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Momentum, kinetic energy, and bullets

  1. Dec 9, 2003 #1
    Hi all. My understanding of physics is...well not so good.

    I have a question or three that I'll try to relate with my limited knowledge, so bare with me.

    With bullets, a rating of KE is generally given by the bullets known mass(given in grains/1/7000th pounds) and velocity(given in fps). The momentum, however is not given. Momentum is mass*velocity correct? I've been told that the KE is the abilty to do work. Bullet energy(from manufacturers) is usually measured in FT/LBS. How does KE relate to momentum? From what I see there is virtually no relation. For instance, a projectile may have much more momentum while having much less KE and vice versa. It seems to me that having much greater momentum per it's Kinetic energy would give more "abilty to do work". An example, a 100 grain bullet at a velocity of 1500fps has nearly identical KE to a 250 grain bullet moving at a velocity of 950fps, yet the 250 grain bullet will have over 50% more momentum. How does the extra momentum not translate into more work, since it will have more resistance to change in velocity and direction on impact? Maybe...the high vs. low momentum bullets are using available energy in different ways? Given that the total kinetic energies of the two projectiles are equal I know that thier total abilities are equal, and I am not questioning that, just trying to gain an understanding.
    BTW, are there any units of measurement for momentum that can be found with a known mass and velocity?

    Thanks in advance for any replies. I've not had the chance to study physics and am admittedly ignorant on this subject, but have a desire to learn. Please go easy on me if I'm way off base here.

    Last edited: Dec 9, 2003
  2. jcsd
  3. Dec 9, 2003 #2


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    Kinetic Energy is expressed as

    [tex] KE = \frac 1 2 mv^2 [/tex]

    note that the velocity is squared.

    Momentum is expressed

    [tex] p = mv [/tex]

    While both of these quanities are functions of the mass and velocity they have different meanings and differnt uses. The Kinetic energy is an expression of the ability to do work, why? Well work is given as

    [tex] W = F d [/tex]

    F here is force, which is given by

    [tex] F = {mass}* {acceleration} [/tex]

    acceleration has units of

    [tex] a = \frac {Length} {{Time}^2}[/tex]

    now if we but these last pieces together we can see that the fundamental units of work are
    [tex] W = {mass} \frac {Length} {{Time}^2} * {Length} [/tex]

    combine the 2 lengths to get

    [tex] W = {mass} ({\frac {Length} {Time}})^2 [/tex]

    This is the same units as those of Kinetic energy because length over time is velocity so the last term is the above expression is the velocity squared.

    A good thing to learn about Physisc is that you can check your answer by simply computing the units, this is called dimensional analysis, if the units are correct chances are you have the correct solution.

    Speaking of units I see that you have writen your Kinetic Energy as

    [tex] KE = \frac {ft} {lbs}[/tex]

    This cannot be correct! KE must be given in Ft-Lbs. Do you see the difference? Where did ft lbs come from? Look at the above expression for work, as force times distance. The english unit for mass is actually SLUGS, pounds is a unit of force so ft-lbs is the correct unit for energy or work.

    Now why cannot you use momentum to compute the work done by a bullet? Simply because it is not the work done! Momentum is a different quanity with different units, like apples and oranges work and momentum cannot be compared.
    Last edited: Dec 9, 2003
  4. Dec 30, 2003 #3
    KE or momentum?

    so can can you use acceleration as a substitute? once fired and out of the barrel, the bullet is no longer accelerating.
  5. Dec 30, 2003 #4
    It's useful to consider Impulse when considering momentum and how it relates to kinetic energy.

    Impulse represents a change in momentum just as work represents a change in energy. Mathematically impulse is defined as

    [tex]I = F t[/tex] or [tex] I = m a t [/tex] since [tex]F = m a[/tex].

    Both Impulse and momentum are measured in Newton*seconds and you can analyze their dimensions to verify this.

    The change in momentum (impulse) and Kinetic energy are related in a specific, mathematical way.

    [tex]E_k = \frac{I^2}{2 m}[/tex]

    Given the same impulse, it can be seen how the kinetic energy varies for objects of different mass.

    The same impulse applied to an object twice as massive will result in a change in kinetic energy which is one half. The same impulse applied to an object with half the mass will gain twice the kinetic energy.

    Kinetic energy and mass are inversely proportional with respect to impulse.

    Given the same mass and varying impulse, an object which receives twice the impulse that another equal mass receives will gain four times the kinetic energy. If the impulse changes by one half, the kinetic energy will decrease by a factor of four or increase by a factor of 1/4.

    Kinetic energy varies by the square of the impulse with respect to mass.
    Last edited: Dec 30, 2003
  6. Dec 31, 2003 #5


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    That's a good example. Since we're talking about bullets, the relevant thing is how the energy is converted as the bullet is decelerated to a stop when it hits something. As Integral points out, KE is proportional to mass times speed squared, so indeed, the 2 cases you give have almost identical KE. The amount of energy is definitive, so the two cases will do roughly the same damage provided of course we compare cases when all the KE is used up. One difference that will be easily apparent is that the gun in the heavy-bullet case will have more recoil.

    One can think of energy absorption as force times distance, and momentum absorption as force times time. Hence, the heavier but slower bullet with the same energy will travel the same distance in the absorbing material, but because of larger momentum, will take a longer time doing it. It will thereore also impart a greater "kick" to the absorber object.
  7. Dec 31, 2003 #6
    yeah that makes sense. thanks. so when one would like to stop a bullet in flight, which one to use, the kinetic energy or the momentum?
  8. Jan 25, 2004 #7
    Hi guys, noobie here and my first post. I thought since this thread is about bullets my question will be more appropriate than starting a new thread. Hope Scoob doesn't mind. :)

    I was just wondering between a lighter and heavier bullet of the same "power factor" (bullet wt(in grains) X velocity (in feet/sec)), which one does more work to the slide and cause it to move faster rearward?

    Sorry for the very basic question but these sort of things totally elude me.

  9. Jan 26, 2004 #8


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    Two things happen to an object hit by a bullet; it's messed up, and it's thrown back. The first one requires WORK to break chemical bonds and such and that's proportional to ENERGY, in this case kinetic energy. The kick back is produced by conservation of momentum; every bit of momentum (mass times speed) that the bullet has has to be transferred to the target.

    Mass of bullet X its speed at impact = Mass of target X its kickback speed
  10. Jan 30, 2004 #9
    impact result

    ok so let me get this straight.... the distance that the bullet can travel is ofcourse governed by the kinetic energy and trajectory of the bullet. but what about when the bullet reaches its target... how do you determine the force at which it hits the target and the effect it has on the target?
  11. Jan 31, 2004 #10


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    The bullet in flight has energy and momentum. It does not have "force" in the same sense.

    One way to calculate the force necessary to stop a bullet is:
    Knowing the speed and mass of the bullet, calculate its kinetic energy. You must use that much energy to stop it. The formula for energy due to a force is E= Fx where F is the force and x the distance over which the force is applied: F= E/x. Notice that you cannot simply ask how much force is required to stop the bullet. If you fire a bullet into a block of wood, the bullet will penetrate further and experience less force, for a greater distance, than if you fire the bullet at a sheet of steel.
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