Answer:
a) v_{2f} = 12.0 m / s, b) v_{2f} = 9.6 m / s
Explanation:
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 crash
p₀ = m v₁₀ + m v₂₀
where we use index 1 for the green ball and index 2 for the blue ball.
Final moment. After the crash
p_f = m v_{1f} + m v_{2f}
the moment is preserved
p₀ = p_f
m v₁₀ + m v₂₀ = m v_{1f} + m v_{2f}
they tell us that the blue ball is at rest before the crash
m v₁₀ = m v_{1f} + m v_{2f}
a) it is indicated that the green ball stops after the collision v1f = 0
m v₁₀ = m v_{2f}
v_{2f} = v₁₀
v_{2f} = 12.0 m / s
b) the speed of the green ball is v_{1f} = 2.4 m / s
m v₁₀ = m v_{1f} + m v_{2f}
v_{2f} = v_{1o}- v_{1f}
v_{2f} = 12.0 - 2.4
v_{2f} = 9.6 m / s
