Building on Jack's analysis and setting this problem up on a spreadsheet
(takes only a couple minutes) and making all measurements in the original
frame of reference--you have
Grampa's Velocity = Previous velocity + Mass(spit) x v(spit)/M(grandpa)
But v(spit) in fixed reference frame is 5 - v(grandpa)
TYPICAL SPREADSHEET
A B
1 0 5 - a1
2 A1+.001*B1/50 5 - A2
3 A2+.001*B2/50 5 - A3
etc.
Gramps hits 1 m/s between step 11,160 and 11,161 in this analysis--of
course the velocity of the spit relative to the fixed frame is now down to
4 m/s. I think this makes sense. Obviously even under the artificial
condition of gramps having a fixed mass, this is not a simple problem and I
suspect the authors of the problem wanted the 10,000 answer. However, it
is a good thought problem because one can reason that because the later
'spits' do not have a full .005 kgm/s worth of momentum relative to the
fixed frame AND for gramps to gain 1 m/s in velocity then the total
momentum of the spit must be 50 kgm/s and 10,000 spits will give 50 kgm/s
ONLY if each spit has the .005kgm/s of momentum relative to the fixed
frame. Therefore, it is clear there must be more than 10,000 spits (and
more than 10 kg of spit--yuccch!)
rick
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+++++++++
Richard W. Tarara
Department of Chemistry & Physics Free Physics & Energy Instructional
Software
Saint Mary's College available at:
Notre Dame, IN 46556
219-284-4664 http://estel.uindy.edu/aapt/rickt/software/
rtarara@saintmarys.edu http://www-hpcc.astro.washington.edu/mirrors/tarara/
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+++++++++