# Homework Help: Bullet Force

1. Oct 11, 2008

### PhysicslyDSBL

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

A bullet of mass 0.0022 kg initially moving at 504 m/s embeds itself in a large fixed piece of wood and travels 0.72 m before coming to rest. Assume that the acceleration of the bullet is constant.

What force is exerted by the wood on the bullet?

F=N

2. Relevant equations

Ok, so I need to find acceleration to plug into the equation for force, which is F=m*a. The equation I am using is x=x0 + v0*t + .5*a*t^2.

3. The attempt at a solution

x=x0 + v0*t + .5*a*t^2 so this equation need to be set to solve for a. This is what I am plugging in:

x0 = 0
v0 = 504
t = 0.0014 (from d/s = t)

My answer is a = 0.706

The equation vf^2 = v0^2 + 2a (delta x), and set to solve for a but I am a bit confused about delta x? In this equation, vf = 0.72 and v0 = 504

I am either having trouble setting up the first equation for a, or I am using the wrong one in which case, I am not sure by what is meant by delta x in the second. Some direction is appreciated!

2. Oct 11, 2008

### Rake-MC

It was looking good, but you said this:

"x0 = 0
v0 = 504
t = 0.0014 (from d/s = t)"

This assumes constant velocity, which it's not.

You have to use this formula instead : $$v^2 = u^2 + 2as$$ where v = final velocity
u = initial velocity
a = acceleration
s = displacement.

Solve for acceleration, then multiply by mass.

3. Oct 11, 2008

### Staff: Mentor

You could use this method if you found the right time. To find the time you need to use the average speed, not the initial speed.

This equation is a better choice, but 0.72m is the distance (Δx) not the final speed. (The final speed is zero, of course.)

4. Oct 12, 2008

### PhysicslyDSBL

Thanks for the help!

I got the correct answer of 388.08 N by taking the equation v^2=u^2 + 2as and solving for a and then multiplying by mass. Both of your suggestions were very helpful in helping me to understand these concepts better!

5. Oct 12, 2008

### Andrew Mason

It often is helpful in this kind of problem to use a graph. A graph of speed vs. time makes it very easy. The slope is the acceleration (constant -) and the distance is the area under the graph.

AM