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Barukh
Uberpuzzler      Gender: Posts: 2276 Elastic Collisions   « on: Nov 1st, 2015, 12:36am » Quote Modify

On a horizontal line, two balls are placed: the first ball has unit mass, and the second ball  which is to the right of the first ball - has mass M. To the left of the first ball there is a vertical wall.

The second ball starts moving towards the first ball with constant speed, until they collide. This starts a process of collisions of balls with each other and vertical wall. It is assumed that all the collisions are perfectly elastic, and no energy is lost in any of the movements.

Denote by c(M) the total number of collisions between balls in the above experiment. Find lim c(M)/ M when M -> .

Source: G. Galperin IP Logged
markr
Junior Member   Gender: Posts: 90 Re: Elastic Collisions   « Reply #1 on: Nov 2nd, 2015, 10:23pm » Quote Modify

No proof, but my spreadsheet model seems to be converging to pi/2. IP Logged
SWF
Uberpuzzler      Posts: 879 Re: Elastic Collisions   « Reply #2 on: Dec 2nd, 2015, 9:54pm » Quote Modify

Conserving momentum and energy gives relation between velocity after n collisions and n+1 collisions:
u(n+1) = (M-1)*u(n)/(M+1) - 2*M*v(n)/(M+1)
v(n+1) = (M-1)*v(n)/(M+1) + 2*u(n)/(M+1)
where v(n) is velocity of ball of mass M after n collisions, and u(n) is velocity of ball of unit mass after n collisions.  u(n) is the positive going velocity (after its most recent collision being with the wall).

Eliminate u from the equation for v, and v from equation for u to get:
v(n+2) - 2*(M-1)*v(n+1)/(M+1) + v(n) = 0
u(n+2) - 2*(M-1)*u(n+1)/(M+1) + u(n) = 0

Solve difference equations by assuming solution of form exp(n*s) and determine the two values of s that work. This results in complex numbers and solution to each equation is of the form:
A*exp(i*n*p) + B*exp(-i*n*p)  where cos(p)=(M-1)/(M+1)

Using initial conditions of u(0)=0 and v(0)=-V results in:
u(n)=sqrt(M)*V*sin(n*p)
v(n)=-V*cos(n*p)

Collisions will end for the lowest n where u(n)<v(n) which requires that
sqrt(M)*sin(n*p) < -cos(n*p)

For large M, this occurs near where n*p=pi, and using cos(p)=(M-1)/(M+1) approx equal to 1-p*p/2 gives p approximately equal to 2/sqrt(M+1). S
o n is approximately pi/2*sqrt(M+1)  and dividing by sqrt(M) and taking limit results in pi/2. IP Logged

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