# Kinetic Energy/Velocity problem (not sure what I'm doing wrong)

1. Dec 5, 2007

### KatieLynn

1. The problem statement, all variables and given/known data

A 10.89kg bowling ball moves at 3.46 m/s. How fast must a 40.3g golf ball move so that the two balls have the same kinetic energy?

2. Relevant equations

Kinetic Energy = (1/2)V^2

3. The attempt at a solution

I set plugged the information into the kinetic energy problem and set them equal to each other to solve for V.

(1/2)(10.89kg)(3.46m/s)^2=(1/2)(.0403kg)(V^2)
V=238.3 m/s

Thats not the answer in the back of the book though, I'm not sure what I'm doing wrong.

2. Dec 5, 2007

### malty

You may want to recheck that calculation, I did it and got 56.88 m/s, but then again I may also have to check mine :)

3. Dec 5, 2007

### KatieLynn

Ahh I get that now, and thats the answer in the back of the book, I'm not sure how I was doing the algebra wrong, thanks:)!

4. Dec 5, 2007

### dotman

Hello,

Your formula is off (you're missing the m), but it looks like you used the correct form in the calculation.

Your calculation is off, you need to recheck. Are you sure your only squaring the velocity?

Incidentally, another way to do this is to solve for the velocity of the golf ball before you substitute numbers in:

$$\frac{1}{2}m_{b}{v_{b}}^2 = \frac{1}{2}m_{g}{v_{g}}^2 \Rightarrow m_{b}{v_{b}}^2 = m_{g}{v_{g}}^2 \Rightarrow \frac{m_{b}{v_{b}}^2}{m_{g}} = {v_{g}}^2 \Rightarrow v_{g} = \sqrt{\frac{m_{b}}{m_{g}}} \cdot v_{b}$$

Hope this helps.

Edit: You solved it before I could get the LaTeX down :-) Good job

Last edited: Dec 5, 2007