# Bullet passes through block - conservation of momentum

1. Dec 6, 2013

### danielamartins

1. The problem statement, all variables and given/known data
There's a stationary block (m1=1,5kg) in a table (Coefficient of Friction=0,4). A bullet passes through the block and hits a person. The block moved 1,2m. Calculate the velocity of the bullet (mb=0,0079kg and vib=709,88m/s) when it hit the person.

2. Relevant equations
W(f) = ΔKE

Pix=Px

3. The attempt at a solution

W(f) = ΔKE

v1 - velocity of the block after being hit by the bullet
μ - coefficient of friction = 0,4
d - distance - 1,2 m

- m1×g×μ×d=0- 1/2×m1×v1^2

v1= 3.098 m/s

Then, mb= 0,0079kg and vib=709,88 m/s:
By conservation of momentum:
Pix=Px
p(ix,b)+p(ix,1)= p(x,b)+ p(x,1)
mb×vib+m1×vi1=mb×vb+m1×v1

vb= 121,65 m/s

So, basically, this is what I did. I had to make up the values for the mass of the block and for the distance it traveled.
The thing is, is this analysis right? Because slowing down that much after passing through a block of wood doesn't seem right. Any thoughts?

2. Dec 6, 2013

### TSny

Hello.

Your work looks correct to me.

3. Dec 6, 2013

### danielamartins

Thanks! I appreciate it ;)

4. Dec 6, 2013

### PeroK

I assume you're using g = 10 ms^-2. I get 121.58 ms^-1.

5. Dec 6, 2013

### danielamartins

Yes, I'm using g=10m/s^2

6. Dec 6, 2013

### haruspex

Since the given data are expressed to several significant figures, I would use a more precise value for g. The calculation involves taking the difference of two large numbers to obtain a rather smaller one, so a small error in an input can produce a relatively large error in the result. With g = 9.8 ms-2 I get 127.5m/s.
The answer could indeed be that low. A bullet might not go right through a 1.5kg block of wood.

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