# Homework Help: Stopping a bullet

1. Feb 6, 2010

### mybrohshi5

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

A .22 rifle bullet, traveling at 357 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.131 m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.65 g. Assume a constant retarding force.

What is the average speed of the bullet while it is being stopped by the wood?

2. Relevant equations

F=m(a)
V=d/t
d=x_o + V_o*(t) + 1/2at^2

3. The attempt at a solution

I thought the average speed while it is beings stopped would just be the 357 m/s but its not :(

any help on how to find this? i dont really like the wording of this question. it's kind of confusing to me.

Thanks

2. Feb 6, 2010

### kbaumen

Since the retarding force is constant, the deceleration is constant as well. What is the average speed for a constant acceleration (deceleration)? (357 m/s would definitely not be the average speed. That is the initial speed, while the final speed is 0 m/s)

Also, note that:

$$\vec{F}t=\Delta \vec{p}$$

where $$\vec{p}$$ is momentum.

3. Feb 6, 2010

### mybrohshi5

So i found the time using the equation:
$d = \frac{V_i + V_f}{2)} * t$
0.131 = 357/2 * t

t=0.00073389 seconds

then i found the speed using the equation:

speed = distance/t

speed = 0.131/0.00073389

speed = 178.5 m/s

Does this look correct?

Thank you

Last edited: Feb 6, 2010
4. Feb 6, 2010

### kbaumen

Yep. Looks fine.

You could've solved it a bit quicker though. For constant acceleration:

$$d = v_{avg}t$$