Each croquet ball in a set has a mass of 0.50 kg. The green ball, traveling at 12.0 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:

Question

a. The green ball stops moving after it strikes the blue ball.
b. The green ball continues moving after the collision at 2.4 m/s in the same direction.

Answer 1

a. Since the balls are sliding on a frictionless surface, momentum is conserved in this system. Therefore, we can use the equation:

m1v1_initial + m2v2_initial = m1v1_final + m2v2_final

where
m1 = 0.50 kg (mass of green ball)
v1_initial = 12.0 m/s (initial velocity of green ball)
m2 = 0.50 kg (mass of blue ball)
v2_initial = 0 m/s (initial velocity of blue ball)
v1_final = 0 m/s (final velocity of green ball)
v2_final = final velocity of blue ball

Plugging in the values, we get:

(0.50 kg * 12.0 m/s) + (0.50 kg * 0 m/s) = (0.50 kg * 0 m/s) + (0.50 kg * v2_final)
6.0 kg m/s = 0.50 kg * v2_final
v2_final = 12.0 m/s

Therefore, the final speed of the blue ball is 12.0 m/s after the green ball stops moving.

b. Using the same equation as above, we can write:

(0.50 kg * 12.0 m/s) + (0.50 kg * 0 m/s) = (0.50 kg * 2.4 m/s) + (0.50 kg * v2_final)
6.0 kg m/s = 1.2 kg m/s + 0.50 kg * v2_final
4.8 kg m/s = 0.50 kg * v2_final
v2_final = 9.6 m/s

Therefore, the final speed of the blue ball is 9.6 m/s after the green ball continues moving at 2.4 m/s in the same direction.