Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 24

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of \(64.0^{\circ}\) above its initial direction of motion and oxygen nucleus recoils at an angle of \(51.0^{\circ}\) below this initial direction. The final speed of the oxygen nucleus is \(1.20 \times 10^{5} \mathrm{~m} / \mathrm{s}\). What is the final speed of the alpha particle? (The mass of an alpha particle is \(4.00 \mathrm{u}\) and the mass of an oxygen nucleus is \(16.0 \mathrm{u}\).)

Short Answer

Expert verified
To get the exact numerical speed of the alpha partice after collision, calculate using the values from the information given in the exercise. The short answer can't be given without those exact values.

Step by step solution

01

Analyze the problem and diagram

Initially, only the alpha particle is moving and the oxygen is at rest. After the collision, both particles are moving at different angles. Draw the initial and final positions of the particles with their corresponding velocities and angles.
02

Apply conservation momentum in X-Direction

The total momentum along the x-axis before the collision is equal to the total momentum along the x-axis after the collision. In the x direction, the initial momentum is \(4.00u * v_{\alpha}\) (where \(v_{\alpha}\) is the initial speed of the alpha particle) and the final momentum is \(4.00u * v_{\alpha}'cos(64) - 16.0u * v_{O}'cos(51) \) (where \(v_{\alpha}'\) is the final speed of the alpha particle and \(v_{O}'\) is the speed of the oxygen nucleus). By substituting the values \(v_{O}' = 1.20 \times 10^{5} m/s\), and \(16.0u = 4 * 4.00u\), and simplifying, we obtain a relation between \(v_{\alpha}'\) and \(v_{\alpha}\).
03

Apply conservation momentum in Y-Direction

The total momentum along the y-axis before the collision is equal to the total momentum along the y-axis after the collision. In the y direction, the initial momentum is 0 (as both are moving horizontally initially) and the final momentum is \(4.00u * v_{\alpha}'sin(64) + 4.00u * v_{O}'sin(51)\). Simplifying this equation gives a another relation between \(v_{\alpha}'\) and \(v_{\alpha}\).
04

Solve the equations

Now we have two equations, one from step 2 and the other from step 3. Solve these equations to find the speed of the alpha particle after collision (\(v_{\alpha}'\)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pellet gun fires ten \(2.14\) -g pellets per second with a speed of \(483 \mathrm{~m} / \mathrm{s}\). The pellets are stopped by a rigid wall. (a) Find the momentum of each pellet. (b) Calculate the average force exerted by the stream of pellets on the wall. \((c)\) If each pellet is in contact with the wall for \(1.25 \mathrm{~ms}\), what is the average force exerted on the wall by each pellet while in contact? Why is this so different from \((b)\) ?

A \(2500-\mathrm{kg}\) unmanned space probe is moving in a straight line at a constant speed of \(300 \mathrm{~m} / \mathrm{s}\). A rocket engine on the space probe executes a burn in which a thrust of \(3000 \mathrm{~N}\) acts for \(65.0 \mathrm{~s}\). What is the change in momentum (magnitude only) of the probe if the thrust is backward, forward, or sideways? Assume that the mass of the ejected fuel is negligible compared to the mass of the space probe.

In the laboratory, a particle of mass \(3.16 \mathrm{~kg}\) moving at \(15.6 \mathrm{~m} / \mathrm{s}\) to the left collides head-on with a particle of mass \(2.84 \mathrm{~kg}\) moving at \(12.2 \mathrm{~m} / \mathrm{s}\) to the right. Find the velocity of the center of mass of the system of two particles after the collision.

How fast must an 816 -kg Volkswagen travel to have the same momentum as \((a)\) a \(2650-\mathrm{kg}\) Cadillac going \(16.0 \mathrm{~km} / \mathrm{h} ?(b)\) a 9080 -kg truck also going \(16.0 \mathrm{~km} / \mathrm{hr}\) ?

A 195-lb man standing on a surface of negligible friction kicks forward a \(0.158\) -lb stone lying at his feet so that it acquires a speed of \(12.7 \mathrm{ft} / \mathrm{s}\). What velocity does the man acquire as a result?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Solid State Physics

Read Explanation

Work Energy and Power

Read Explanation

Particle Model of Matter

Read Explanation

Oscillations

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free