# Bullet hits block

1. Mar 13, 2005

### UrbanXrisis

the question is here

I got part A, but for part B I'm haveing trouble. Here's 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?

Last edited: Mar 13, 2005
2. Mar 13, 2005

### aek

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

AEK

3. Mar 13, 2005

### cepheid

Staff Emeritus
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?

4. Mar 13, 2005

### UrbanXrisis

not quite sure

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

do I subtract them?

5. Mar 13, 2005

### 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?

6. Mar 13, 2005

### cepheid

Staff Emeritus
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.

7. Mar 13, 2005

### UrbanXrisis

thank you for the help :)