(a) Draw a sketch of the situation before the collision and after the collision (label everything).
(b) Total momentum is conserved. See equation lines 8.33 and 8.34 in section 8.4 as a guide. Substitute any known values and simplify the expression as much as possible.
(c) Label this result as Equation #1.
(d) Total kinetic energy is conserved in an elastic collision. See equation line 8.35 in section 8.4 as a guide. Substitute any known values and simplify the expression as much as possible.
(e) Label this result as Equation #2.
(f) Looking at Equation #1 and Equation #2 you will notice that you have two equations and two unknowns.
(g) Solve Equation #1 for v2′ ( gives you the value of v2′ in terms of v1′ ).
(h) Substitute your expression for v2′ (from g above) into Equation #2. Simplify the expression as much as possible.
(i) Rearrange your result from (h) into the form: a v1′ 2 + b v1′ + c = 0.
(j) Use the quadratic formula to solve for v1′. Note: as usual, the quadratic formula gives two answers.
(k) Use the values obtained from (j) and Equation #1 to determine values for v2′.
(l) Notice what the pair of results for v1′ and v2′ show you: one result is the same as the initial set of velocities, and the other is the actual set of final velocities.
Answer(s): v1′ = 1.33 m/s, v2′ = 5.33 m/s, both moving east
 Two equal mass billiard balls undergo a perfectly elastic head-on collision. Before the collision, the first ball moves with a speed of 2.00 m/s along the positive x axis while the second ball moves in the opposite direction with a speed of 3.00 m/s. What are the velocities of each ball after the collision?
Follow the setup and approach used in the previous problem.
Answer(s): v1′ = -3.00 m/s, v2′ = 2.00 m/s
need this done asap