On Wed, 2 Nov 2005 23:32:07 -0500, with neither quill nor qualm, Ned
Simmons quickly quoth:

In article ,
says...
On Wed, 02 Nov 2005 23:54:36 GMT, Lew Hartswick wrote:
I think Dave needs to take a basic Physics course to get

You give no context, maybe you're talking about me? If so, pardon me
for having decades between physics and today, and perhaps you could
explain in English how my statements are factually inaccurate. For the
record, I contend that the same amount of force will push on your
shoulder, as is imparted to the bullet - but your shoulder is not
injured because the force is more spread out in both area, and in time.
What part of that is wrong, exactly, and why (assuming your
non-attributed response was in fact to me)?


Momentum is equal to mass times velocity. Conservation of
momentum says that the total momentum of gun + bullet is
the same before and after the bullet is fired. In order for
momentum to be conserved, the ratio of the speeds (absolute
velocities) of the bullet and gun must be inversely
proportional to their masses.

But kinetic energy is 1/2 mass x velocity squared. It's
that velocity squared term that leads to the unequal
division of kinetic energy between gun and bullet.

As an example, the gun weighs 100 units of mass to the
bullet's 1 unit. Momentum is conserved by the gun recoiling
at 1/100 the speed of the bullet.

KE of the bullet is
1/2 * 1m * (100v)^2 = 5000 mv^2

For the gun
1/2 * 100m * (1v)^2 = 50 mv^2

IOW, if one thinks a rifle recoil is forceful (spread out over
a larger area of your chest and shoulder) imagine the force on
that tiny point of skin as a bullet hits you.

Ouch! That HAS to hurt.

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