Simple math question proportions

Shjanzey

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Lets say you have:

KE = 1/2mv^2 and then you double KE. The velocity increases by a factor of square root 2. Can someone show the algebraic steps to this. Thank you

The equation for KE is K = (1/2)*mv^2

So you solve for v

v1 = sqrt(2K/m)

Now double K (kinetic energy) to see what happens

v2 = sqrt((2*2K)/m) --> sqrt(4K/m)

If you compare the two, v2 has increased by sqrt(2).
 

gettheleadout

MD
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I would think about it like this:

You can't increase mass, so you must increase velocity, right? We want the total KE doubled, so:

KE = (1/2)*m*(v)^2

needs to become

KE = 2*(1/2)*m*(v)^2

This means we need to put something into the parentheses that contain "v" that, when squared (and thus pulled out from the parentheses under the exponent), will give us a coefficient of 2. What can we put in as a coefficient of "v" that will be squared to yield 2? The square root of 2 of course.
 
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Hemorrage

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I would think about it like this:

You can't increase mass, so you must increase velocity, right? We want the total KE doubled, so:

KE = (1/2)*m*(v)^2

needs to become

KE = 2*(1/2)*m*(v)^2

This means we need to put something into the parentheses that contain "v" that, when squared (and thus pulled out from the parentheses under the exponent), will give us a coefficient of 2. What can we put in as a coefficient of "v" that will be squared to yield 2? The square root of 2 of course.

Very intuitive method
 

MD9

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Jan 27, 2013
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Hopefully this is what you're looking for. If you need any more explanation, please let me know!

ZltsWv8.jpg


You can see that instead of doubling the K.E., if we generalize and say we multiply it by some number n, then the result will be that the new velocity, v2, is equal to the initial velocity, v1, multiplied by the square root of n (i.e. just replace "2" in "2K.E." with the variable "n" to get n*K.E. and the result should be v2=sqrt(n)*v1).
 
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