How much momentum would a bullet need to have in order to cause James Bond to fall? An investigation by Zaid Syed and Nirav Agarwal

So, would this be actually possible, neglecting air resistance?

Newton's third law of motion dictates that every action has an equal and opposite reaction. When Eve shoots the gun, the momentum is conserved—the momentum of the bullet leaving the gun is equal and opposite to the momentum of the recoil going in the opposite direction.

Eve uses an Olympics Arms K23B Tactical assault rifle. The mass of one bullet is 0.04kg, and the muzzle velocity is 853m/s.

The momentum of the bullet is p = 0.04 x 853m/s, which is 34.12 kg m/s. The recoil would be roughly the same — the shooter, Eve, would have roughly the same momentum due to the recoil of the gun. Assuming her weight is 60kg, the velocity by which she moves back is 0.568 m/s.

If James Bond were to move with this same momentum (as momentum is conserved), as his mass is 80kg, his velocity would be 34.12/80 = 0.4265m/s, a lot less than the velocity by which Eve moved. Thus, Daniel Craig would not fall off the train due to as small a velocity as this.

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