MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics IC-W08D2-11 Jumping Off as Flatcar Solution

Transcription

1 N eole, eah o mass MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deartment o Physis Physis 8.01 IC-W08D2-11 Juming O as Flatar Solution m, stand on a railway latar o mass m. They jum o one end o the latar with veloity u relative to the ar. The ar rolls in the oosite diretion without rition. a) What is the inal veloity o the ar i all the eole jum at the same time? b) What is the inal veloity o the ar i the eole jum o one at a time? ) Does ase a) or b) yield the largest inal veloity o the lat ar. Solution: We begin by hoosing a reerene rame at rest with reset to the ground and identiy our system as the latar and all the eole. Sine there are no external ores in the horizontal diretion, the horizontal omonent o the momentum o the system is onstant =. (1.1) x, i x, We an use this at to solve or the inal seed v o the latar when all the eole jum o together. We need to be areul to use the at that the seed o eah jumer relative to ground is given by u v. We take as our initial state the ar and eole at rest. The inal state is immediately ater all the eole have jumed o. The shemati momentum diagram below shows these states. Then the initial x-omonent o the momentum is xi, = 0. (1.2)

2 The inal x-omonent o the momentum is x, = mv+ Nm( u v). (1.3) Substituting Eq. (1.2) and Eq. (1.3) into Eq. (1.1) yields 0 = m v + Nm ( u v ). (1.4) We an solve Eq. (1.4) or the inal veloity o the ar, Nm v = u. (1.5) (b) i the eole jum o one at a time, we need to be more areul. Again the momentum o the system is onstant but we have N jums. Beore the irst jum, the momentum is still zero. Immediately ater the irst erson jumed, the x-omonent o the momentum is = (( N 1) m + m ) v + m ( u v ). (1.6) x,,1,1,1 Sine the x-omonent o the momentum is onstant we have that 0 = (( N 1) m + m ) v + m ( u v ). (1.7),1,1 We an solve this equation or the seed o the ar ater the irst jum and ind that v,1 = m u. (1.8) Note that this is 1/ N o the seed ound when the eole all jumed at one (Eq. (1.5). Now let s onsider the seond jum.

3 The x-omonent o the momentum beore the jum is = (( N 1) m + m ) v (1.9) xi,,2,1 The x-omonent o the momentum immediately ater the seond erson jumed is = (( N 2) m + m ) v + m ( u v ). (1.10) x,,2,2,2 Again alying the at that the x-omonent o the momentum is onstant yields (( N 1) m + m ) v = (( N 2) m + m ) v + m ( u v ). (1.11),1,2,2 We an rewrite this equation as (( N 1) m + m ) v = (( N 1) m + m ) v + m u. (1.12),1,2 Ater dividing through by (( N 1) m + m ) and rearranging Eq. (1.12) beomes m v,2 = v,1 + u. (1.13) ( N 1) m + m Substituting Eq. (1.8) into Eq. (1.13) yields the seed o the ar immediately ater the seond erson jumed o m m v,2 = u+ u. (1.14) ( N 1) m + m Notie that the seond term on the right hand side o Eq. (1.13) is larger than the irst 2 term on the right hand side, so the seed is now larger than v,2 > 2v,1 = v. N

4 By indution, the seed ater the jth erson jumed o is m m m v, j= u+ u+ + u. (1.15) ( N 1) m + m ( N ( j 1)) m + m Hene the seed ater the last erson (the Nth) erson jumed o is m m m v, N= u+ u+ + u. (1.16) ( N 1) m + m m + m () To omare the answers to the revious two arts, note that eah term in Eq.(1.16) is larger than the revious one, so we an onlude that Nm v, N = u. (1.17) Without doing the alulation, we an alternatively use a roo by ontradition to understand why juming one at a time rodues a larger inal veloity or the latar. Consider ase A to be the everybody-jum-at-one ase, and ase B the one-at-a-time ase. Let v and v, N be the inal seed o the latar in ases A and B, resetively. Then, sine eah jumer is seiied to have a seed u relative to the latar's seed immediately ater his jum, in ase A every jumer ends with an x-omonent o the veloity u v. Now suose that v, N. Then eah jumer in ase B has a inal x- omonent o veloity greater than or equal to u v, N, and hene larger than the x- omonent o the jumers in ase A, whih is u v. Thus the total x-omonent o the momentum o the jumers in ase B is greater than in ase A, so the magnitude o x- omonent the reoil momentum o the latar must also be greater in ase B (we need to take the magnitude o the x-omonent o the reoil momentum beause the reoil is in the negative x-diretion and so the x-omonent is negative). Thus we have ontradited our hyothesis. Similarly, i we suose that v = v, N, we ould onlude that all the jumers exet the last in ase B would have an x-omonent o momentum larger than eah jumer in ase A, so again we would have a ontradition. Thus, v, N is the only ossibility.

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