Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 33

A proton moving at \(6.9 \mathrm{Mm} / \mathrm{s}\) collides elastically head-on with a second proton moving in the opposite direction at \(11 \mathrm{Mm} / \mathrm{s}\). Find their subsequent velocities.

Short Answer

Expert verified
The subsequent velocities of the protons are \(8.21 \mathrm{Mm/s}\) (to the right) and \(-12.31 \mathrm{Mm/s}\) (to the left) respectively.

Step by step solution

01

Apply the Conservation of Momentum

According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after. \[(6.9m) + (-11m) = (m)(v_1) + (m)(v_2)\] Since the masses cancel out we get: \[v_1 + v_2 = -4.1 \mathrm{Mm/s}\] This is our first equation.
02

Apply the Conservation of Kinetic Energy

According to the conservation of kinetic energy, the total kinetic energy before the collision is equal to the total kinetic energy after. \[\frac{1}{2} m (6.9)^2 + \frac{1}{2} m (11)^2 = \frac{1}{2} m (v_1)^2 + \frac{1}{2} m (v_2)^2\] If masses and 0.5 cancel out, we have: \[(6.9)^2 + (11)^2 = (v_1)^2 + (v_2)^2\] or \(v_1^2 + v_2^2 = 193.61 \mathrm{(Mm/s)}^2\] This is the second equation.
03

Solve the System of Equations

Now you have a system of two equations which you can solve to get the final velocities \(v_1\) and \(v_2\). This can be done by substitution or elimination method. Since these are quadratic equations, the solutions will have two possible values - one negative and one positive. Given the premise of the problem, the proton that was initially moving to the right (positive direction) will continue moving to the right and the one that was moving to the left will continue moving to the left. Hence, \(v_1\) will be a positive value and \(v_2\) a negative one.
04

Find the Final Velocities

Solving the system of equations gives us \(v_1=8.21 \mathrm{Mm/s}\) and \(v_2= -12.31 \mathrm{Mm/s}\). This means the first proton is moving to the right at \(8.21 \mathrm{Mm/s}\) and the second proton is moving to the left at \(12.31 \mathrm{Mm/s}\).

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

Find the center of mass of a uniform slice of pizza with radius \(R\) and angular width \(\theta\).

Physicians perform needle biopsies to sample tissue from internal-organs. A spring-loaded gun shoots a hollow needle into the tissue; extracting the needle brings out the tissue core. A particular device uses 8.3 -mg needles that take 90 ms to stop in the tissue, which exerts a stopping force of \(41 \mathrm{mN}\). (a) Find the impulse imparted by the tissue. (b) How far into the tissue does the needle penetrate?

A pitched baseball moves no faster than the pitcher's hand. But a batted ball can move much faster than the bat. What's the difference?

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.

An hourglass is inverted and placed on a scale. Compare the scale readings (a) before sand begins to hit the bottom; (b) while sand is hitting the bottom; and (c) when all the sand is on the bottom.
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Linear Momentum

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Dynamics

Read Explanation

Torque and Rotational Motion

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