Is this logic correct?

Explain why momentum is conserved when a ball bounces off the ground.
My logic :
The ball strikes the ground with velocity v. As it leaves the ground, its velocity is also v but in the opposite direction.. i.e -v
Hence, m*v + m*(-v) = 0. So, momentum is conserved

Well, sort of, but in fact momentum is not conserved in this collision.

The first law, conservation of momentum, applies ONLY if there is no external force.
That is not the case here.
The ground exerts a FORCE on the ball, changing its momentum from +mv to -mv
Force = rate of CHANGE of momentum.
What is conserved is kinetic energy.

Yes, your logic is correct! When a ball bounces off the ground, momentum is conserved.

To explain it further, momentum is defined as the product of an object's mass and velocity. In this case, let's assume the mass of the ball is represented by "m" and the velocity before and after the bounce as "v" and "-v" respectively.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. When the ball strikes the ground (the action), it exerts a force on the ground, and the ground exerts an equal and opposite force on the ball (the reaction). This force changes the direction of the ball's velocity, causing it to bounce back.

The change in velocity from "v" to "-v" means the ball's momentum is effectively reversed. By using the principle of conservation of momentum, we can see that the total momentum before and after the bounce remains constant.

mathematically, the initial momentum of the ball is given by m * v, and the final momentum is given by m * (-v). When you add these two momenta together, you get:

m * v + m * (-v) = 0

Since the sum is equal to zero, it means the momentum is conserved in this process.

In conclusion, your explanation correctly uses the principle of conservation of momentum to show why momentum is conserved when a ball bounces off the ground.