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

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Homework Help Overview

The problem involves calculating the velocity required for a golf ball to have the same kinetic energy as a bowling ball, given their respective masses and the bowling ball's velocity. The subject area is kinetic energy in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting the kinetic energy equations equal to each other to solve for the golf ball's velocity. There are questions about the accuracy of calculations and the proper use of the kinetic energy formula.

Discussion Status

Some participants have provided guidance on rechecking calculations and suggested alternative methods for solving the problem. There is acknowledgment of differing results, with one participant noting a potential error in the original poster's algebra.

Contextual Notes

There is mention of a discrepancy between the calculated answer and the answer provided in the textbook, prompting participants to question the setup and calculations involved.

KatieLynn
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Homework Statement



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?

Homework Equations



Kinetic Energy = (1/2)V^2

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.
 
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KatieLynn said:

Homework Statement



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?

Homework Equations



Kinetic Energy = (1/2)V^2

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.

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

[tex]\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}[/tex]

Hope this helps.

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

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