Each croquet ball in a set has a mass of 0.53 kg. The green ball, traveling at 14.4 m/s, strikes the blue ball, which is at rest. Assuming that the balls slide on a frictionless surface and all collisions are head-on, find the final speed of the blue ball in each of the following situations: a) The green ball stops moving after it strikes the blue ball. Answer in units of m/s.b) The green ball continues moving after the collision at 2.4 m/s in the same direction. Answer in units of m/s. c) The green ball continues moving after the collision at 0.9 m/s in the same direction. Answer in units of m/s.

The problem can be solved by using the law of conservation of total momentum; in fact, the total initial momentum must be equal to the final total momentum:

[tex]p_i = p_f[/tex]

So we have:

[tex]m_g u_g + m_b u_b = m_g v_g + m_b v_b[/tex] (1)

where

[tex]m_b = m_g = m = 0.53 kg[/tex] is the mass of each ball

[tex]u_g = 14.4 m/s[/tex] is the initial velocity of the green ball

[tex]u_b = 0[/tex] is the initial velocity of the blue ball

[tex]v_g=0[/tex] is the final velocity of the green ball

[tex]v_b[/tex] is the final velocity of the blue ball

Simplifying the mass in the equation and solving for [tex]v_b[/tex], we find

[tex]v_b = u_g = 14.4 m/s[/tex]

b) 12.0 m/s

This time, the green ball continues moving after the collision at

[tex]v_g = 2.4 m/s[/tex]

So the equation (1) becomes

[tex]u_g = v_g + v_b[/tex]

And solving for [tex]v_b[/tex] we find

[tex]v_b = u_g - v_g = 14.4 m/s-2.4 m/s=12.0 m/s[/tex]

c) 13.5 m/s

This time, the green ball continues moving after the collision at

[tex]v_g = 0.9 m/s[/tex]

So the equation (1) becomes

[tex]u_g = v_g + v_b[/tex]

And solving for [tex]v_b[/tex] we find

[tex]v_b = u_g - v_g = 14.4 m/s-0.9 m/s=13.5 m/s[/tex]