Applied Maths - Mechanics 28

((a) Let the speed of P and hoop be Vk and Sk after the kth impact
mV = mV1 + nmS1 = > V = V1 + nS1 … (1)
S1 – V1 = e(V – 0) => S1 = eV + V1 … (2)
Sub into (1), V = V1 + enV + nV1
V(1 – en) = V1(1 + n)
V1 = (1 – en)V/(1 + n)
Sub into (2) S1 = eV + (1 – en)V/(1 + n)
S1 = (1 + e)V/(1 + n)
(b) Note that the momentum is always conserved, so that mVr + nmSr = mV
Vr + nSr = V … (3)
Sr – Vr = e(Vr-1 – Sr-1)
(Sr – Vr)/(Sr-1 – Vr-1) = -e
[(Sr – Vr)/(Sr-1 – Vr-1)][(Sr-1 – Vr-1)/(Sr-2 – Vr-2)]...[(S2 – V2)/(S1 – V1)] = (-e)r-1
(Sr – Vr)/(S1 – V1) = (-e)r-1
Use (2)
S1 – V1 = eV => Sr – Vr = -V(-e)r
Use (3)
Sr – (V – nSr) = -V(-e)r
(1 + n)Sr = V – V(-e)r
Sr = V[1 - (-e)r]/(n + 1)
(c) Relative speed of P and hoop is | Sr – Vr |
Time between impact = 2a / | Sr – Vr |
Distance traveled by hoop = | 2aSr / (Sr – Vr) |
= | 2aV[1 - (-e)r]/(n + 1)/[-V(-e)r] |
= 2a[1 - (-e)r]/[(n+1)er]
= [2a/(1+n)][1/er - (-1)r]
Distance moved by the hoop between the moment of the first impact and that of the (2k+1)th impact is
[2a/(1+n)][1/e + 1 + 1/e2 -1 + 1/e3 + 1 +... 1/e2k -1]
= [2a/(1+n)][1/e + 1/e2 + 1/e3 + ... + 1/e2k]
= [2a/(1+n)][(1/e)2k+1 – 1]/[1/e -1]
= [2a/(1+n)][(1/e)2k – e]/(1 – e)
= [2a/(1+n)][(1 – e2k+1]/e2k(1 – e)
= 2a(1 – e2k+1)/[(n+1)(1 – e)e2k]


2009-08-04 01:00:59 補充：
Mistakes correction
[2a/(1+n)][(1/e) + (1/e)^2 + (1/e)^3 + ... + (1/e)^(2k)]
= [2a/(1+n)](1/e)[(1/e)^(2k) – 1]/[1/e -1]
= [2a/(1+n)][(1/e)^(2k) – 1]/(1 – e)
= [2a/(1+n)][(1 – e^(2k)]/e^(2k)(1 – e)
= 2a[1 – e^(2k)]/[(n+1)(1 – e)e^(2k)]

2009-08-04 01:01:09 補充：
Mistakes correction
[2a/(1+n)][(1/e) + (1/e)^2 + (1/e)^3 + ... + (1/e)^(2k)]
= [2a/(1+n)](1/e)[(1/e)^(2k) – 1]/[1/e -1]
= [2a/(1+n)][(1/e)^(2k) – 1]/(1 – e)
= [2a/(1+n)][(1 – e^(2k)]/e^(2k)(1 – e)
= 2a[1 – e^(2k)]/[(n+1)(1 – e)e^(2k)]