Question

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.

Answer 1

To find the final speeds of the blue ball after the green ball collides with it, the conservation of momentum principle is used, considering the green ball's various final speeds post-collision for each situation. The final speeds of the blue ball are 14.4 m/s, 12.0 m/s, and 13.5 m/s, respectively.

Step-by-step explanation:

The problem involves conservation of momentum during collisions, as the surface is considered frictionless, and energy is conserved during perfectly inelastic collisions (although the question does not specify if the collisions are perfectly elastic or inelastic).

To solve for the final speed of the blue ball in each given situation, we'll apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Calculating the Final Speeds

1. When the green ball stops: Conservation of momentum implies that the momentum of the green ball before the collision is transferred entirely to the blue ball. Therefore, the blue ball's final speed will be the same as the initial speed of the green ball, i.e., 14.4 m/s.
2. When the green ball continues at 2.4 m/s: The momentum lost by the green ball is gained by the blue ball. By conserving momentum, the final speed of the blue ball can be found and is calculated to be 12.0 m/s.
3. When the green ball continues at 0.9 m/s: Again, by conserving momentum, the final speed of the blue ball is calculated to be 13.5 m/s.

Answer 2

a) 14.4 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:

`p_i = p_f`

So we have:

`m_g u_g + m_b u_b = m_g v_g + m_b v_b` (1)

where

`m_b = m_g = m = 0.53 kg` is the mass of each ball

`u_g = 14.4 m/s` is the initial velocity of the green ball

`u_b = 0` is the initial velocity of the blue ball

`v_g=0` is the final velocity of the green ball

`v_b` is the final velocity of the blue ball

Simplifying the mass in the equation and solving for
`v_b`, we find

`v_b = u_g = 14.4 m/s`

b) 12.0 m/s

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

`v_g = 2.4 m/s`

So the equation (1) becomes

`u_g = v_g + v_b`

And solving for
`v_b` we find

`v_b = u_g - v_g = 14.4 m/s-2.4 m/s=12.0 m/s`

c) 13.5 m/s

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

`v_g = 0.9 m/s`

So the equation (1) becomes

`u_g = v_g + v_b`

And solving for
`v_b` we find

`v_b = u_g - v_g = 14.4 m/s-0.9 m/s=13.5 m/s`