Question

A white billiard ball with mass m_w = 1.54 kg is moving directly to the right with a speed of v = 3.31 m/s and collides elastically with a black billiard ball with the same mass m_b = 1.54 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θ_w = 60° and the black ball ends up moving at an angle below the horizontal of θ_b = 30°.

1)What is the final speed of the white ball?

2)What is the final speed of the black ball?

3)What is the magnitude of the final total momentum of the system? (kg-m/s )

4)What is the final total energy of the system?

Answer 1

1.54*3.31 + 1.54*0 = 1.54*V_1*cos(60) + 1.54*V₂*cos(-30)

3.31 = 0.5*V₁ + 0.866*V₂ .........(eq. 1)

0 = 1.54*V₁*sin(60) + 1.54*V_2*sin(-30)

0 = 0.866*V₁ - 0.5*V₂

V₁ = 0.577*V₂ ........... (eq. 2)

Substitute (equation 2) in (equation 1)

3.31 = 0.5*0.577*V₂ + 0.866*V₂

3.31 = 1.15*V₂

V₂ = 2.87 m/s ............(ans.2)

Substitute that value in (equation 2)

V₁ = 0.577*2.87

V₁ = 1.65 m/s .........ans.1

(3)

Its the same as the initial momentum, so

p = 1.54*3.31 + 1.54*0

p = 5.09 kgm/s

And we check that the final momentum is the same:

px = 1.54*1.65*cos(60) + 1.54*2.87*cos(-30) = 5.09

py = 1.54*1.65*sin(60) + 1.54*2.87*sin(-30) = 0

p = ( 5.09^2 + 0^2 ) = 5.09 ....................same answer 3

(4)

It's the same as the initial total energy, so

E = 1.54 * 3.31² / (2 + 1.54 * 0^2 / 2 )

E = 8.43 J  

Andwe check that the final energy is the same:
E = 1.54 * 1.65² / (2 + 1.54 * 2.87² / 2 ) = 8.43 J ..........same answer 4