# Bullet Stopping in Wood

1. Feb 17, 2012

### mandabox

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

A bullet of mass 0.0022 kg moving at 488 m/s embeds itself in a large fixed piece of wood and travels 0.56 m before coming to rest. Assume that the deceleration of the bullet is constant.

What force is exerted by the wood on the bullet?
F = N
1.01 NO

HELP: Find the deceleration of the bullet.
HELP: Use Newton's Second Law

2. Relevant equations

velocity & displacement:
v2=v02+2aΔx

Newton's 2nd Law:
F⃗net = ΣF⃗ = ma⃗

3. The attempt at a solution

Well clearly in the help options it told me to do what I had already done based on instinct.
I plugged v = 0 , v0 = 488m/s , and Δx = 0.56 m. I figured that at the end of 0.56 m the final velocity would have to be 0 and obviously when the bullet begins its deceleration it had hit the wood. So I plugged in 02 = (488m/s)2 + 2a(0.56m) and came out with the result a = -461.116657m/s2.

I plugged this into the formula for force and got an answer of -1.01. I tried both -1.01 and +1.01 and neither attempt worked and I got the same answer no matter the number of sigfigs I did my calculations with. Am I getting it wrong because I've overlooked something? This problem seemed easy, what am I doing wrong? Doesn't the wood exert a force that's equal and opposite on the force of the bullet? I am assuming the bullet is traveling parallel to the surface of the earth along the positive x-axis.

2. Feb 17, 2012

### mandabox

I just went even simpler and used v = Δd/Δt to find time and then a = Δv/Δt to find acceleration. But the time to the last decimal shown on my calculator is 0.001147541 seconds and the acceleration is an enormous number in the four hundred thousands! And depending on how many sigfigs I take from time dictates how large or small my acceleration will be and I've gotten a force ranging from 930 to 976 and none of them seem right. I've dealt with this TYCHO system before and I KNOW that it is picky about your numbers. But whether I keep all the numbers to the end or round it off to two sigfigs at EVERY step, it's wrong! I'm getting really frustrated with this.

3. Feb 17, 2012

### mandabox

After much deliberation and calculations I've discovered that the formula to use is in Chapter 8 of my text book. 1/2mv2=Fd.

A bullet of mass 0.0022 kg moving at 488 m/s embeds itself in a large fixed piece of wood and travels 0.56 m before coming to rest. Assume that the deceleration of the bullet is constant.
What force is exerted by the wood on the bullet?
F = N *
467.78 OK

HELP: Find the deceleration of the bullet.
HELP: Use Newton's Second Law

I must have been plugging it into the original equations wrong because they also worked after I did a reverse calculation. I'm not sure how someone can make the same error for an hour but I guess that's what happened.

Last edited: Feb 17, 2012
4. Feb 17, 2012

### Shootertrex

You should be able to easily use the Work-Energy Theorem for this problem:

$\Sigma W=\Delta KE$

You know the change in kinetic energy, and you know the distance that the force is applied over.