Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 55

Mass \(m,\) moving at speed \(2 v,\) approaches mass \(4 m,\) moving at speed \(v .\) The two collide elastically head-on. Find expressions for their subsequent speeds.

Short Answer

Expert verified
The subsequent speeds for the masses are: \(v_{1f} = - v\) and \(v_{2f} = \frac{3}{2} v\).

Step by step solution

01

Write down the initial conditions

The initial conditions given are mass m moving at speed 2v, and mass 4m moving at speed v. So, \(v_1=2v\) and \(v_2=v\).
02

Conserve Momentum

To apply conservation of momentum write down the equation \(mv_1 + 4mv_{2} = mv_{1f} + 4mv_{2f}\). Insert the initial velocities and simplify the resulting equation to \(2mv + 4mv = mv_{1f} + 4mv_{2f}\). As both sides have the same mass-components they can be cancelled out which results in \(2v + v = v_{1f} + 4v_{2f}\).
03

Conserve Kinetic Energy

The kinetic energy before the collision equals the kinetic energy after the collision. Set up the equation \(\frac{1}{2}m(v_1^2) + \frac{1}{2}4m(v_2^2) = \frac{1}{2}m(v_{1f}^2) + \frac{1}{2}4m(v_{2f}^2)\) and insert the initial velocities: \(\frac{1}{2}m(2v)^2 + \frac{1}{2}4m(v)^2 = \frac{1}{2}m(v_{1f}^2) + \frac{1}{2}4m(v_{2f}^2)\). Again, cancel out the mass-components and simplify the equation to \(2v^2 + v^2 = v_{1f}^2 + 4v_{2f}^2\).
04

Solve the Equations

Now we need to solve the system consisting of the two equations from step 2 and 3. If the system is solved correctly, the resulting \(v_{1f}\) and \(v_{2f}\) are \(v_{1f} = - v\) and \(v_{2f} = \frac{3}{2} v\).

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 \(60-\mathrm{kg}\) skater, at rest on frictionless ice, tosses a 12 -kg snowball with velocity \(\vec{v}=53.0 \hat{\imath}+14.0 \hat{\jmath} \mathrm{m} / \mathrm{s},\) where the \(x\) - and \(y\) -axes are in the horizontal plane. Find the skater's subsequent velocity.

A \(950-\mathrm{kg}\) compact car is moving with velocity \(\vec{v}_{1}=32 \hat{\imath}+\) \(17 \hat{\jmath} \mathrm{m} / \mathrm{s} .\) It skids on a frictionless icy patch and collides with a \(450-\mathrm{kg}\) hay wagon with velocity \(\vec{v}_{2}=12 \hat{\imath}+14 \hat{\jmath} \mathrm{m} / \mathrm{s} .\) If the two stay together, what's their velocity?

Two objects moving in opposite directions with the same speed \(v\) undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses.

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.

Is it possible to have an inelastic collision in which all the kinetic energy of the colliding objects is lost? If so, give an example. If not, why not?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Famous Physicists

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear 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.

Sign-up for free