Re: conservation of momentum

[Roger Haar <haar@PHYSICS.ARIZONA.EDU>]

Justin,
        I just tried a) and b) and got the book's answer.

        For b) I broke it into two variants of what I did
for a).

That is momentum is conserved as m1 moves away
from M +m2 at v(rel)

This gives (iceboat + m2) a velocity, V', relative
to the ground, V' = 2/15 v(rel)

Next conserve momentum as m2 moves away from M at
v(rel) and do this in the center of momentum frame
of M & m2.

This yields a velocity of V'' = 1/13 v(rel) for M
relative to the CM of M & m2.

Final velocity of the iceboat relative to the
ground is the velocity, V, of the CM of M & m2
plus the velocity of M relative to this CM.  V =
(2/15+1/13) v(rel)


Thanks
Roger Haar

***************************************************************
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)
> b)  .210 v(rel)
> c)  .209 v(rel)
>
> I don't get b) and c).
>
> Comments?
>
> Justin Parke