Tough Energy problem Law of conservation of energy

[Senjai]

Homework Statement

Bill throws a 10.0g (0.0100kg) straight down froma height of 2.0m. The ball strikes the floor at a speed of 7.5 m/s.
a) what was the original speed of the ball?
b) if 30% of the balls energy is transformed in thermal energy during the collision with the floor, find the new height reached by the ball.

Homework Equations

$$\frac{1}{2}mv^2 + mgh = \frac{1}{2}mv'^2 + mgh'$$
$$\sum{E} = \sum{E'}$$

The Attempt at a Solution

A) i solved this fairly easily and got the right answer.

$$v = \sqrt{\frac{2\left(\frac{1}{2}mv'^2 - mgh\right)}{m}}$$

v initial = 4.13 m/s

B) this is the toughy...

The answer is supposed to be 2.0 m. i attempted to rearrange the question including the initial kinetic energy + initial potential = end potential and solve for h. but i didnt get it. A friend of mine told me i don't need to include work done by NC forces in this equation. i think i got 1.7m, i know that if i just have initial potential energy = to kinetic energy @ the point of collision.. when losing 30% of energy, it isn't supposed to reach the same height correct? Which leads be to believe that the person throwing the ball supplied the additional 30% energy to return it to the 2.0m height. I don't understand how to mathmatically show how to solve to get h' = 2.0m..

Thanks,
Senjai

 

[Delphi51]

I get 2.01 m for (b). I just did
E = .7*.5*m*v^2 = .7*.5*.01*7.5^2 = .1969 J
.28125 = mgh

 

[tomcue]

I would approach this problem by first considering the law of conservation of energy. This law states that energy cannot be created or destroyed, only transformed from one form to another. In this case, we know that the ball initially has gravitational potential energy due to its height and kinetic energy due to its motion. When it strikes the floor, some of this energy is transformed into thermal energy, but the total amount of energy in the system remains the same.

Using this concept, we can set up an energy equation:

Initial kinetic energy + initial potential energy = final kinetic energy + final potential energy + thermal energy

We know that the initial kinetic energy is equal to 1/2mv^2, and the initial potential energy is equal to mgh. We also know that the final kinetic energy is equal to 1/2mv'^2, and the final potential energy is equal to mgh'. Therefore, we can rewrite the equation as:

1/2mv^2 + mgh = 1/2mv'^2 + mgh' + thermal energy

We also know that the thermal energy is equal to 30% of the initial energy, so we can substitute that in:

1/2mv^2 + mgh = 1/2mv'^2 + mgh' + 0.3(1/2mv^2 + mgh)

Now, we can solve for the new height, h':

h' = (1/2mv^2 + mgh - 1/2mv'^2 - 0.3(1/2mv^2 + mgh))/mg

Plugging in the values we know, we get:

h' = (0.5(0.01)(4.13)^2 + 0.01(9.8)(2) - 0.5(0.01)(7.5)^2 - 0.3(0.5(0.01)(4.13)^2 + 0.01(9.8)(2)))/(0.01)(9.8)

Simplifying, we get:

h' = 2.0 m

This shows that the ball will reach the same height of 2.0m after the collision, as the energy lost due to thermal energy is supplied by the person throwing the ball.