Задача id11953

A gun, rigidly attached to an aircraft at rest, imparts kinetic energy mv2/2 to a projectile of mass m. Determine the energy of the projectile relative to the Earth if the shot is fired from an airplane flying horizontally with speed v, in the direction of its flight. The students gave different answers to the question in the problem. 1st student: Before the shot, the projectile together with the plane was flying at a speed v relative to the Earth and, therefore, had kinetic energy Exn = mv2/2. When fired from a cannon on a flying aircraft, the kinetic energy of the projectile changed in the same way as when fired from an aircraft at rest, i.e. ΔEkсн=mv2/2, therefore, the kinetic energy of the projectile relative to the Earth after being fired from a flying plane is: E,kсн=Ekсн + ΔEkсн = mv2/2 + mv2/2 = mv2 2nd student: The speed of the projectile before the shot in a flying plane was equal to v relative to the Earth. After the shot, the speed of the projectile increased by v. According to the law of addition of velocities, the speed of a projectile after a shot relative to the Earth is equal to 2v, and, therefore, its kinetic energy is equal to: E,kсн=m(2v)2/2=2mv2 Which reasoning is correct?

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