Re: conservation of momentum

[Rick Tarara <rtarara@SAINTMARYS.EDU>]

By taking v(rel) to be the velocity of the stone relative to the MOVING
iceboat.  That is, v(rel) has the magnitude of the sum of the speeds of the
two objects as you've solved the problem.  I also get your 1/4 for what I
think is the more conventional interpretation of the problem--now expunged
from the 6th edition.

Rick

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Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, IN 46556
rtarara@saintmarys.edu

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----- Original Message -----
From: "John J. Trammell" <trammell@TRAMMELL.DYNDNS.ORG>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, November 15, 2001 1:44 PM
Subject: Re: conservation of momentum


> On Thu, Nov 15, 2001 at 11:56:29AM -0500, Justin Parke wrote:
> > If you will bear with me, I have another question about a homework
problem where the book and I disagree.  This one is from Halliday, Resnick,
and Walker Fundamentals of Physics, 5th edition.  p. 211 number 47.
> > "You are on an iceboat on frictionless, flat ice; you and the boat have
a combined mass M.  Along with you are two stones of masses m1 and m2 such
that M=6m1=12m2.  To get the boat moving, you throw the stones rearward,
either in succession or together, but in each case with a cetain speed
v(rel) relative to the boat.  What is the resulting speed of the boat if you
throw the stones
> > a)  simultaneously
> > b)  m1 then m2
> > c)  m2 then m1?
> >
> > book:
> > a)  .2 v(rel)  (I agree)
>
> Okay, perhaps I'm being blind here, but:
>
>  M = 6 * m1 = 12 * m2
>
> so
>
>  m1 = 2 * m2
>
> and
>
>  M' = M + m1 + m2 = 15 * m2
>
> Initial momentum is zero.
>
> Final momentum = M * v + (m1+m2) * vrel = 0
>
>  |v| / |vrel| = (m1 + m2) / M
>
>  = (3 * m2) / (12 * m2)
>  = 1/4
>
> How is everyone else getting 1/5?
>
> Thanks,
> J