# How come shooter does not receive same energy as bullet (Newton's third law)?

So for example, a bullet weighting 7 g and going out of barrel at 420 m/s has ~617 joules of kinetic energy. So I'd think same would apply to shooter (or his hand specifically) - a 80 kg shooter would be tossed back at ~3.9 m/s? Obviously that doesn't happen even with bigger and faster bullets.

Where am I wrong?

Because Newton's third law isn't about energy. It's about momentum. Momentum's a vector quantity, so in the situation where all the momentum goes into the shooter and the bullet, in the reference frame where the momentum is zero to start with and there's nothing else in the system the bullet departs at 420 m/s, and the shooter is pushed in the opposite direction at 0.037 m/s. Momentum is conserved.

Looking at it from the energy position, the bullet has 617 Joules of kinetic energy. upon firing, which came from the energy of its propellant system. Depending on how the gun in question works, this could be stored mechanical, chemical (typical with firearms), or heat energy, or some other type. But energy is a scalar quantity, it doesn't have a direction.

Some of that propellant energy went into heat energy, some of it went into sound, some of it went into the bullet, and some of it went into the gun. But there's no requirement that any individual object mentioned here received an equal amount of energy as any other, and energy doesn't have a direction, just that the total of all energy equal the energy generated by the propellant system. Energy is conserved.

• Oh, so I should use V1*m1 = V2*m2 ? I remember learning this in school, but I never understood why shouldn't we use kinetic energy. One of tasks was "two cars going into each other, measure the speed after hit. Don't account the deformation" - in this problem I should use the vector formula (the one I typed here) or kinetic energy formula? Commented Aug 18, 2023 at 11:32

As already pointed out, Newton’s 3rd law is not about conservation of energy.

As for conservation of momentum, the shooter would need to be standing on a frictionless surface in order to acquire momentum equal and opposite to that of the bullet.

Momentum is only conserved for a system (in this case the shooter-bullet system) when there are no external forces acting on the system. In this case the friction force (static or kinetic depending on whether or not the shooter slips) exerted by the ground on the shooter’s feet is an external force on the shooter/bullet system. If the Earth is included in the system, momentum is conserved.

Hope this helps.

The following may be a worthwhile addition to the answer by contributor notovny.

The following simplified model is an instance of energy distributing in such a way that we can think of it as an even distribution.

Let's way we have a vessel filled with a diatomic gas. From outside we introduce heat. The molecules of the gas have a velocity distribution. Adding heat means the average of the velocity distribution goes up. The fact that the content is a diatomic gas makes a difference. There is the kinetic energy of spinning of the diatomic molecules. Some of the molecule-molecule collisinons are glancing collisions, and in such a collicion linear kinetic energy of motion in a straight line can transition to rotational kinetic energy of a molecule spinning.

Generally, the energy will tend to flow from high to low. So if a particular molecule happens to be spinning very fast, and it happens to nudge another slow molecule, then the most likely outcome is that the two molecules fly apart.

What tends to happen is that when a system has multiple degrees of freedom, then the energy content of that system tends to become redistributed in such a way that all degrees of freedom have the same level of energy.

Comparison: the case of communicating vessels. Fill any one of the set of communicating vessels, and over time the liquid becomed redistributed, such that across all vessels the level is equal.

Returning to energy:
An example on 1-to-1 distribution:
If you spin a liquid around the surface assumes a concave shape. At the perimeter the following two energies are larger than at the center:
-At the perimeter the kinetic energy is larger
-At the perimeter the liquid is higher, so the potential energy is larger

In fact, the outcome is that when the liquid is uniformly rotating then at every distance to the center of rotation the amount of kinetic energy (relative to the center of rotation) is equal to the amount of potential energy (relative to the center of rotation)

So there are cases of equipartition of energy, but the case of a projectile fired from a barrel is a different case altogether.

To avoid confusion, I would have preferred Newton's third law worded as "for every force, there is an equal and opposite force". If you multiply the speed by the mass of each, they are the same. Energy is a combination of force x time x speed. The gun and the bullet are exposed to the same force for the same time, but the speed is different.