacement is downward. Hence, the work it does is negative: $$ F \cos \theta=(m g)(h)\left(\cos 180^{\circ}\right)=-m g h $$ The force of gravity acting on the object is also \(m g\), but it is directed downward in the same direction as the displacement. The work done on the object by the force of gravity is therefore positive: $$ F \operatorname{scos} \theta=(m g)(h)\left(\cos 0^{\circ}\right)=m g h $$

Compute the power output of a machine that lifts a 500 -kg crate through a height of \(20.0 \mathrm{~m}\) in a time of \(60.0 \mathrm{~s}\).

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