Collision Problems

2-body

R & r coordinates

We have already defined the center of mass coordinate, R, as: R = Σm_ir_i / Σm_i .
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:

F₁^int + F₁^ext = m₁r₁’’

F₂^int + F₂^ext = m₂r₂’’

where F₁^int is the internal force on particle 1 due to particle 2, and F₁^ext 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:

F₁^int + F₁^ext + F₂^int + F₂^ext = m₁r₁’’ + m₂r₂’’ .

We now note that Newton’s Third Law of Motion says that F₁^int = - F₂^int , and we recall that (m₁r₁’’ + m₂r₂’’) = MR’’. This simplifies the above equation to become:

F₁^ext + F₂^ext = 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:

m₂F₁^int + m₂F₁^ext – m₁F₂^int - m₁F₂^ext = m₁m₂r₁’’ - m₁m₂r₂’’ .

We again use Newton’s Third Law, F₁^int = - F₂^int , and if we can assume that m₂F₁^ext ≈ m₁F₂^ext , then we get:

(m₁ + m₂) F₁^int = m₁m₂ (r₁’’ – r₂’’) = m₁m₂r’’ .

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

F₁^int = 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 m₂F₁^ext ≈ m₁F₂^ext .

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 v²

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 r_min = 363,300 km and r_max = 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-q_o) .

1/r_min = B + A and 1/r_max = B – A , so

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

A = ½ (1/r_min - 1/r_max) = 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 (r_min + r_max) = 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*M_earth*M_moon and m = M_moon ;

M_earth = 5.974 x 10²⁴ kg, and M_moon = .0123 * M_earth = 7.347 x 10²² kg ; so

L = [M_moon * G * M_earth * M_moon / B]^1/2 = 2.871 x 10³⁴ kg*m²/s .

Therefore, T = 2*M_moon *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 M_moon with m = {M_earth*M_moon /(M_earth + M_moon) = .98785 * M_moon , but we don’t replace it inside the K.

Note that the m = M_moon 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 m₂F₁^ext ≈ m₁F₂^ext hold here? The magnitude of F₁^external = G*M_sun*M_earth / r_earth-sun² and the magnitude of F₂^external = G*M_sun*M_moon / r_moon-sun² so that

m₂F₁^external ≈ m₁F₂^external becomes M_moon*G*M_sun*M_earth / r_earth-sun² ≈ M_earth* G*M_sun*M_moon / r_moon-sun² . There is a slight difference between r_earth-sun and r_moon-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 r_earth-sun is the same as r_moon-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(½mv_o²)( m²v_o²s²)}] = K/mv_o²s .

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, r’ = v, will be deflected in the scattering by an angle, Θ, where Θ can be determined from the relation:

tan(Θ/2) = K/μv_I²s

where v_I is the magnitude of the initial relative velocity: v_I = (dr/dt)_I = [d(r₁ – r₂)/dt]_I = v_1I – v_2I .

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