Collision Problems

2-body

R & r coordinates

We have already defined the center of mass coordinate, R, as:     R = Σmiri / Σmi .
For a two-body problem, this becomes:

R  =  (m1r1 + m2r2) / (m1 + m2) .

If we take the second derivative of each side, we get:

R  =  (m1r1 + m2r2) / (m1 + m2),  or

MR  =  (m1r1 + m2r2)  ,

where M = m1 + m2 .

We will now define the relative coordinate, r, as:     r  =  r1  r2 .

To see how these coordinates are useful, we start with Newtons Second Law applied to each particle:

F1int + F1ext   =   m1r1

F2int + F2ext   =   m2r2

where F1int is the internal force on particle 1 due to particle 2, and F1ext is the sum of all external forces on 1 (not including the force from particle 2).

If we now add the two equations of motion above, we get:

F1int + F1ext + F2int + F2ext   =   m1r1 + m2r2 .

We now note that Newtons Third Law of Motion says that F1int  =  - F2int , and we recall that (m1r1 + m2r2)  =  MR.  This simplifies the above equation to become:

F1ext + F2ext   =   MR .                    ***

This says that the external forces cause the center of mass to accelerate, but that the internal forces do not.

To get an equation for the relative coordinate, r, we need to multiply the equation of motion for m1 by m2, and multiply the equation of motion for m2 by m1.  We can then subtract the two equations to get:

m2F1int + m2F1ext  m1F2int - m1F2ext    =    m1m2r1  - m1m2r2 .

We again use Newtons Third Law, F1int  =  - F2int , and if we can assume that m2F1ext   m1F2ext , then we get:

(m1 + m2) F1int    =   m1m2 (r1  r2)   =   m1m2r .

If we further define a quantity we will call the reduced mass, m, as:  μ = m1m2/(m1+m2) ,  we get:

F1int   =  μr .                                              ***

This indicates that the motion of particle 1 as seen from particle 2 (that is what the vector, r, indicates) is the same as if particle 2 were stationary and particle 1 had the reduced mass, μ, instead of its regular mass, m1.  This is just a single particle equation of motion, and we can apply all the techniques we learned in parts 1 and 2 to this type of problem.

Please remember that to obtain the above equation we had to assume that   m2F1ext   m1F2ext .

If we are able to solve the problem using the Rr system, we should be able to go back to the inertial system, r1 and r2, to get a solution in this system.

To do this, we need to develop the inverse transformation equations.  We start with our definitions of R and r:

R  =  (m1r1 + m2r2) / (m1 + m2)      and       r  =  r1  r2 .

Knowing R and r, we have two equations for our two unknowns: r1 and r2 .

Start with     r2 = r1  r,    and use this in the R equation to get:

R   =   {m1r1 + m2(r1  r)}/(m1+m2)   =   r1  m2r/(m1+m2),

or

r1   =   R + m2r/(m1+m2),      or        r1  =  R + (μ/m1)r,

In a similar way, we get

r2   =   R  m1r/(m1+m2),    or      r2   =   R  (μ/m2)r .

You should now be able to do Homework #26 (see below) which asks how KE, L, and p look when expressed in the Rr coordinates.

From this homework problem the kinetic energy is

KE   =   ½m1v12 + ½m2v22   =    ½μv2 + ½MV2

where v = v, and V = R.

If there are no external forces, V is constant and so the only energy available in an inelastic collision between m1 and m2 is the ½μv2 amount.

Homework Problem #26:

a) Starting from KE = ½ m1 v1² + ½ m2 v2²  show that KE = ½ M V² + ½ m

where M is the total mass and m is the reduced mass; V is the

center‑of‑mass velocity and v is the relative velocity.

b)    Starting from L = m1 (r1 x v1) + m2 (r2 x v2)  show that   L = MR x V + mr x v.

c)    Starting from p = m1 v1 + m2 v2  show that   p = M V.

