/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 An object collides elastically w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 72

An object collides elastically with an equal-mass object initially at rest. If the collision isn't head-on, show that the final velocity vectors are perpendicular.

Short Answer

Expert verified
The final velocity vectors are perpendicular when an elastic collision occurs between two equal-mass objects not in a direct head-on collision. This is due to the conservation laws of momentum and kinetic energy, which follow the pattern of the Pythagorean theorem indicating perpendicularly.

Step by step solution

01

Setup

Let's denote the initial velocity of the moving object as \(v_1\) and let's set the initial velocity of the stationary object as \(v_2=0\). We can represent the final velocities after the collision as \(v_1'\) for the initial moving object and \(v_2'\) for the initial stationary object.
02

Applying Conservation of Momentum

Because the system experiences an elastic collision, the principle of conservation of momentum applies here. This principle can be expressed by the equation: \(mv_1+mv_2 = mv_1'+mv_2'\). Given \(v_2=0\) as the initial condition, the equation simplifies to \(v_1 = v_1' + v_2'\).
03

Applying Conservation of Kinetic Energy

Since this collision is elastic, the conservation of kinetic energy also applies. We can express this principle as: \(\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}mv_1'^2+\frac{1}{2}mv_2'^2\). Given \(v_2=0\) as the initial condition, the equation simplifies to \(v_1^2 = v_1'^2 + v_2'^2\).
04

Comparing the Two Equations

Upon comparing the equations obtained from Steps 2 and 3, we notice that this follows the pattern of the Pythagorean theorem, where \(v_1\) is the hypotenuse and \(v_1'\) and \(v_2'\) are the remaining sides of a right-angled triangle; hence the vectors \(v_1'\) and \(v_2'\) are perpendicular.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Model rocket motors are specified by giving the impulse they "provide, in \(\mathrm{N} \cdot \mathrm{s}\), over the entire time the rocket is firing. The table below shows the results of rocket-motor tests with different motors used to launch rockets of different masses. Determine two data-based quantities that, when plotted against each other, should give a straight line and whose slope should allow you to determine \(g .\) Plot the data, establish a best-fit line, and determine \(g\). Assume that the maximum height is much greater than the distance over which the rocket motor is firing, so you can neglect the latter. You're also neglecting air resistance- -but explain how that affects your experimentally determined value for \(g\). $$\begin{array}{|l|r|r|r|r|r|} \hline \text { Impulse, } J(\mathrm{N} \cdot \mathrm{s}) & 4.5 & 7.8 & 4.5 & 7.8 & 11 \\ \hline \begin{array}{l}\text { Rocket mass }(\mathrm{g}) \\ \text { (including motor) }\end{array} & 180 & 485 & 234 & 234 & 485 \\ \hline \begin{array}{l}\text { Maximum height } \\\\\text { achieved (m) }\end{array} & 22 & 13 & 19 & 51 & 23 \\\\\hline\end{array}$$

A proton (mass 1 u) moving at 6.90 Mm/s collides elastically head-on with a second particle moving in the opposite direction at \(2.80 \mathrm{Mm} / \mathrm{s}\). After the collision, the proton is moving opposite its initial direction at \(8.62 \mathrm{Mm} / \mathrm{s}\). Find the mass and final velocity of the second particle.

Wildlife biologists fire \(20-\mathrm{g}\) rubber bullets to stop a rhinoceros charging at \(0.81 \mathrm{m} / \mathrm{s} .\) The bullets strike the rhino and drop vertically to the ground. The biologists' gun fires 15 bullets each second, at \(73 \mathrm{m} / \mathrm{s}\), and it takes \(34 \mathrm{s}\) to stop the rhino. (a) What impulse does each bullet deliver? (b) What's the rhino's mass? Neglect forces between rhino and ground.

The momentum of a system of pool balls is the same before and after they are hit by the cue ball. Is it still the same after one of the balls strikes the edge of the table? Explain.

Roughly where is your center of mass when you're standing?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Atoms and Radioactivity

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation

Thermodynamics

Read Explanation

Translational Dynamics

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.