A billiard ball traveling at 8.0 m/s has an elastic collision with a second billiard ball of equal mass that is at rest. After

the collision, the first ball is at rest. What is the velocity of the second ball after the collision?

Anything that hits something of the same mass will transfer the velocity to the other object while coming to a rest itself so the velocity of 8 would be given to the other billboard ball. additionally within the question itself elastic collision is mention and elastic collision means a collision where both the momentum and kinetic energy is conserved meaning that the momentum doesn't change

Well, it looks like we've got a classic game of billiards here! Let's see what happened. We have one billiard ball zipping along at 8.0 m/s and another one just chilling, minding its own business. They have an elastic collision, which means they bounce off each other without losing any kinetic energy.

Now, after the collision, our first ball comes to a complete stop. I guess it just wanted to take a break or something. But what about the second ball? Well, it must have gained all the momentum the first ball had before the collision, since momentum is conserved!

So, if the first ball had a velocity of 8.0 m/s, and it comes to a stop, the second ball will have to take on that velocity. Therefore, the velocity of the second ball after the collision will be 8.0 m/s in the same direction as the first ball originally had. Talk about an energetic exchange of momentum! Keep those balls rollin'!

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the velocity of the second ball after the collision as "v2".

Given:
- Mass of both balls: m
- Initial velocity of the first ball: v1 = 8.0 m/s
- Final velocity of the first ball: vf1 = 0 m/s (at rest)
- Initial velocity of the second ball: v2(initial) = 0 m/s

Using the conservation of momentum, we can write the equation as:
(mass of the first ball * initial velocity of the first ball) + (mass of the second ball * initial velocity of the second ball) = (mass of the first ball * final velocity of the first ball) + (mass of the second ball * final velocity of the second ball)

(m * v1) + (m * v2(initial)) = (m * vf1) + (m * v2)
(m * 8.0 m/s) + (m * 0 m/s) = (m * 0 m/s) + (m * v2)

Since the mass of both balls is the same (m), we can simplify the equation to:
8.0 m + 0 m = 0 m + v2

Now we can solve for v2:
8.0 m = v2

Therefore, the velocity of the second ball after the elastic collision is 8.0 m/s in the same direction as the initial velocity of the first ball.

To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.

The momentum of an object is calculated by multiplying its mass and velocity. Since the masses of both billiard balls are equal, we can simplify our calculation.

Let's denote the initial velocity of the first ball as v₁ and the final velocities of the first and second balls as v₁f and v₂f, respectively.

1. The momentum before the collision is equal to the momentum after the collision, so we have:
initial momentum = final momentum
(mass of the first ball * initial velocity of the first ball) = (mass of the first ball * final velocity of the first ball) + (mass of the second ball * final velocity of the second ball)
(m * v₁) = (m * v₁f) + (m * v₂f)

2. We are given that the first ball is initially traveling at 8.0 m/s and comes to rest after the collision, so v₁ = 8.0 m/s and v₁f = 0 m/s.

Substituting these values into the momentum equation, we get:
(m * 8.0) = (m * 0) + (m * v₂f)

3. Now we can solve for v₂f. By canceling out the mass of the balls (since they are equal), we find:
8.0 = 0 + v₂f

4. Therefore, the velocity of the second ball after the collision is 8.0 m/s in the opposite direction of its initial velocity.

So, the velocity of the second ball after the collision is -8.0 m/s (opposite direction to its initial motion).

v1=8m/s, v2=0 , m1=m2->m , v=?

m1*v1+m2*v2=(m1+m2)*v
m*8+m*0=(m+m)*v
8m=2m*v
8m/2m=v
4=v

v=4m/s