Conservation of kinetic energy
½m 1 vf1^2 + ½ m 2 vf2^2 = ½ m 1 vi1^2 + ½ m 2 vi2^2
Step-by-step solution
First, we use the conservation of momentum to find an equation where the only unknown values are the two final velocities. Since all the
motion takes place on a horizontal line, we use sign to indicate direction.
The conservation of kinetic energy gives us another equation with these two unknowns.
We substitute the expression for the first ball's final velocity found in equation 3 into the quadratic equation, and solve. This gives us the second
ball's final velocity. Then we use equation 3 again to find the first ball's final velocity. One velocity is negative, and one positive í one ball
moves to the left after the collision, the other to the right.
A quick check shows that the total momentum both before and after the collision is 10 kg·m/s. The kinetic energy is 25 J in both cases. This
verifies that we did the computations correctly.
There is a second solution to this problem: You can see that vf2 = 0 is also a solution to the quadratic equation in step 9, and then using step
3, you see that vf1 would equal 5.0 m/s. This solution satisfies the conditions that momentum and KE are conserved, and it describes what
happens if the balls do not collide. In other words, the purple ball passes by the green ball without striking it.