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Bullet in block of wood.

by TylerK
Tags: block, bullet, wood
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TylerK
#1
Oct3-10, 10:47 AM
P: 4
1. The problem statement, all variables and given/known data

A rifle with a barrel length of 56 cm fires a 10 g bullet with a horizontal speed of 350 m/s. The bullet strikes a block of wood locked in position and penetrates to a depth of 13 cm.

What resistive force (assumed to be constant) does the wood exert on the bullet?

How long does it take the bullet to come to rest?

2. Relevant equations

f = m * a
d = v * t
vf = vi + a * t

3. The attempt at a solution

Force of wood on bullet = 4700 (was confirmed correct)
f = m * a
a = -471154 N

How long does it take the bullet to come to rest?

I tried
D = v * t1 for the barrel
.55 = 350 * t1
t1 = 350 / .55 = .0016 seconds

For the wood
vf = vi + a * t2
0 = 350 - 471154 t2
t2 = 350 / 471154 = .000743 seconds

t1 + t2 = .002343 seconds which is wrong.
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Borek
#2
Oct3-10, 11:12 AM
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If the barrel was twice longer, would it change time needed to stop the bullet?

You wrote

Quote Quote by TylerK View Post
a = -471154 N
Are you sure N is a correct unit of acceleration?
TylerK
#3
Oct3-10, 11:51 AM
P: 4
Yes it is m/s^2 I mistakenly typed N. I tried to brake the problem up into 2 parts.

Part 1: Time of bullet in barrel, with 0 acceleration so

d = v * t
.56m = 350 m / s * t
t = .56 m * s / 350 m = .0016 seconds

Part 2: Time it enters the wood until the bullet stops.

vf =vi + a * t
0 = 350 m/s + (-471154 m / s^2) t
[U]
-350 m/s * (1/-471154 m/s^2) = t
t = .000743 seconds

I apparently do not add these 2 times together, as the combined result is incorrect.

If the barrel was twice as long, wouldn't it double the amount of time the bullet was in the barrel, as accounted for in the first part? (assuming no friction in the barrel)

Borek
#4
Oct3-10, 12:29 PM
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P: 23,391
Bullet in block of wood.

You are missing the point.

Is the time spent in the barrel in any way relevant?


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