Answer:
(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)
