Applied Maths - Mechanics 30

After first impact: y1 – x1 = ev
After second impact y2 – x2 = e(x1 – y1)
| y2 – x2 | = e| x1 – y1 | = e2v
After third impact | y3 – x3 | = e| x2 – y2 | = e3v
After n-1th impact | yn-1 – xn-1 | = en-1v
Time between first and second impact = a / | x1 – y1 | = a/(ev)
Time between second and third impact = a / | x2 – y2 | = a/(e2v)
Time between n-1th and nth impact = a / | xn-1 – yn-1 | = a/(en-1v)
Total time between first and nth impact = (a/v)[1/e + 1/e2 + 1/e3 + … + 1/en-1]
= (a/v)(1/e)[(1/e)n-1 – 1]/(1/e – 1)
= (a/v)[(1/e)n-1 – 1]/(1 – e)
= (a/v)(1 – en-1)/[en-1(1 – e)]
Initial momentum = mv which remains constant throughout no matter how many impacts
The centre of mass of the system is always mid-way between the centre of mass of the tube and the bead and the velocity of the centre of mass = mv/(m + m) = v/2
Distance traveled by the centre of mass = (a/v)(1 – en-1)/[en-1(1 – e)](v/2)
= a(1 – en-1)/[2en-1(1 – e)]

2009-08-04 20:45:25 補充：
where x and y are the velocities

2009-08-04 20:45:36 補充：
where x and y are the velocities

What I want to say is:
If there's only a minor mistake, just express the opinion here and no need to make the second answer cos this will give a feel to the first answerer that he is being 針對.

六呎將軍:
I think the correct answer should be:
Time between (n-1)th and nth impact:
=a/[ue^(n-1)]

Total time taken between first and nth impact:
=a/ue + a/ue^2 + ...+a/[ue^(n-1)]
=a[1 - e^(n-1)]/[ue^(n-1) (1-e)]

2009-08-04 20:47:03 補充：
As center of mass moves with speed u/2 all the time,
distance travelled by the centre of mass:
=a[1 - e^(n-1)]/[ue^(n-1) (1-e)] x u/2
=a[1 - e^(n-1)]/[2(1- e)e^(n-1)]
which is independent of the fact that n is even or odd.