7 ms·

With an actual firearm and bullets, it might not be; bullets don't have a lot of mass.

• hga 12y ago
  For this purpose, isn't it force, not mass, that matters? F = mv^2, so the velocity of bullet is a very large factor, especially when the terminal ballistics allow for most or all of the energy to be dumped into the target. And the history of cartridges, the ability to "reload", make your own custom load, and the ability to stand off to let air resistance bleed off velocity, means you'd likely be able to apply the exact amount of energy you'd want.
  • chronial 12y ago
    No, impulse is what matters here, and that is mv. And force is ma. You must be thinking of energy.
    • hga 12y ago
      Yeah, I'm thinking of energy, but in this case I'm assuming the terminal ballistics will dump all the bullet's energy into the block that's hit (that it won't penetrate all the way through the block). Impulse is force over a period of time; in this case, doesn't it equals the same thing?
• lotsofmangos 12y ago
  What you have to think about is the force of the hit, the friction and inertia of the block and the material properties of wood and metal. This would let you calculate the impulse of the hit, which should let you work out if the block will slide out before expanding much vertically as the bullet trashes it.