Question

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?

Answer 1

Answer:

The motion of the two balls is  in the positive x direction at the speed of 0.5 m/s

Explanation:

Since this is a typical case of an inelastic collision, the total momentum is conserved.

This means that total momentum before collision = total momentum after collision.

$mv_{1} +mv_{2} =(m+m)v_{3}$

since the masses are the same, we have that

$m(v_{1}+v_{2} ) = 2mv_{3}$

The masses will cancel out

$v_{1}=2m/s\\ v_{2}=-1m/s$

$m(2-1)=2mv_{3}$

$m=2mv_{3}$

$v_{3} =\frac{1}{2} m/s$

This will be in the positive x direction.

Answer 2

Answer:

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!!!