# Why does the bullet have greater KE than the rifle?

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.

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.

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.