Ykhjuy

Firstly, some guns do give quite a kick! So the effect you are thinking of is real.

However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.

The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.

Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.

Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.

Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.

Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.

One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"

"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."

There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.

By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.

If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.

Lets nail down some figures with a random choice of compatible equipment with specs publicly available.

Gun https://us.glock.com/products/G19

Gun Weight with empty magazine 670 g | 23.63 oz

Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)

Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum

Bullet mass 8.04 g (124 gr) Federal FMJ

Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)

We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.

The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)

The gun weighs a lot (83 times) more than the bullet.

It is coupled to a human that weighs very much more.(180lb = 10155 times)

It is coupled over a large, soft, energy absorbing surface(hand).

The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.

Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.

Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.

But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".

Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.

So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.

Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.