Math session 10 Nov 11

Morin 5.15

In Morin 5.15 you are asked to consider propelling a car by throwing snowballs (or superballs).
Between time t and time t+Δt a snowball of mass Δm moving at velocity u collides with a car of mass M moving at v(t).
In the car frame the speed of the incoming snowball is u-v.  Assuming the collision is elastic (not so realistic for a snowball),
the outgoing speed is (v-u) in the car frame, and therefore 2v - u in the ground frame.  Conservation of momentum then reads:

Δp = (M (v+Δv) + Δm (2v -u)) - (M v +Δm u)

s=Solve[Δp==0,Δv][[1]]

Next divide by Δt to get the equation of motion:

Δv/Δt /. s

To finish, we need to express Δm in terms of Δt.  The trickiest part of this might be realizing that the answer is not σ Δt, but some other σ' times Δt.
(One quick way of putting it: if you throw snowballs with frequency ⋁, snowballs hit with Doppler shifted frequency ⋁' = ⋁ (u-v)/u.  Another one: the density of the snow in the air is σ/u, and it lands with relative speed (u-v).)
However you get there, the differential equation we want to solve is:

eqn515 = v'[t]==( Δv/Δt /. s/.{Δm->(σ(u-v[t])/u) Δt,v->v[t]}//Simplify)

Your Task for Homework

Use DSolve on this differential equation, starting from rest at t=0.
Then Integrate to find x[t] (supposing x[0]==x0).
Likewise DSolve the coupled set of differential equations you need for Morin 5.16:

eqn516m = m'[t]== (u-v[t])σ/u
eqn516mbc= m[0]== m0
eqn516v = m[t] v'[t] == (u-v[t])^2 σ/u 
eqn516vbc = v[0]==v0

Finally choosing numerical parameters (e.g. here is a set:)

p = {m0->1,v0->0,u->1,σ->1}

and Plot[] the resulting x[t] for parts 5.15 and 5.16.  Which did  you expect to be more "efficient," the superballs or the open window?
PRINT the solution cells and plot to hand in with the homework.

Optionally, animate the motion!
If you want an example of the incantations for animation, here is the solution to 5.9.

Morin 5.9

If a particle moves in one dimension under the influence of a force F(x) = -V'(x), then at any place where the force drops to zero (i.e. any x0 such that F(x0)=0) it is a possible "motion" for the particle to stay put (i.e x(t) = x0).  If the particle is displaced a small amount from x0, though, what happens?
Expanding F(x) in a Taylor series near x0: F(x) = F(x0) + F'(x0)(x-x0) + ... = 0 + F'(x0)(x-x0) + ...
We obtain the force law of a spring with equilibrium point x0 and spring constant -F'(x0).
So a very common question to be asked is: given V(x) where are its minima (i.e. find the x0 such that V'(x0)=0), and for each minimum what is the frequency of small oscillations about that minimum
(i.e. what is k_eff = V''(x0), and then ω=Sqrt[k_eff/m]).

Here we are asked these questions for

V[x_] = - C x^n Exp[-α x]

First let's plot this thing:

p = {C->1,n->3,α->1}

plotV = Plot[V[x]/.p,{x,0,10}]

Next locate the minimum:

eqn0 = V'[x]==0

soln0 = Solve[eqn0,x]

And compute the 2nd derivative there:

V''[x]/. soln0[[1]]

keff = Simplify[%]

So to answer the question posed, the (angular) frequency of small oscillations is

ω0 = Sqrt[keff/m]

And the period would then be

T0 = 2 π / ω0

So let's choose parameters and actually find the motion.  First atttempt:

p = {C->1,n->3,α->1,m->1}

eqnFma = m x''[t]==-V'[x[t]]

eqnFma/.p

Wait as long as you like for this one to come back, then hit Alt-plus to abort the calculation:

DSolve[{eqnFma/.p},x[t],t]

Even when the differential equation is too hard for Mathematica, we can get a numerical solution.
Let's launch the motion from the equilibrium point with some choice of initial velocity:

x0 = x/. soln0[[1]] /.p

sN = NDSolve[{eqnFma/.p, x[0]==x0,x'[0]==.1},x[t],{t,0,2T0/.p}]

x1[t_] = x[t]/.sN[[1]]

Plot[x1[t],{t,0,2T0/.p}]

So for a "small" velocity (.1) this looks pretty much like a sine.
Let's find the exact period and compare

FindRoot[x1[t]==x0,{t,9}]

T0/.p//N

For larger energies, the motion will not look so much like a sine:

sN2 = NDSolve[{eqnFma/.p, x[0]==x0,x'[0]==1},x[t],{t,0,2T0/.p}]

x2[t_] = x[t]/.sN2[[1]]

Plot[x2[t],{t,0,2T0/.p}]

Let's animate the 1D motion on top of the graph of the potential.
For maximum artistry, let's choose to draw this motion at the appropriate energy level:

E0 = (V[x0] + (1/2)m x2'[0]) /.p

plotE0 = Plot[E0,{x,1,9}]

Animate[Show[plotV,plotE0,ListPlot[{{x2[t],E0}},PlotStyle->{Red,PointSize[.02]}]],{t,0,2T0/.p}]

Collisions

Morin 5.73 asks about a particular collision problem with given masses and velocities.
Let's solve the general problem: mass 1 comes in with velocity v1, mass 2 has initial velocity v2 (perhaps some -v in some particular problem).
Your job: complete the following equations and Solve for v1f and v2f.

eqnp = m1 v1 + m2 v2 ==

eqnK = (1/2) m1 v1^2 + ...

Why are there two solutions?  Which one do we want?
Then get 5.73 by specializing to the case

rule573= {m1->2m, v1->v, m2->m, v2->-v}

And if there is time, you are of course free to either debug or cross-check your work for Morin 5.80, where an m1=2m collides with an m2=m at rest in *two* dimensions, and it just so happens that final velocity vectors make the same angle θ with the initial velocity.
Equation-wise one way to express it is:

eqn580px = m v0 == m v1f Cos[θ] + 2m v2f Cos[-θ]
eqn580K = 0 == ()+()
eqn580K = (1/2) m v0^2 == ()+()