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Re: conservation of momentum



John,
In your eq for final momentum it should be
M * V_M + (m1 + m2 ) * v_m = 0
V_M + v_m = v(rel)

Thanks
Roger Haar


*****************************
"John J. Trammell" wrote:

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