# What is the kinetic energy now?

1. Feb 25, 2009

### genu

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

Initially a body moves in one direction and has kinetic energy K. Then it moves in the opposite direction with three times its initial speed. What is the kinetic energy now?

2. Relevant equations

k =1/2mv^2

3. The attempt at a solution

k = 1/2m(3v)^2
k=1/2m(9v^2(
k=(m9v^2)/2
2k=m9v^2
2/9k=mv^2
...
?

2. Feb 25, 2009

### swuster

You've actually found the answer within the first few steps.

As you've said, $$K = 1/2*mv^2$$

And afterwards, $$K' = 1/2*m(-3v)^2$$

so that $$K' = 9/2*mv^2$$

What do you notice about K and K' when comparing them?

K' = 9K.

3. Feb 25, 2009

### genu

I'm not understanding how you've arrived to that last step...what happened to all the other variables?

4. Feb 25, 2009

### LowlyPion

You found that the final KE was 9 * (½mv²)

If you divide by the initial KE you get

Final KE = 9 * (Initial KE)

The suggestion was to encourage you to recognize the initial KE in the final KE expression.