🚀 go-pugleaf

In article <333@mtnmath.UUCP>, paul@mtnmath.UUCP (Paul Budnik) writes...
>In article <1992Nov2.124821.9376@daimi.aau.dk>, snil@daimi.aau.dk (Sven Nielsen) writes:
>> The law of action and reaction holds. Another rule that is valid is : F=m*a
>> If you assume that the mass of the bullet < mass of the gun + mass of man 
>> holding it you'll see that the bullet is accelerated to a higher velocity.
>> 
>> The bullet will in fact get a much larger part of the kinetic energi released
>> as the gun is fired. (K=m*v^2 /2). You can also see this if you apply the rule
>> of conservation of the linear momentum the the problem.
> 
>You have given the best answer to this question so far. I'm sorry I did not

I don't think so.  It's not the kinetic energy which knocks you down - it's
the linear momentum.  And the bullet has no more momentum than the gun, at
least if it doesn't have an exploding shell.  I think that the best answer
is that the gunman is prepared for the recoil.  I recently fired a 30-0-6 of 
WWII vintage, and nearly got knocked right over because I expected (in spite 
of the warning of my friend) no more recoil than from a typical shot gun.  
But after the first shot, I had no problem keeping my balance, though I 
paid the price on the next day with a sore shoulder.  (I also punched myself
in the face a couple of times with my thumb, which I consistently forgot
to keep out of the way.)  Clearly, I'm a better physicist than marksman.
(Of course, I'm probably a better French chef than I am a marksman...)

-Scott
--------------------
Scott I. Chase			"It is not a simple life to be a single cell,
SICHASE@CSA2.LBL.GOV		 although I have no right to say so, having
				 been a single cell so long ago myself that I 
				 have no memory at all of that stage of my 
				 life." - Lewis Thomas