Each croquet ball in a set has a mass of 0.45

kg. The green ball, traveling at 12.6 m/s,
strikes the blue ball, which is at rest.
Assuming that the balls slide on a friction-
less surface and all collisions are head-on, find
the final speed of the blue ball in each of the
following situations:
b) The green ball continues moving after the
collision at 2.5 m/s in the same direction.
Answer in units of m/s.
c) The green ball continues moving after the
collision at 0.8 m/s in the same direction.
Answer in units of m/s.

To find the final speed of the blue ball in each situation, we can use the principle of conservation of linear momentum.

According to the principle of conservation of linear momentum:
Total initial momentum = Total final momentum

Let's denote the initial speed of the blue ball as vb(initial) and the final speed of the blue ball as vb(final). Since the blue ball is initially at rest, vb(initial) = 0.

Let's denote the mass of the green ball as mg = 0.45 kg, and the mass of the blue ball as mb = 0.45 kg.

b) In this situation, the green ball continues moving after the collision at a speed of 2.5 m/s in the same direction.

Using the principle of conservation of linear momentum:
mg * vg(initial) + mb * vb(initial) = mg * vg(final) + mb * vb(final)

Since the blue ball is initially at rest, the first term on the left side of the equation becomes 0.

mg * vg(initial) = mb * vb(final)

Substituting the given values:
0.45 kg * 12.6 m/s = 0.45 kg * vb(final)

Simplifying the equation:
5.67 kg m/s = 0.45 kg * vb(final)

Dividing both sides of the equation by 0.45 kg:
vb(final) = 5.67 kg m/s / 0.45 kg
vb(final) ≈ 12.6 m/s

Therefore, the final speed of the blue ball in situation b) is 12.6 m/s.

c) In this situation, the green ball continues moving after the collision at a speed of 0.8 m/s in the same direction.

Using the principle of conservation of linear momentum:
mg * vg(initial) + mb * vb(initial) = mg * vg(final) + mb * vb(final)

Since the blue ball is initially at rest, the first term on the left side of the equation becomes 0.

mg * vg(initial) = mb * vb(final)

Substituting the given values:
0.45 kg * 12.6 m/s = 0.45 kg * vb(final)

Simplifying the equation:
5.67 kg m/s = 0.45 kg * vb(final)

Dividing both sides of the equation by 0.45 kg:
vb(final) = 5.67 kg m/s / 0.45 kg
vb(final) ≈ 12.6 m/s

Therefore, the final speed of the blue ball in situation c) is 12.6 m/s.

To solve this question, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.

The momentum of an object is calculated by multiplying its mass by its velocity. Let's represent the mass of the green ball as m₁, the velocity of the green ball before the collision as v₁, the mass of the blue ball as m₂, and the final velocity of the blue ball as v₂.

For part (b) of the question, the momentum before the collision can be calculated as:

Initial momentum = (mass of green ball) * (velocity of green ball) + (mass of blue ball) * (velocity of blue ball)

Since the blue ball is at rest initially, its velocity is 0. Plugging the given values:

Initial momentum = 0.45 kg * 12.6 m/s + 0.45 kg * 0 m/s
= 5.67 kg·m/s

The momentum after the collision can be calculated as:

Final momentum = (mass of green ball) * (final velocity of green ball) + (mass of blue ball) * (final velocity of blue ball)

Since the green ball continues moving after the collision at 2.5 m/s in the same direction, the final velocity of the blue ball is equal to the velocity of the green ball after the collision (v₂ = 2.5 m/s). Plugging the given values:

Final momentum = 0.45 kg * 2.5 m/s + 0.45 kg * 0 m/s
= 1.125 kg·m/s

According to the conservation of momentum principle, the initial momentum should be equal to the final momentum:

Initial momentum = Final momentum

5.67 kg·m/s = 0.45 kg * 2.5 m/s + 0.45 kg * v₂

Let's solve this equation to find the final velocity of the blue ball in part (b):

5.67 kg·m/s = 1.125 kg·m/s + 0.45 kg * v₂
4.545 kg·m/s = 0.45 kg * v₂
v₂ = 4.545 kg·m/s / 0.45 kg
v₂ = 10.1 m/s

Therefore, the final speed of the blue ball in part (b) is 10.1 m/s.

Now, let's move on to part (c) of the question. We follow the same steps as before, plugging in the given values:

Initial momentum = 0.45 kg * 12.6 m/s + 0.45 kg * 0 m/s
= 5.67 kg·m/s

The final velocity of the blue ball is now given as 0.8 m/s (same as the velocity of the green ball after the collision).

Final momentum = 0.45 kg * 0.8 m/s + 0.45 kg * 0 m/s
= 0.36 kg·m/s

Applying conservation of momentum:

5.67 kg·m/s = 0.45 kg * 0.8 m/s + 0.45 kg * v₂
4.87 kg·m/s = 0.45 kg * v₂
v₂ = 4.87 kg·m/s / 0.45 kg
v₂ ≈ 10.82 m/s

Therefore, the final speed of the blue ball in part (c) is approximately 10.82 m/s.