# If a rubber ball of mass 1.0 kg is dropped from a height of 2.0 m and rebounds on the first...

## Question:

If a rubber ball of mass 1.0 kg is dropped from a height of 2.0 m and rebounds on the first bounce to 0.75 of the height from which it was dropped, how much energy is dissipated during the collision with the floor?

## Conservation of Energy:

The conservation of energy is one of the laws of physics. This law tells us that energy can neither be created nor can it be destroyed. What energy can do though is transform into different forms like electrical energy, mechanical energy, thermal energy, etc. When we have a closed system, energy cannot flow in or out of the system. Therefore, the total energy must always be constant.

Given:

• {eq}\displaystyle m = 1\ kg {/eq} is the mass of the ball
• {eq}\displaystyle h_1 = 2\ m {/eq} is the initial height
• {eq}\displaystyle h_2 = 0.75h_1 = 1.5\ m {/eq} is the final height

If we treat our rubber ball as a closed system, then we can determine the lost energy by using the conservation of energy. The initial gravitational potential energy must be equal to the sum of the lost energy and the final gravitational energy. That is,

{eq}\displaystyle mgh_1 = mgh_2 + E {/eq}

Now let us isolate the energy loss:

{eq}\displaystyle E = mgh_1 - mgh_2 {/eq}

Or:

{eq}\displaystyle E = mg(h_1 - h_2) {/eq}

We substitute:

{eq}\displaystyle E= (1\ kg)(9.8\ m/s^2)(2\ m - 1.5\ m) {/eq}

We will thus get the lost energy:

{eq}\displaystyle \boxed{E = 4.9\ J} {/eq}