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A rifle is fired, and the bullet moves much faster than the rifle, and momentum is conserved.

My question is whether the kinetic energy of the bullet is greater than the kinetic energy of the rifle because the mass of it is smaller, therefore the force acting on it (for the same time it acts on the rifle) results in greater acceleration, and thus greater final velocity (basically, the change in momentum needs to be equal and in opposite direction to the rifle, which has greater mass). On the other hand, kinetic energy depends on work done, which is force times displacement. Therefore, we can say that the force acted on the bullet for a greater distance traveled.

Are these explanations stating the same thing, or are they causally different? (From the expression of the momentum formula as m. dx)

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Conservation of momentum implies

$$mv+MV=0$$ where $m$ and $v$ are the mass and speed of the bullet, $M$ and $V$ of the rifle. Of course, it is $0$ in total as the system is at rest at the beginning.

This implies

$$|v|=MV/m$$ so the smaller the bullet, the bigger its speed.

In terms of KE

$$ke={1\over 2} m v^2={1\over 2} M^2V^2/m$$ for the bullet whereas

$$KE={1\over 2} MV^2$$ for the rifle.

Thus the ratio is given by $$ke/KE={M/m}$$

Now, a gun/rifle is in the 5-10 kg range whereas for a bullet we have 10 g (rounding up) so you - just by momentum conservation - get a bullet's kinetic energy 500-1000 times bigger than the rifle's.

(Notice that from this perspective we can only extract the ratio between the two energies. The absolute values would be given by information about the energy generated by the gunpowder. But that energy is partitioned between bullet and rifle according to momentum's conservation.)

Now, because - as you mention - the change in kinetic energy is the same as the work done, this indeed means that the work done on the bullet is greater than the work done on the rifle.

To see it, assume the explosion generates a constant force $F$ on both the rifle and the bullet, but in different directions (because of action-reaction), and that the explosion lasts for $t_0$ seconds (this is of course an approximation because $F$ is in general a non-constant force but we approximate it as a mean constant force applied for a mean time $t_0$)).

The bullet position over time will be $$x={1\over 2}at^2={1\over 2}{F\over m}t^2$$ where I used $a=F/m$.

Thus at the end of the interaction ($t=t_0$) the bullet will have travelled a distance $$d={1\over 2}{F\over m}t_0^2$$ If we do the same for the rifle, we get $$D={1\over 2}{F\over M}t_0^2$$ so you see that despite $F$ and $t_0$ being constants, the bullet moves more as it is smaller ($m\ll M$ so $d\gg D$).

Now, because work is force times distance, this means that the work done on the bullet ($w=Fd$) is bigger than the work done on the rifle ($W=FD$) by the same ignition. Their ratio is of course $$w/W=Fd/FD=d/D=M/m$$ which is the same ratio of the kinetic energy (and it has to be as work done = change in kinetic energy).

(Again, we only know the ratio as we don't know the value of $F$ - we just know that because of action-reaction it will be the same in magnitude on both bullet and rifle.)

So a simple mass difference makes it so that the bullet accelerates more hence gets more kinetic energy in the same time.

To this you need to add the fact that:

  1. as you said, one usually holds the rifle fixed so indeed the bullet might get more of the ignition force transformed into momentum.
  2. bullets are small and their energy is "condensed" into a minimal surface making them very good at perforation. Also, they have all sort of structural details made to decrease air friction, etc.

But from a purely momentum conservation point of view you can understand that the bullet, just by being smaller, absorbs a bigger part of the total energy and therefore moves at high speed with respect to the rifle.

