Consider two particles that experience a mutual force but no external forces. The classical equation of motion for particle 1 is v˙1=F2on1/m1, and for particle 2 is v˙2=F1on2/m2, where the dot means a time derivative. Show that these are equivalent to v˙cm=constant, and v˙rel=FMutual/μ .Where, v˙cm=(m1v˙1+m2v˙2)/(m1m2),FMutual=-Fion2andμ=m1m2(m1+m2).

In other words, the motion can be analyzed into two pieces the center of mass motion, at constant velocity and the relative motion, but in terms of a one-particle equation where that particle experiences the mutual force and has the “reduced mass” μ.

Short Answer

Expert verified

The given equations are equivalent to,v˙cm=0or v˙cm=constant.

Also,

v˙rel=1μFMutual

Where, μ=m1m2m1+m2

Step by step solution

01

A concept:

Centre of massis the point on a structure which characterizes the motion of the object if the whole mass of that object is concentrated on that point.

m1v˙1=F2on1 …..(1)

And

m2v˙2=F1on2 ….. (2)

02

Showing v˙cm=0 or v˙cm=constant :

As you know,

-F2on1=F1on2 ….. (3)

Hence, by adding equations (1) and (2) and then using (3), you get,

m1v1?+m2v2?=0 ….. (4)

If you divide eq. (4) by m1+m2, you get,

v˙cm=0or v˙cm=constant

03

Finding vrel:

Relative motion of an object is its motion with respect to any other moving or stationary object.

v˙rel=v2˙-v˙1

If you rearrange and subtract equation (1) from equation (2), you get,

v˙2=-v˙1=F1on2m2-F2on1m1=1m2+1m1F1on2v˙rel=m1+m2m1m2FMutual=1μFMutual

Where, μ=m1m2m1+m2

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 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

See all solutions

Recommended explanations on Physics Textbooks

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