HELP-Elastic Collision Problem!!?

Question

A 0.440 kg ice puck, moving east with a speed of 3.04 m/s, has a head-on collision with a 0.950 kg puck initially at rest. Assume a perfectly elastic collision. Use east as the positive axis.

What is the speed of the lighter puck?

What is the speed of the heavier puck?

How would you answer this hard question?

Answer 1

Here we need to conserve energy. An elastic collision is one in which no energy is lost, so the initial kinetic energy must be equal to the final.
The formula for kinetic energy is (.5)(m)(v^2) Where m is the mass and v the velocity.
The initial kinetic energy is
KEi = (.5)(.44kg)(3.04m/s)^2 = 2.03J
The final kinetic energy is
KEf = (.5)(.44kg)(final speed of the lighter puck)^2 + (.5)(.95kg)(final speed of the heavier puck)^2.
Setting these equal to each other, we have
2.03J = (.5)(.44kg)(final speed of the lighter puck)^2 + (.5)(.95kg)(final speed of the heavier puck)^2.

We will note that there are two things we don't know, the final speed of the lighter puck and the final speed of the heavier puck, and only one equation. Now we have to conserve momentum. The formula for momentum is
momentum = (m)(v).
In our case
(.44kg)(3.04m/s) = (.44kg)(final velocity of lighter puck) + (.95kg)(final velocity of heavier puck)
3.04m/s = (final velocity light puck) + 2.16(final velocity heavy puck)

Now we have two equations and two unknowns, perfect! We can solve for one and plug it into the other (solve by substitution).
(Vf light puck) = 3.04 m/s - 2.16(Vf heavy puck)
. Now plug into the other equation and solve.
2.03J = (.22kg)(3.04-2.16(Vf heavy puck))^2 + (.4755kg)(Vf heavy puck)^2
Now solve for Vf heavy puck.
2.03J = 2.03 - 2.9(Vf heavy puck) + 1.5(Vf heavy puck)^2
0 = 1.5(Vf heavy puck)^2 - 2.9(Vf heavy puck)
Vf heavy puck = 1.93 m/s
Vf of the light puck would then be.
3.04m/s - 2.16(1.93m/s) = -1.13 m/s, meaning it will rebound backwards at a rate of 1.13 m/s