Solving Math Mistake: What's Wrong with \frac{m}{M+m}?

[UrbanXrisis]

the question is http://home.earthlink.net/~urban-xrisis/clip002.jpg

I got part A, but for part B I'm haveing trouble. http://home.earthlink.net/~suburban-xrisis/clip001.jpg my work for part B

My book gives me an answer of $$\frac{M}{M+m}$$

however my own answer is $$\frac{m}{M+m}$$

where is my mistake?

 

[aek]

i'll only supply you with a hint,
you have other 2 mistakes.
pm me if your still troubled.

AEK

 

[cepheid]

Huh? I don't think so. Your work looks fine. Your only error was not reading the question carefully enough. It asks what fraction of the kinetic energy was LOST in the collision. What you have is the ratio of the final kinetic energy to the initial kinetic energy, which is what fraction of the KE was left over. Given what you have, the fraction left over, how would you calculate the fraction lost? Hint: what should the two fractions add up to?

 

[UrbanXrisis]

not quite sure

KE initial is $$.5mv^2$$
KE final is $$\frac{m^2v^2}{2M+2m}$$

do I subtract them?

 

[UrbanXrisis]

$$\frac{m}{M+m}+KE_{lost}=1$$
$$KE_{lost}=\frac{M+m}{M+m}-\frac{m}{M+m}$$
$$KE_{lost}=\frac{M}{M+m}$$

Is this what you mean?

 

[cepheid]

Why are you going back to the energies? It's the fractions lost and left over we're worried about. Now, you already derived an expression for the fraction of the original KE left over. For illustration, let's say the masses of the bullet and block respectively are 10g and 90g.

According your formula, (10g) / ( 100g) = 0.1 = 10% of the KE is left over

If only 10% of the KE is left over, how much was lost? 90% of it. How did I calculate that?

I hope that makes it clearer...sometimes it helps to see a numerical example.

EDIT: Yeah, you posted again with the right answer while I was typing. Nice work.

 

[UrbanXrisis]

thank you for the help :)