Question

A billiard ball of mass m hits another one of the same mass. The first ball moves off at 30 degrees. For an elastic collision what are the velocities of both balls after the collision. In the lab frame where the second ball starts at rest, what angle compared to the direction of initial velocity of the first ball does the 2nd ball move?

Answer 1

Answer:

v₁ = u₁/2√3 ≈ 0.866u₁

v₂ = u₁/2      = 0.5u₁

θ = 60°

Explanation:

let u₁ be the initial velocity of the first ball

let v₁ be the final velocity of the first ball

let v₂ be the final velocity of the second ball

For elastic collisions, the angle between the departing masses is 90°

assume the first ball initially moves along the x axis in the positive direction

conservation of momentum

In the y direction, initial momentum is zero

After the collision

mv₁sin30 = mv₂sin60

½v₁ = ½√(3)v₂

v₁ = √(3)v₂

in the x direction,

mu₁ = mv₁cos30 + mv₂cos-60

u₁ = v₁cos30 + v₂cos60

u₁ = (√(3)v₂)½√(3) + ½v₂

u₁ = 2v₂

v₂ = u₁/2

v₁ = √(3)v₂ = √(3)(u₁/2)