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?

Question

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.

Answer 1

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)

Answer 2

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