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    $\begingroup$ Nice explanation. Indeed, rifle passes some recoil to human body holding it, so in principle in momentum conservation equation $M=M_{rifle}+M_{human}$. And it may sound fantastic, but it's true. Human is fixed at Earth surface by his foots, when shooting, so part of shoot recoil could be passed by human legs to Earth as change in Earth angular velocity, because it must be conserved too. This may be a ridiculously small effect (if any), but imagine ALL world people shooting at the same time & at the same direction, then maybe it could affect Earth spin considerably. $\endgroup$ Feb 16 at 8:07
  • $\begingroup$ Interesting question would be if damage done if proportional to momentum or energy. $\endgroup$
    – lalala
    Feb 16 at 11:17
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    $\begingroup$ @lalala Damage done is most definitely linked to Kinetic Energy in the projectile. Not quite linearly, as higher energy projectiles tend to go though their target, wasting some of their energy. But most definitely kinetic energy, not momentum.(There's also a factor of kinetic energy per impact area to consider, smaller area does more damage from the same energy. But that gets very complicated very fast) $\endgroup$
    – PcMan
    Feb 16 at 12:54
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    $\begingroup$ @AgniusVasiliauskas Unless the bullets (or the combustion gases) escape Earth's gravity, there would be no long-term change. The bullet's momentum would be re-absorbed by the Earth, partly via the atmosphere and the rest when the bullet landed back on the ground. Angular momentum is conserved in a closed system. $\endgroup$
    – J...
    Feb 16 at 18:59
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    $\begingroup$ @PcMan Indeed, it's even messier than that - smaller area does more damage per energy up to a point, but at the point at which you're overpenetrating, you want more area, to minimise how much energy goes out the back of the target. There's also some non-physics questions about exactly what you mean by "damage" (if the target is a battleship, penetration matters more than width of damage, because the outer layers are just armour. If the target has all of its valuable bits on the surface, the opposite is true). $\endgroup$ Feb 17 at 1:32
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In effect you are asking what the nature of kinetic energy is.

After the rifle has been fired the bullet and the rifle have the same momentum - in opposite direction. In that sense we can say that momentum is equally shared.

To get a closer look at what kinetic energy is I propose the following demonstration: you set up a series of ribbons that can be snapped relatively easily. Think the kind of ribbon at the finish line of a runners event. The ribbon is strong enough to span the width of the road, but it snaps easily.

Let's say you set up a series of ribbons like that, spaced equally. You set up an object with an initial velocity, such that that mass is decelerated by stretching ribbon after ribbon. For simplicity assume that as each ribbon snaps the object is in touch with the next ribbon so that the deceleration is fairly constant.

Let's say the deceleration is 1 unit of velocity per unit of time.

Now compare: initial velocity 1 unit of velocity versus 2 units of velocity. When the initial velocity is twice as large it takes twice as much time to come to a standstil. But distance traveled during acceleration/deceleration is proportional to the square of the duration of the acceleration/deceleration.

That is: the object with twice the initial velocity will snap four times as much ribbons.

Let's divide the deceleration in two phases, from 2 units of velocity to 1 unit of velocity, and then from 1 unit down to zero.
Starting with 2 units of velocity the object is snapping ribbons, and by the time the velocity is down to 1 unit of velocity the object has snapped a lot of ribbons. During the deceleration from 1 unit of velocity to zero velocity the amount of change of velocity is the same as from 2 to 1, but it's all with a smaller velocity, so less ribbons are snapped.


That is what is expressed with the concept of kinetic energy.

The more velocity an object has the more damage it will do upon hitting something, in proportion to the square of the velocity.

This quadratic relation comes from the fact that during acceleration/deceleration the distance travelled is proportional to the square of the duration.

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The bullet and rifle have the same, but oppositely directed momentum, say p. The kinetic energy of each object is $p^2/2m$. The smallest mass has the highest kinetic energy.

On a pedantic note, I assumed that the rifle was at rest before it fired. For a sensible rifle this should hold. If someone holds the rifle the mass of that person should be added and transient effects will occur. Dissipation by friction is also ignored.

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Same time interval does not mean same distance. It means the same impulse which results in the same change in momentum (in magnitude ) As you say, the momentum is conserved. On the other hand, the distances travelled by gun and bullet depend on the accelerations of the two objects and these are not the same, as you already mentioned. So the work done on the bullet is much higher than the work on the gun, even if the gun is not supported by the shoulder.

