# Bullet in block of wood.

by TylerK
Tags: block, bullet, wood
 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.
P: 21,728
If the barrel was twice longer, would it change time needed to stop the bullet?

You wrote

 Quote by TylerK a = -471154 N
Are you sure N is a correct unit of acceleration?
 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)
The text passage contains a physics problem that involves calculating the resistive force exerted by a block of wood on a bullet and the time it takes for the bullet to come to rest after penetrating the wood. The passage includes the problem statement, relevant equations, and an attempt at a solution, with a discussion on the correctness of the solution.
single_question_with_answer

the passage contains a single mathematical question (how long it takes the bullet to come to rest) and provides an answer to this question, although the correctness of the answer is being discussed.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Bullet in block of wood.

by TylerK
Tags: block, bullet, wood
 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.
P: 21,728
If the barrel was twice longer, would it change time needed to stop the bullet?

You wrote

 Quote by TylerK a = -471154 N
Are you sure N is a correct unit of acceleration?
 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)
user
The text passage contains a physics problem that involves calculating the resistive force exerted by a block of wood on a bullet and the time it takes for the bullet to come to rest after penetrating the wood. The passage includes the problem statement, relevant equations, and an attempt at a solution, with a discussion on the correctness of the solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question (how long it takes the bullet to come to rest) and provides an answer to this question, although the correctness of the answer is being discussed.