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Answer:a) v_{2f} = 12.0 m / s, b) v_{2f} = 9.6 m / sExplanation:This is an exercise in conservation of momentum, let's start by defining a system formed by the two balls, so that the forces during the collision have been internal and the moment is preserved.Initial instant. Before the crashp₀ = m v₁₀ + m v₂₀where we use index 1 for the green ball and index 2 for the blue ball.Final moment. After the crashp_f = m v_{1f} + m v_{2f}the moment is preservedp₀ = p_fm v₁₀ + m v₂₀ = m v_{1f} + m v_{2f}they tell us that the blue ball is at rest before the crashm v₁₀ = m v_{1f} + m v_{2f}a) it is indicated that the green ball stops after the collision v1f = 0m v₁₀ = m v_{2f}v_{2f} = v₁₀v_{2f} = 12.0 m / sb) the speed of the green ball is v_{1f} = 2.4 m / sm v₁₀ = m v_{1f} + m v_{2f}v_{2f} = v_{1o}- v_{1f}v_{2f} = 12.0 - 2.4v_{2f} = 9.6 m / s...