Homework Problem #27:  In part 2, the period of orbit, t, for an inverse square law force was expressed by the relation:  t2 = 4p2a3½m/K½ where a is the semimajor axis distance and K for gravity is GMm.  (This is a generalization of one of Keplars laws.)  This applies as long as m is much smaller than M.  Work out a correction for the case where M = m.

Example:

The moon orbits the earth in an elliptical orbit with rmin = 363,300 km and rmax = 405,500 km.  What should the period be, and how will this be modified by the fact that the earth is not actually stationary but does wobble a bit due to the moon pulling on it?

We start by defining the orbit:  1/r = B + A cos(q-qo) .

1/rmin = B + A   and   1/rmax = B  A  ,  so

B = ½ (1/rmin + 1/rmax) = ½ (1/[3.533 x 108 m] + 1/[4.055 x 108 m])  =  2.60932 x 10-9 m-1  and

A = ½ (1/rmin - 1/rmax)  = ½ (1/[3.533 x 108 m] - 1/[4.055 x 108 m])  =  1.43227 x 10-10 m-1  , so

e  =  A/B  =   1.43227 x 10-10 m-1 /  2.60932 x 10-9 m-1  =  .05489 .

a  =  ½ (rmin + rmax)  =  3.844 x 108 m,  and

b  =  a[1 - e2]1/2  =  3.8382 x 108 m.

S (area) = pab  =  (L/2m)T.  We have a and b, now we need L:

B = -mK/L2    so  L = [-mK/B]1/2 ,  where K = -G*Mearth*Mmoon and  m = Mmoon  ;

Mearth = 5.974 x 1024 kg,  and  Mmoon = .0123 * Mearth  =  7.347 x 1022 kg  ;  so

L = [Mmoon * G * Mearth * Mmoon / B]1/2  =  2.871 x 1034 kg*m2/s .

Therefore, T = 2*Mmoon *p*a*b / L  =  2.372 x 106 seconds = 27.46 days (without taking the wobbling earth into account).

To take the wobble of the earth into account, we need to replace Mmoon with m = {Mearth*Mmoon /(Mearth + Mmoon) = .98785 * Mmoon , but we dont replace it inside the K.

Note that the m = Mmoon is inside the L, but it goes as the square root of m.  The m is in the numerator of the T relation and the L is in the denominator of the T relation, so the net effect is the period is reduced by the square root of the .98785 factor, so the new T becomes 2.372 x 106 seconds * [.98785]1/2 = 2.358 x 106 seconds = 27.29 days.

The experimental period is given as 27.3 days.  Note that the month is from full moon to full moon and is 29.5 days, but the extra time is due to the earth (and moon) orbiting the sun.  The 27.3 days is the period with respect to the stars rather than the sun.

Does the condition that  m2F1ext   m1F2ext  hold here?  The magnitude of F1external = G*Msun*Mearth / rearth-sun2  and  the magnitude of F2external = G*Msun*Mmoon / rmoon-sun2  so that

m2F1external   m1F2external   becomes  Mmoon*G*Msun*Mearth / rearth-sun2    Mearth* G*Msun*Mmoon / rmoon-sun2  .  There is a slight difference between rearth-sun and rmoon-sun both as to magnitude and direction, so this condition does not exactly hold, but it is pretty close, especially when you consider that on average rearth-sun is the same as rmoon-sun .

Scattering

From Part 2, in the section on Scattering from an inverse square law force, we obtained the expression:

tan(Θ/2)  =  [mK2/2EL2]½   =  [mK2/{2(½mvo2)( m2vo2s2)}]  =  K/mvo2s .

Since the equation of motion for the relative position, r, is the same as for the above case with the one change that the m of the particle becomes the reduced mass, μ = m1m2/(m1+m2), we can see that the relative velocity, = v, will be deflected in the scattering by an angle, Θ, where Θ can be determined from the relation:

tan(Θ/2)  =  K/μvI2s

where vI is the magnitude of the initial relative velocity:  vI  =  (dr/dt)I  =  [d(r1  r2)/dt]I  =  v1I  v2I  .