# Problem about the Conservation of Energy

1. Nov 28, 2007

### aquamarine08

[SOLVED] Problem about the Conservation of Energy

1. The problem statement, all variables and given/known data
When a certain rubber ball is dropped from a height of 1.25m onto a hard surface, it loses 18% of its mechanical energy on each bounce. (a) How high will the ball bounce on the first bounce? (b) How high will it bounce on the second bounce?(c) With what speed would the ball have to be thrown downward to make it reach its original height on the first bounce?

2. Relevant equations
Eq.1 ~ $$V_{1}^{2}$$= $$V_{0}^{2}$$+2a($$d_{1}$$-$$d_{0}$$)
Eq.2 ~ E= U+K (E is mechanical energy, U is potential energy, and K is kinetic energy)
Eq.3 ~ U=mgh
Eq.4 ~ K= $$\frac{1}{2}$$m$$v^{2}$$

3. The attempt at a solution

I first solved for velocity using Eq.1 ...
$$V_{1}^{2}$$= $$V_{0}^{2}$$+2a($$d_{1}$$-$$d_{0}$$)
$$0^{2}$$= $$V_{0}^{2}$$+ 2(-9.8)(1.25-0)
0=$$V_{0}^{2}$$+24.5
-24.5 = $$V_{0}^{2}$$
4.95 m/s = $$V_{0}$$

and then I thought I would use this value to solve Eq.2...
E= U+K
E=mgh+$$\frac{1}{2}$$m$$v^{2}$$

then I saw a problem...I don't have the mass of the ball!!

What I was planning on doing before seeing this mass situation...was to solve for the mechanical energy and then take off 18% of its energy and use the same Eq.2 to solve for the height and repeat this again for the second bounce.

Last edited: Nov 28, 2007
2. Nov 28, 2007

### Kurdt

Staff Emeritus
What will the potential energy be at the maximum height after the first bounce?

3. Nov 28, 2007

### aquamarine08

Im not sure how to figure that out...don't I need the mass of the ball to figure out the potential energy because of this equation U=mgh?

4. Nov 28, 2007

### BlackWyvern

mgh = (0.5)mv^2

If you examine the kinetic energy equation, which says that the energy at the height, will have turned entirely into kinetic energy at the bottom of the fall, you will notice that both sides have mass, it's the same mass, so you can eliminate it as a variable. Always look for this in tests and hard questions, because without making this simplification, you couldn't solve the problem. You can only find the height that the ball bounces to from this, not it's energy.

5. Nov 28, 2007

### aquamarine08

ok thanks so much for your help!

Last edited: Nov 28, 2007
6. Nov 29, 2007

### Kurdt

Staff Emeritus
What I was getting at is the fact that you know 18% of the energy is dissipated whne the ball strikes the ground. Thus at the new height after the bounce the kinetic energy will have been converted to potential but that potential energy will be 18% less than the starting potential. You could then have eliminated the mass and found the height in terms of the original height.

7. Nov 29, 2007

### D H

Staff Emeritus
You don't need a value for the mass. Simply carry the mass through as the symbol m. It will drop out in the end if you do the math correctly.

8. Nov 30, 2007