Stop a .22 Bullet: Average Speed Calculation

[mybrohshi5]

Homework Statement

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?

Homework Equations

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

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 don't really like the wording of this question. it's kind of confusing to me.

Thanks

 

[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:

<br /> \vec{F}t=\Delta \vec{p}<br />

where \vec{p} is momentum.

 

[mybrohshi5]

So i found the time using the equation:
<br /> d = \frac{V_i + V_f}{2)} * t<br />
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

 

[kbaumen]

Yep. Looks fine.

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

<br /> d = v_{avg}t<br />