How Does Wood Resistance Affect Bullet Penetration and Stoppage Time?

[TylerK]

Homework Statement

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?

Homework Equations

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

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.

 

[Borek]

If the barrel was twice longer, would it change time needed to stop the bullet?

You wrote

a = -471154 N

Are you sure N is a correct unit of acceleration?

 

[TylerK]

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

-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]

You are missing the point.

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

 

[rwisz]

It is important to note that the equations used in this attempt at a solution are only valid for motion in a straight line with constant acceleration. In this scenario, the bullet experiences a resistive force from the wood that is not constant and also changes direction as it enters the block of wood. Therefore, the equations used may not accurately represent the motion of the bullet in this situation.

To accurately determine the time it takes for the bullet to come to rest, a more complex analysis would be needed, taking into account the varying resistive force and changes in direction. Additionally, the type of wood and density of the block may also affect the resistive force and therefore impact the calculation.

It would also be important to consider the conservation of energy in this scenario. The kinetic energy of the bullet before impact would be equal to the work done by the resistive force of the wood in stopping the bullet. This could potentially provide a more accurate estimate of the time it takes for the bullet to come to rest.

Further experimentation and data analysis would be necessary to accurately determine the resistive force and time it takes for the bullet to come to rest in this scenario.