Two billiard balls with the same mass m move towards one another. Ball one moves in the positive x-direction with a speed of v1i, and ball two moves in the negative x-direction with a speed of v2i. The two balls collide elastically, and both balls change direction after the collision. If the initial speeds of the balls were v1i=2.0m/s and v2i =1.0m/s, what would be the final speed and direction of ball two, v2f, in m/s?

The final speed and direction of ball two, v₂բ = 2 m/s in the positive x-direction.

Explanation:

mass of ball one = mass of ball two = m

velocity of ball one before collision = (v₁ᵢ) m/s = 2 m/s

velocity of ball two before collision = (-v₂ᵢ) m/s = - 1 m/s (minus sign because the ball is moving in the negative x-direction)

velocity of ball one after collision = (v₁բ) m/s

velocity of ball two after collision = (v₂բ) m/s

In an elastic collision, both the momentum and kinetic energy are conserved.

First of, conservation of Momentum

Momentum before collision = Momentum after collision

Momentum of ball one before collision = (m)(v₁ᵢ) = (m)(2) = (2m) kgm/s

Momentum of ball two before collision = (m)(-v₂ᵢ) = (m)(-1) = (-m) kgm/s

Momentum of ball one after collision = (m)(v₁բ) = (mv₁բ) kgm/s

Momentum of ball two after collision = (m)(v₂բ) = (mv₂բ) kgm/s

Conservation of Momentum

2m - m = mv₁բ + mv₂բ

m = m(v₁բ + v₂բ)

(v₁բ + v₂բ) = 1

Conservation of kinetic energy

Kinetic energy before collision = kinetic energy after collision

Kinetic energy of ball one before collision = (1/2)(m)(v₁ᵢ)² = (mv₁ᵢ²)/2 = (m×2²)/2 = 2m

Kinetic energy of ball two before collision = (1/2)(m)(v₂ᵢ)² = (m×1²)/2 = m/2

Kinetic energy of ball one after collision = (1/2)(m)(v₁բ)² = (mv₁բ²)/2

Kinetic energy of ball two after collision = (1/2)(m)(v₂բ)² = (mv₂բ²)/2

Kinetic energy balance

2m + (m/2) = [(mv₁բ²)/2] + [(mv₂բ²)/2]

Multiplying through by (2/m)

4 + 1 = v₁բ² + v₂բ²

v₁բ² + v₂բ² = 5

From the momentum balance,

(v₁բ + v₂բ) = 1

v₁բ = (1 - v₂բ)

v₁բ² + v₂բ² = 5

(1 - v₂բ)² + v₂բ² = 5

1 - 2v₂բ + v₂բ² + v₂բ² = 5

2v₂բ² - 2v₂բ + 1 = 5

2v₂բ² - 2v₂բ - 4 = 0

v₂բ² - v₂բ - 2 = 0

Solving the quadratic equation

v₂բ = 2 m/s or v₂բ = -1 m/s

when v₂բ = 2 m/s,

v₁բ = 1 - 2 = -1 m/s

when v₂բ = -1 m/s

v₁բ = 1 - (-1) = 2 m/s

(v₁բ, v₂բ) = (-1, 2) or (2, -1)

But in the question, we're told that both balls change direction after the collision, hence, the correct answer is

(v₁բ, v₂բ) = (-1, 2) (since ball one was moving with positive velocity initially and ball two was moving with negative velocity)

Hence, the final speed and direction of ball two, v₂բ = 2 m/s in the positive x-direction.

Hope this Helps!!!

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.