What Are the Final Speeds of Billiard Balls After a Perfectly Elastic Collision?

[VanKwisH]

Homework Statement

Two Billiard balls of equal mass undergo a perfectly elastic head on collision
(Energy is conserved) . If the speed of one ball was initially at 2.0 m/s
and the other 3.0 m/s in the opposite direction, what will be their speeds after the
collision??

Homework Equations

m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2 m2mv2^2 = 1/2m1v1 '^2 + 1/2m2v2 '^2

The Attempt at a Solution

so far i have simplified both formulas so that it's

v1' = v1 + v2 - v2'
v2' = v1^2 + v2^2 - v1 '^2

That's as far as a can go but I'm stuck what should my next step be...

The answer is v1 = 3 m/s & vs = 2 m/s if it helps anyone

 

[kplooksafterme]

Well, you've done the bulk work of the problem already. You have two unknowns (v1 and v2 final) in two separate expressions. Use your algebraic skillz to solve (i usually substitute when there's only 2 unknowns, but you can eliminate too).

 

[ogb p]

.

As a scientist, your next step would be to apply the conservation of energy and momentum principles to solve for the final velocities of the two billiard balls. First, you would set up the equations using the given information:

Initial momentum: m1v1 + m2v2 = m1v1' + m2v2'
Initial kinetic energy: 1/2 m1v1^2 + 1/2 m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

Then, you would substitute in the values for mass and initial velocities:

m1(2.0 m/s) + m2(3.0 m/s) = m1v1' + m2v2'
1/2 m1(2.0 m/s)^2 + 1/2 m2(3.0 m/s)^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

Since the masses are equal, we can simplify to:

2.0 m/s + 3.0 m/s = v1' + v2'
1/2 (2.0 m/s)^2 + 1/2 (3.0 m/s)^2 = 1/2 v1'^2 + 1/2 v2'^2

Solving for v1' and v2', we get:

v1' = 3.0 m/s
v2' = 2.0 m/s

Therefore, the final velocities of the two billiard balls after the perfectly elastic collision will be v1 = 3.0 m/s and v2 = 2.0 m/s, as given in the answer. This shows that energy and momentum are conserved in the collision, and the speeds of the two balls are simply reversed.