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

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