How to calculate rebound speed of ball hitting a wall?

[MattDutra123]

Homework Statement

A ball of mass 0.075 is traveling horizontally with a speed of 2.20 m/s. It strikes a vertical wall and rebounds horizontally. Due to the collision with the wall, 20% of the ball's initial kinetic energy is dissipated.
Show that the ball rebounds from the wall with a speed of 1.97 m/s.

Homework Equations

Impulse = F*t
Kinetic Energy = 1/2mv²

The Attempt at a Solution

I attempted to use conservation of energy by having the initial kinetic energy (0.075*2.20²) be equal to 20% final kinetic energy (0.075*v²*1/5) This gave me a result of 2.2 m/s. I don't know how to reach the desired result of 1.97 m/s.
Apologies in advance for bad formatting.

 

[haruspex]

20% of the ball's initial kinetic energy is dissipated.

That means 20% was lost.

the initial kinetic energy (0.075*2.202) be equal to 20% final kinetic energy

That would be the ball gaining energy, ending with five times what it had to start with.

 

[MattDutra123]

That means 20% was lost.

That would be the ball gaining energy, ending with five times what it hard to start with.

So if I divide the right hand side by 1/5 as opposed to multiplying it as I did, would my approach work?

 

[RPinPA]

So if I divide the right hand side by 1/5 as opposed to multiplying it as I did, would my approach work?

No, because the final energy is not 1/5 of the initial energy.

20% of the ball's initial kinetic energy is dissipated.

It lost 20% of its energy. That means 80% is left. The final energy is 80% of the original energy.

 

[MattDutra123]

No, because the final energy is not 1/5 of the initial energy.

20% of the ball's initial kinetic energy is dissipated.

It lost 20% of its energy. That means 80% is left. The final energy is 80% of the original energy.

Thank you. Very basic misunderstanding.