Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of objects 1 and 2? Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express the velocities in terms of v.

(v₁, v₂) = [ (v/3), (4v/3) ]

Or

(v₁, v₂) = (v, 0)

Explanation:

In elastic collisions, the momentum and kinetic energy is usually conserved.

The momentum before collision = momentum after collision

And

Kinetic energy before collision = Kinetic energy after collision

Momentum of object 1 before collision = (2m) v = 2mv

Momentum of object 2 before collision = (m) (0) = 0

Momentum of object 1 after collision = (2m) (v₁) = 2mv₁

Momentum of object 2 after collision = (m) (v₂) = mv₂

So, we have

2mv = 2mv₁ + mv₂

2v = 2v₁ + v₂

v₂ = 2v - 2v₁ (eqn 1)

Kinetic energy of object 1 before collision = (1/2) (2m) (v²) = mv²

Kinetic energy of object 2 before collision = (1/2) (m) (0²) = 0

Kinetic energy of object 1 after collision = (1/2) (2m) (v₁²) = mv₁²

Kinetic energy of object 2 after collision = (1/2) (m) (v₁²) = (mv₂²/2)

So, we have,

mv² = mv₁² + (mv₂²/2)

v² = v₁² + (v₂²/2)

2v² = 2v₁² + v₂² (eqn 2)

Substitute (v₂ = 2v - 2v₁) from (eqn 1) into (eqn 2)

2v² = 2v₁² + (2v - 2v₁) ²

2v² = 2v₁² + 4v² - 8vv₁ + 4v₁²

6v₁² - 8vv₁ + 2v² = 0

6v₁² - 6vv₁ - 2vv₁ + 2v² = 0

6v₁ (v₁ - v) - 2v (v₁ - v) = 0

(6v₁ - 2v) (v₁ - v) = 0

6v₁ = 2v or v₁ = v

v₁ = (v/3) or v₁ = v

If v₁ = (v/3)

From (eqn 1)

v₂ = 2v - 2v₁

v₂ = 2v - 2 (v/3)

v₂ = 2v - (2v/3)

v₂ = (4v/3)

If v₁ = v,

From eqn 1,

v₂ = 2v - 2v₁

v₂ = 2v - 2v = 0

(v₁, v₂) = [ (v/3), (4v/3) ]

Or

(v₁, v₂) = (v, 0)