I believe that you are asking if the bullet is what causes so much damage to something it strikes, then why doesn't the shooter suffer an equivalent amount of damage? The answer does have a lot to do with momentum. I think that there may be some confusion between the conservation of momentum, which is the essence of Newton's Third Law, and the definition of kinetic energy.
The magnitude of the momentum of the bullet just as it leaves the barrel of a gun is equal to, but in the opposite direction of, the magnitude of the momentum of the gun plus any momentum imparted to the shooter's body.
I apologize if you already know much of my answer, but I will show the equations for momentum and kinetic energy. Momentum is a vector because it has both direction and magnitude, and the equation for momentum is p = m•v, where p is momentum, m is the object's mass in kg, and v is the object's velocity in meters/sec. (The vector quantities are in bold. I'm using the units prescribed in the kms system.)
The reason the bullet causes damage to whatever it strikes is due to the bullet's kinetic energy, which is defined by the equation Ek=1/2mv2, where m is the mass of the bullet in kg and v is the bullet's speed in meters/sec. The fact that the kinetic energy is proportional to the square of the velocity is of paramount importance. Just like momentum, energy is also conserved. Energy (or the combination of energy and mass in the cases of nuclear fission and nuclear fission) is never lost; it can only change forms. All the energy released from every bullet, bomb, and shell used in World War II, or any other war, is still present somewhere in the universe, although almost all of it is still present on the Earth in some form. Thus, the kinetic energy of a bullet fired from a gun is initially exactly equal to the kinetic energy imparted to the backward-moving gun, the gun's report, the flash of light from the muzzle, the heating of the gun's receiver, chamber, and barrel, the heating of the bullet and of whatever it hit, the frictional heating of the air through which the bullet passed, and the energy expended in damaging what the bullet struck.
The reason the shooter of the gun is not severely damaged (although anyone who has ever fired a very powerful rifle such as a 375 H&H or a 460 Wetherby Magnum might disagree) is because the energy of the bullet is proportional to the square of the velocity. Because the mass of the gun is so much greater than the mass of the bullet, the rearward velocity of the gun is very much less than the velocity of the bullet such that the gun cannot cause significant damage to the shooter.
Third law: When a first body exerts a force F1 on a second body, the second body simultaneously exerts a force F2 = −F1 on the first body. This means that F1 and F2 are equal in magnitude and opposite in direction.
Before the shot, total momentum of the rifle/bullet system is zero. Momentum is conserved, so must total zero after the shot. Magnitude of momentum = m V (mass, speed); we'll take care of direction independently. Momentum of the rifle: m V = (3.8) x (2.4) = 9.12 kg-m/sec backwards. We need momentum of the bullet = 9.12 kg-m/sec forward m V = 9.12 ===> V = ( 9.12 / m ) = ( 9.12 / 0.013 ) = 701.54 m/s forward
50 g = 0.05 kgF = m AA = F/m = 4000/0.05 = 80,000 meters per second2If the bullet starts from rest and no other forces act on it, thenSpeed = (acceleration) x (time) = (80,000) x (0.01) = 800 meters per second
the conversion of momentum law states that the total momentum of twos systems must be equal therefore M1V1 = M2V2 i am assuming the mass of the bullet is 0.0050 kg and not 50kg so (0.0050 * 250) = ( 9 * X) X = (1.25 / 9) X = 0.139 You can't answer these kind of questions with so few parameters. The bullet diameter, barrel length, powder burn rate all greatly effect the answer. The recoil is caused mainly by the gas exiting the barrel, hence muzzle brakes work.
No. Assuming you have a gun that can fire in space, you would travel in the opposite direction, conserving momentum, but because you are more massive than the bullet, your speed in the opposite direction would be less. Momentum is mass*velocity, so your speed should be less than the bullet's by the same factor as the ratio of the bullet's mass to your own. You would also start rotating unless your shot was perfectly aligned with your center of mass.If that's not convincing, consider what happens if you shoot a bullet straight into the air. Does the earth move in the opposite direction at the same speed?
The mass of a bullet is nowhere near the mass of a gun. A bullet weighs at most a few hundred grains. Most guns weigh at least a couple of pounds, some weigh several pounds (talking about handguns and rifles).
Yes, the bullet fired from a recoiling rifle typically has a greater velocity compared to the rifle itself.
The total momentum of the bullet and the gun before firing is zero, as the gun and the bullet are at rest. Momentum is the product of mass and velocity, and since both the gun and the bullet are not moving, their momentum is zero.
If the gun is stationary before the shot, then the momentum of the gun and the momentum of the bullet are equal and opposite after the shot.
Momentum = mass x velocity A bullet has a high momentum because its velocity is really high.
By Newton's third law of motion, for every action, there is an equal and opposite reaction. When a gun fires a bullet, the bullet gains momentum in one direction, causing the gun to recoil in the opposite direction with an equal amount of momentum. So the momentum of the bullet is equal in magnitude to the momentum of the gun, but in opposite directions.
The momentum of the bullet fired from a rifle is greater than the momentum of the rifle's recoil. This is because the bullet has a higher velocity and mass compared to the rifle, resulting in a greater momentum.
This is conservation of momentum. You have the hot gases from the explosive charge along with the bullet moving out the barrel, away from the person. Momentum is mass times velocity. While the mass of the bullet and gases are small, the velocity is very high. So nothing was moving before the trigger was pulled, so net momentum is zero. After the trigger is pulled, the momentum is still net zero. Any momentum away from the gunner will have an equal momentum (the gun recoiling) toward him. Since the gun has much more mass than the bullet, the velocity is much less.
Newton said "For every action, there is an equal and opposite reaction." The action is pushing the bullet in one direction. Reaction is the gun being pushed in the other. The energy is speed times mass. The gun masses much more than the bullet, and so will move proportionately slower than the bullet. Which is a very good thing- or the recoil of the gun would kill the shooter.
A bullet fired from a gun has more momentum than a train at rest because momentum is the product of an object's mass and velocity. The bullet, despite being smaller in mass compared to the train, can have a significantly higher velocity, resulting in a greater momentum.
Third law: When a first body exerts a force F1 on a second body, the second body simultaneously exerts a force F2 = −F1 on the first body. This means that F1 and F2 are equal in magnitude and opposite in direction.
Before the shot, total momentum of the rifle/bullet system is zero. Momentum is conserved, so must total zero after the shot. Magnitude of momentum = m V (mass, speed); we'll take care of direction independently. Momentum of the rifle: m V = (3.8) x (2.4) = 9.12 kg-m/sec backwards. We need momentum of the bullet = 9.12 kg-m/sec forward m V = 9.12 ===> V = ( 9.12 / m ) = ( 9.12 / 0.013 ) = 701.54 m/s forward