Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If the speed of ball 1 was initially 5.00 m/s, and that of ball 2 was 7.00 m/s in the opposite direction, what will be their speeds after the collision?

I figured out how to calculate the velocity when on eobject is at rest but I am confused when it comes to both objects moving at once. I know I need the mv=mva'+ mvb' equation and the va-vb=vb'-va' equation but I am stumped as to solving them.

When the balls are moving in opposite directions the magnitudes of their velocities are

v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂)
v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)

To solve the problem, you can use the principles of conservation of momentum and conservation of kinetic energy. First, let's define some variables:

- m1 and m2 are the masses of the two balls (assuming they are equal).
- v1 and v2 are the initial velocities of the two balls.
- v1' and v2' are the final velocities of the two balls after the collision.

Using the conservation of momentum, we can write:

m1v1 + m2v2 = m1v1' + m2v2'

Since the masses of the balls are equal (m1 = m2), the equation simplifies to:

v1 + v2 = v1' + v2' ..........(Equation 1)

Now, using the conservation of kinetic energy, we have:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

Since the balls undergo a perfectly elastic collision, kinetic energy is conserved, so the equation simplifies to:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2 ..........(Equation 2)

Now we have two equations (Equation 1 and Equation 2) with two unknowns (v1' and v2'). We can solve these equations simultaneously to find the final velocities.

Let's substitute the given values into the equations:

m1 = m2 (equal masses)
v1 = 5.00 m/s (velocity of ball 1)
v2 = -7.00 m/s (velocity of ball 2 in the opposite direction)

Plugging these values into Equation 1, we get:

5.00 - 7.00 = v1' + v2'

Simplifying further:

-2.00 = v1' + v2' ..........(Equation 3)

Now let's plug the values into Equation 2:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

(1/2)(5.00)^2 + (1/2)(7.00)^2 = (1/2)(5.00)^2 + (1/2)(7.00)^2

(1/2)(25.00) + (1/2)(49.00) = (1/2)v1'^2 + (1/2)v2'^2

12.50 + 24.50 = (1/2)v1'^2 + (1/2)v2'^2

37.00 = (1/2)v1'^2 + (1/2)v2'^2 ..........(Equation 4)

We have now reduced the problem to a system of two equations (Equation 3 and Equation 4) with two unknowns (v1' and v2'). To solve this system, you can use algebraic methods such as substitution or elimination.

Substituting -2.00 for v1' + v2' from Equation 3 into Equation 4, we get:

37.00 = (1/2)v1'^2 + (1/2)(-2.00)^2

37.00 = (1/2)v1'^2 + 2.00

35.00 = (1/2)v1'^2

Multiplying both sides by 2:

70.00 = v1'^2

Taking the square root of both sides:

v1' = ±√70.00

Since we are dealing with a head-on collision, the sign of the velocity will depend on the direction of the ball's movement. In this case, the positive root (√70.00) corresponds to the direction of ball 1, and the negative root (-√70.00) corresponds to the direction of ball 2.

Therefore, the final velocities of the two balls after the collision are approximately:

v1' ≈ √70.00 m/s

v2' ≈ -√70.00 m/s

So, ball 1 will have a positive velocity of approximately 8.37 m/s, and ball 2 will have a negative velocity of approximately -8.37 m/s after the collision.

The sum of the initial momentums is equal to the sum of the final momentums:

m*5 - m*7 = m*v1f + m*v2f

The sum of the initial kinetic energy is equal to the sum of the final kinetic energy:

0.5*m*5^2 + 0.5*m*7^2 = 0.5*m*v1f^2 + 0.5*m*v2f^2

where v1f is the final speed of ball 1, and v2f is the final speed of ball 2.
Dividing the equations by common factors:

5 - 7 = v1f + v2f
-2 = v1f + v2f
5^2 + 7^2 = v1f^2 + v2f^2
25 + 49 = v1f^2 + v2f^2
74 = v1f^2 + v2f^2

-2 = v1f + v2f
74 = v1f^2 + v2f^2

Use the above two equations and basic algebra to solve for v1f and v2f