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  • $\begingroup$ I cannot see that the support of a rifle is relevant. It still has kinetic energy. The shock is absorbed, but it did exist. $\endgroup$
    – mckenzm
    Feb 16 at 1:48
  • $\begingroup$ You can think about it as the mass of the rifle is increased. If you rigidly attach the rifle to the Earth almost all the KE goes to to the bullet. If it's firmly pushed against the shoulder it is like adding some your mass to the riffle. Or yu can just think in terms of work. If you reduce the displacement of the rifle, you reduce the work done so you reduce the KE taken by the rifle. See the other answers for quantitative results. $\endgroup$
    – nasu
    Feb 16 at 13:44
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Kinetic energy is not conserved when you fire a bullet. Both the rifle and bullet are at rest until the explosive in the shell provides energy to accelerate the bullet down the barrel. Momentum is, however, always conserved. If we align our coordinates with the barrel and compare the system before the shot (when the system is at rest) and just after,

$$ \begin{align} \text{momentum: } 0 &= m_bv_b + m_rv_r \\ v_b &= -\frac{m_r}{m_b}v_r\\ \text{kinetic energy: } KE_r &= \frac{m_rv_r^2}{2} \\ KE_b &= \frac{m_bv_b^2}{2}\\ KE_b (v_r) &= \frac{m_r^2v_r^2}{2m_b}\\ \frac{m_rv_r^2}{2} & \stackrel{?}=\frac{m_r^2v_r^2}{2m_b}\\ 1 & \stackrel{?}=\frac{m_r}{m_b}\\ \end{align} $$ The bullet has much more kinetic energy because the mass of the rifle is much larger than the mass of the bullet: $1 \ll \frac{m_r}{m_b}$.

Perspective:

It can be shown that the momentum, force and energy pictures (and their more advanced versions) are self-consistent. When a problem may be solved in many pictures, they all give the same result.

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I think the easiest way to work this out is to create an arbitrary model and simplify away some of the complexities that don't directly related to the question. So Let's say we have a bullet that is 10g, a 'gun' that is 10kg. Furthermore, our very simple gun creates a constant acceleration on the bullet while it is firing: 10000m/s^2. After firing is initiated the bullet leaves the muzzle after 0.1 seconds. Also the gun is floating in the vacuum of space.

We can therefore determine that the bullet's muzzle velocity is 1000m/s. We can also calculate the force applied as 0.01kg*10000m/s^2 or 100N.

We know that this force is also applied to the gun as well in the opposite direction. So 100N applied to the 10kg gun -> 100N=10kg*a gives us an acceleration of 10m/s^2. Applying that acceleration over 0.1s means that the gun, after firing, is moving at 1m/s.

Now we calculate the respective kinetic energies:

  • K-bullet = 0.50.01(1000^2) = 5000J

  • K-gun = 0.510(1^2) = 5J

So the upshot here is that the force is equal as is the time it is applied but due to the lower mass, the bullet accelerates to much higher velocity which has a quadratic impact on the kinetic energy than the mass. This is more-or-less your first argument.

Knowing myself, I've probably made a calculation error here. Corrections welcome.

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They key is on the definitions of linear momemtum $\mathbf{p} = m \mathbf{v}$ and kinetic energy $T = \frac{1}{2} m v^2$. Particullary, the fact that KE depends cuadratically on speed while $p$ depends linearly on speed.

If you write the KE in terms of linear momentum $$ T = \frac{p^2}{2m} $$ you realise KE is inversely proportional to mass.

So for the same momentum $p$ (both rifle and bullet have the same momentum since it's conserved), bullet has more KE.

If you defined another magnitud such as $T' = \frac{1}{2} \ln(m)\,v^{3/2}$, you could write it in terms of the momentum as \begin{align*} T' &= \frac{1}{2} \ln(m) \frac{p^{3/2}}{m^{3/2}} \\ T' &= \frac{\ln(m)}{2 m^{3/2}} p^{3/2} \end{align*}

You see that $T'$ isn't inversely proportional to mass. It could behave as crazy as you want based on your definition of KE.

Of course, the definition of KE isn't arbitrary.

Edit: I know this doesn't answer the physical part with the correct definition for KE. Since it was already answered, I just wanted to point out this pretty trivial fact.

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Basic principal of ballistic physics, Once the round leaves the barrel it begins to lose energy. Kinetic energy is energy PUT to work in motion. The bullet may have less mass than the rifle but remember the cartridge stores propellant, that's Chemical energy (Potential) Also most rifles this day have some sort of recoil management to mitigate the energy opposing delivered. The rifle weighs (depending on model) from 5-10 lbs? the energy recoil is mitigated through the body of the rifle.

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  • $\begingroup$ This answer completely misses the point of the question. The OP is asking why the rifle gets a smaller share of the total energy released by the cartridge than the bullet, not how the cartridge stores it. $\endgroup$
    – mathrick
    Feb 17 at 1:35

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