Re: [Phys-l] Ballistics divertissement

[Bernard Cleyet <bernardcleyet@redshift.com>]

However, using maths instead of qualitative reasoning I think I'm very 
wrong.  I thought that w/ o the bullet the low mass POR (products of 
reaction) left w/ high speed.  W/ the bullet much lower speed, but high 
mass, so the p was the same, therefore, the kick the same.

Assuming no friction E the same (no thermodynamic differences) and other 
factors I haven't thought of, and using the two KEs equal to E from the 
charge,  and p = p' ,  I get:

m'v' (the kick) = sqrt(( 2E * m) / (1 + m/m'))

The mass of the bullet makes a huge diff.  Assuming no algebra error or 
missing factor.

bc, chastened

Rauber, Joel wrote:

> | 
> | A firearm w/ out a bullet is a rocket.
> | 
> |
>
> I get what you are saying now, my comments were with respect to the
> firearm with bullet.
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