Two austronauts A and B, Participate in 3 collision experiments in a weightless,

Question

Two austronauts A and B, Participate in 3 collision experiments in a weightless, frictionless environments.

Conservation of momentum in one dimension Two astronauts. A and B. participate in three collision experiments m a weightless frictionless environment. In each experiment astronaut B is initially at rest and astronaut A has an initial momentum of 20 kg m/s to the right. (The velocities of the astronauts are measured w lo a nearby space station.) The astronauts push on each other in different ways so that the outcome of each experiment is different. As shown in the figure at right, astronaut A has a different final momentum in each experiment. Determine the magnitude of the final momentum of astronaut B in each experiment. Explain. Rank the three experiments according to the final kinetic energy of astronaut B. from largest to smallest. Explain. Is the total kinetic energy after the collision in experiment 2 greater than, less than, or equal to that after the collision in experiment 3? (In this ease, total kinetic energy means the sum of the kinetic energies of the two astronauts.) Explain. Consider the following statement: 'The momentum of the system is conserved in each expedient because there t no net force on the system If momentum if conserved, then kinetic energy must also be conserved, because both momentum and kinetic energy are rode up of most end velocity. One of the sentences above is completely correct. Discuss the error(s) in reasoning in the other sentence

Explanation / Answer

"If momentum is conserved, then KE must also be conserved...." This is only the case in elastic collisions.

Consider the completely inelastic collision where a mass of 1kg with velocity 1 m/s collides with an identical mass at rest, after which the two masses stick together and move as one extended object. Conservation of momentum gives p_initial as mv = (1 kg)(1 m/s) = 1 kg m/s, so p_final = 1 kg m/s. But the initial KE of the system is (1/2)(1 kg)(1 m/s)^2 = 0.5 J. Since the final momentum is 1 kg m/s and the final mass is 2 kg, v_final = p_final/m_final = (1 kg m/s)/(2 kg) = 0.5 m/s, and the final KE = (1/2)(2 kg)(0.5 m/s)^2 = 0.25 J.

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