/*! 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 49 Masses \(m\) and \(3 m\) approac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 49

Masses \(m\) and \(3 m\) approach at the same speed \(v\) and undergo a head-on elastic collision. Show that mass \(3 m\) stops, while mass \(m\) rebounds at speed \(2 v\)

Short Answer

Expert verified
After the elastic head-on collision, the larger mass \(3m\) comes to a stop (\(v'_2 = 0\)), and the smaller mass \(m\) rebounds at twice the initial speed (\(v'_1 = 2v\)).

Step by step solution

01

Identify given quantities and apply momentum conservation

We have two masses, \(m\) and \(3m\), moving towards each other with the same initial speed \(v\). Since there is no external force acting on the two masses, the total momentum is conserved before and after the collision. The momentum conservation equation looks like this: \(m \cdot v + 3m \cdot (-v) = m \cdot v'_1 + 3m \cdot v'_2\) where \(v'_1\) and \(v'_2\) are the final velocities of mass \(m\) and \(3m\) respectively.
02

Simplify the momentum conservation equation

Simplifying this equation gives us \(0 = v'_1 - 3v'_2 \). This means that \( v'_1 = 3v'_2 \).
03

Apply kinetic energy conservation

The second principle we apply is the conservation of kinetic energy for elastic collisions. The equation thus becomes: \(0.5m \cdot v^2 + 0.5 \cdot 3m \cdot v^2 = 0.5m \cdot {v'_1}^2 + 0.5 \cdot 3m \cdot {v'_2}^2\). This simplifies to \(2v = {v'_1}^2 + {v'_2}^2\).
04

Solve the system of equations

We now have two equations and two unknowns. We can substitute \( v'_1 = 3v'_2 \) into the second equation to solve for \(v'_2\). Upon solving, we find that \(v'_2 = 0 \), meaning the larger mass has stopped.
05

Substitute v'_2 to find v'_1

Substitute \(v'_2 = 0\) into the equation \( v'_1 = 3v'_2 \) to solve for \(v'_1\). This gives \(v'_1 = 2v\), meaning the smaller mass rebounds at twice the speed.

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!

Key Concepts

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

Momentum Conservation
When studying elastic collisions, understanding momentum conservation is crucial. Momentum, the product of an object's mass and velocity, is always conserved in isolated systems, such as two objects colliding in space without any external forces. This core principle establishes that the total momentum before an impact is equal to the total momentum after the impact, regardless of the objects' interactions with each other.
Kinetic Energy Conservation
Kinetic energy, which is the energy an object has due to its motion, is also conserved in elastic collisions. This conservation separates elastic collisions from inelastic ones, as only elastic collisions preserve both momentum and kinetic energy. In problem solving, this dual conservation allows us to create equations that must be simultaneously satisfied, leading to the specific outcomes of the collision, such as final velocities.
Elastic Collision Problem Solving
Elastic collision problem solving involves applying the conservation of momentum and kinetic energy to set up a system of equations. One must carefully derive and solve these equations to find the unknowns, such as the final speeds of the objects after the collision. Solving these problems typically involves algebraic manipulation and an understanding of how to apply physical laws to predict the outcomes of collisions.

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

How is it possible to have a collision between objects that don't ever touch? Give an example of such a collision.

You're working in mission control for an interplanetary space probe. A trajectory correction calls for a rocket firing that imparts an impulse of \(5.64 \mathrm{N} \cdot \mathrm{s}\). If the rocket's average thrust is \(135 \mathrm{mN}\), how long should the rocket fire?

Playing in the street, a child accidentally tosses a ball at \(18 \mathrm{m} / \mathrm{s}\) toward the front of a car moving toward him at \(14 \mathrm{m} / \mathrm{s}\). What's the ball's speed after it rebounds elastically from the car?

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 \(g\). $$\begin{array}{|l|r|r|r|r|r|} \hline \text { Impulse, } J(\mathrm{N} \cdot \mathrm{s}) & 4.5 & 7.8 & 4.5 & 7.8 & 11 \\ \hline \begin{array}{l}\text { Rocket mass }(\mathrm{g}) \\ \text { (including motor) }\end{array} & 180 & 485 & 234 & 234 & 485 \\ \hline \begin{array}{l}\text { Maximum height } \\\\\text { achieved (m) }\end{array} & 22 & 13 & 19 & 51 & 23 \\\\\hline\end{array}$$

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.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Solid State Physics

Read Explanation

Space Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Fields in Physics

Read Explanation

Further Mechanics and Thermal Physics

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.