In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20°, as shown in the diagram below. Both balls have a mass of 0.6 kg. a. What is the momentum of the system before the collision? (Write it in component form.) (1 point) b. What is the momentum of the system after the collision? (1 point)

c. Write the velocity of the cue ball after the collision in component form. (1 point)

d. What is the x-component of the velocity of the eight ball after the collision? (2 points)

e. What is the y-component of the velocity of the eight ball after the collision? (2 points)

f. At what angle does the eight ball travel after the collision? (2 points)

g. What is the magnitude of the eight ball's velocity after the collision? (1 point)

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

        p₀ = m v₁₀ + 0

        p₀ = 0.6 2

        p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

       p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

        p₀ₓ = 1.2 kg m / s

        p_{fx} = m v1f cos 20 + m v2f cos θ

        p₀ = p_f

       1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

        1.2482 = v_{2f} cos θ

Y axis  

       p_{oy} = 0

       p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

       0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

       0.2736 = v_{2f} sin θ

we write our system of equations

        0.2736 = v_{2f} sin θ

        1.2482 = v_{2f} cos θ

divide to solve

        0.219 = tan θ

         θ = tan⁻¹ 0.21919

         θ = 12.36

let's look for speed

           0.2736 = v_{2f} sin θ

            v_{2f} = 0.2736 / sin 12.36

           v_{2f} = 1.278 m / s

increases it

change tension