Re: recoil & newtons law
- From: "David L. Burkhead" <dburkhead@xxxxxxx>
- Date: Fri, 8 Aug 2008 21:46:38 +0000 (UTC)
"Ed" <Huckleberry@xxxxxxxxxxxxx> wrote in message
news:g7hhs2$qca$1@xxxxxxxxxxxxxxxxxxxxxxxx
#
# #
# # The momentum of the bullet and the gun are the same:
# # (M1)(V1)=(M2)(V2).
# # (mass bullet)(velocity bullet)= (mass gun)(velocity gun).
# #
# # The energy of the bullet and the gun are not the same:
# # 1/2 (M1)(V1)(V1) does not equal 1/2 (M2)(V2)(V2).
# # 1000 ft lb energy bullet does not equal 10 ft lb gun recoil.
# #
#
# So, I think some are going to wonder, where does the energy
# difference go?
#
#
# Ed
Kinetic Energy is not conserved in this kind of situtation (in fact, it's
generally not conserved). What is conserved is momentum.
Momentum is m*v
Kinetic energy is (1/2)*m*v^2
Take bullet of mass "1" (pick whatever units you'd like) moving at speed of
"10" (again, whatever units), it would cause a gun of mass 10 (same units)
to move at speed 1. In a closed system (which we are considering gun and
bullet to be here) any momentum change one way must be matched by an equal
momentum change the other way. Total momentum (gun plus bullet) does not
change.
The kinetic energy of the bullet in this case would be (1/2)*1*10^2 = 50
The kinetic energy of the bun would be (1/2)*10*1^2 = 5
The energy difference doesn't "go" anywhere. The energy comes from the
exploding powder. Part of it goes to moving the bullet, part of it goes to
moving the gun, part of it goes to heating the gun and bullet, part of it
goes to "muzzle blast," part of it goes to report, part of it may go to
cycling the action, etc.
Note, while I don't play a physicist on TV, I am one in real life. ;)
--
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