Two shuffleboard disks of equal mass, one orange and the other green, are involved in a perfectly elastic glancing collision. The green disk is initially at rest and is struck by the orange disk moving initially to the right at oi = 5.90 m/s as in Figure (a) shown below. After the collision, the orange disk moves in a direction that makes an angle of θ = 39.0° with the horizontal axis while the green disk makes an angle of = 51.0° with this axis as in figure (b). Determine the speed of each disk after the collision.

To determine the speed of each disk after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

1. Conservation of Momentum:
The total momentum before the collision is equal to the total momentum after the collision. In this case, since the green disk is initially at rest, the total momentum before the collision is simply the momentum of the orange disk:

m(orange) × v(orange initial) = m(orange) × v(orange final) + m(green) × v(green final)

2. Conservation of Kinetic Energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision. We assume a perfectly elastic collision in which no kinetic energy is lost:

0.5 × m(orange) × v(orange initial)^2 = 0.5 × m(orange) × v(orange final)^2 + 0.5 × m(green) × v(green final)^2

Note that since the disks have equal mass (m(orange) = m(green)), we can simplify the equations further.

Solving the system of equations:
First, calculate the momentum and kinetic energy using the given values:

Momentum of orange disk before the collision = m(orange) × v(orange initial)
= m(orange) × 5.90 m/s

Kinetic energy of orange disk before the collision = 0.5 × m(orange) × v(orange initial)^2
= 0.5 × m(orange) × (5.90 m/s)^2

Let's represent the speeds of the disks after the collision as v(orange final) = vf and v(green final) = vg.

Applying the conservation equations:

m(orange) × v(orange initial) = m(orange) × vf + m(green) × vg
0.5 × m(orange) × v(orange initial)^2 = 0.5 × m(orange) × vf^2 + 0.5 × m(green) × vg^2

Since the mass of both disks is equal (m(orange) = m(green) = m), we can simplify further:

m × 5.90 m/s = m × vf + m × vg
0.5 × m × (5.90 m/s)^2 = 0.5 × m × vf^2 + 0.5 × m × vg^2

The mass cancels out:

5.90 m/s = vf + vg
(5.90 m/s)^2 = vf^2 + vg^2

Now, we have two equations with two unknowns. We need to find both vf and vg.

Using the first equation, we can express vf in terms of vg:
vf = 5.90 m/s - vg

Substituting this expression in the second equation:

(5.90 m/s)^2 = (5.90 m/s - vg)^2 + vg^2

Expand and solve for vg:

34.81 m^2/s^2 = 34.81 m^2/s^2 - 11.8 m/s × vg + vg^2 + vg^2
0 = 2vg^2 - 11.8 × vg

Using the quadratic formula, we can solve for vg:

vg = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = -11.8, and c = 0. Plugging in these values:

vg = (-(-11.8) ± √((-11.8)^2 - 4 × 2 × 0)) / (2 × 2)

vg = (11.8 ± √(139.24)) / 4

vg = (11.8 ± 11.803) / 4

vg = 5.991 m/s or vg = 5.809 m/s

Since the green disk is initially at rest, its speed after the collision (vg) would be 5.809 m/s.

Now substituting this value back into the first equation to find vf:

vf = 5.90 m/s - 5.809 m/s

vf = 0.091 m/s

Therefore, the speed of the orange disk after the collision is 0.091 m/s, and the speed of the green disk after the collision is 5.809 m/s.