Two billiard balls of equal mass undergo a perfectly elastic head-on collision. if the speed of ball 1 was initially 2.00 m/s, and that of ball 2 was 3.00 m/s in the opposite direction, what will be their speeds after the collision? ball 1

The final velocity of the first ball is 3 m/s and the final velocity of the second ball is 2 m/s in opposite direction.

The given parameters;

mass of the billiard balls = minitial speed of the first ball, u₁ = 2 m/sinitial speed of the ball u₂ = 3 m/s

Apply the principle of conservation linear momentum;

Apply one-dimensional velocity;

From the first equation, the values of the final velocity is calculated as follows;

then, v₂ = 5 - 3 = 2 m/s

Thus, the final velocity of the first ball is 3 m/s and the final velocity of the second ball is 2 m/s in opposite direction.

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The collision is elastic, this means that both total momentum and total kinetic energy are conserved after the collision.

- Let's start with conservation of momentum:

Let's take as positive direction the initial direction of ball 1. Let's label with m the mass of the balls (they have same mass), with  the final speed of ball 1, with the final speed of ball 2. The initial velocity of ball 1 is 2 m/s, while the initial velocity of ball 2 is -3 m/s (with a negative sign, because it is going in the opposite direction of ball 1). The conservation of momentum becomes

From which we find


- Let's now write conservation of kinetic energy:

which becomes

Substituting 

as we found in the conservation of momentum, and simplifying, we get


which has two solutions:
(1)
(2)
The second solution is meaningless: in fact, the balls have exactly same velocity and direction as before the collision. This would mean they didn't collide at all, so we can neglect solution 2. And so, the solution of the problem is (1).

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way