# Kinematics problem - Bullet in block of wood

• paytona
In summary: The problem does not say that the bullet stays in the block but perhaps it is implied. I cannot imagine how a 3.5kg block of wood is going to stop a 12g rifle bullet at 2000 feet per second.In summary, a 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. The speed of the bullet immediately before the collision was 612.4684m/s.
paytona

## Homework Statement

A 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. What was the speed of the bullet immediately before the collision?The attempt at a solution
d=1/2at^2
t=*sqrt(2d-vertical-)/a
t=*sqrt[(2)(1.75m)/(9.81m/s^2)
t= 0.597309633s

d= vt
vf = d(horizontal)/t
vf= 1.25m/0.597309633m/s^2
vf= 2.09216961m/s

Conservation of momentum (P1+P2)before = (Psystem) after
(m of bullet * vi of bullet) + (m of block * vi of block) = (m of system * vf of system)
(0.012kg*vi) + (0) = (3.512kg*2.09216961m/s)
(0.012kg)vi = 7.349621967kg*m/s
vi = 7.349621967kg/.012kg
vi = 612.4684m/s

The initial velocity seems a bit high and I'm not sure how else to approach the question. Help please! :)

paytona said:

## Homework Statement

A 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. What was the speed of the bullet immediately before the collision?

The attempt at a solution
d=1/2at^2
t=*sqrt(2d-vertical-)/a
t=*sqrt[(2)(1.75m)/(9.81m/s^2)
t= 0.597309633s

d= vt
vf = d(horizontal)/t
vf= 1.25m/0.597309633m/s^2
vf= 2.09216961m/s

Conservation of momentum (P1+P2)before = (Psystem) after
(m of bullet * vi of bullet) + (m of block * vi of block) = (m of system * vf of system)
(0.012kg*vi) + (0) = (3.512kg*2.09216961m/s)
(0.012kg)vi = 7.349621967kg*m/s
vi = 7.349621967kg/.012kg
vi = 612.4684m/s

The initial velocity seems a bit high and I'm not sure how else to approach the question. Help please! :)
looks good! Bullets travel mighty fast.

1 person
Apart from too many significant figures in the final answer, I do not see anything wrong in your solution. And 612 m/s is not too high for a bullet.

1 person
One cause for concern: it says "topples". That implies a rolling movement, so some of the energy has gone into rotation.

paytona said:

## Homework Statement

A 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. What was the speed of the bullet immediately before the collision?

The attempt at a solution
d=1/2at^2
t=*sqrt(2d-vertical-)/a
t=*sqrt[(2)(1.75m)/(9.81m/s^2)
t= 0.597309633s

d= vt
vf = d(horizontal)/t
vf= 1.25m/0.597309633m/s^2
vf= 2.09216961m/s

Conservation of momentum (P1+P2)before = (Psystem) after
(m of bullet * vi of bullet) + (m of block * vi of block) = (m of system * vf of system)
(0.012kg*vi) + (0) = (3.512kg*2.09216961m/s)
Why are you multiplying by 3.512kg?

AM

Andrew Mason said:
Why are you multiplying by 3.512kg?

AM

Mass of block+bullet?

haruspex said:
Mass of block+bullet?
The problem does not say that the bullet stays in the block but perhaps it is implied. I cannot imagine how a 3.5kg block of wood is going to stop a 12g rifle bullet at 2000 feet per second.

AM

## What is a Kinematics problem involving a bullet in a block of wood?

A Kinematics problem involving a bullet in a block of wood is a physics problem that deals with the motion and velocity of a bullet as it travels through a block of wood. The goal is to analyze the motion of the bullet and determine its final velocity and position.

## What are the key concepts involved in solving a Kinematics problem with a bullet in a block of wood?

The key concepts involved in solving a Kinematics problem with a bullet in a block of wood include Newton's laws of motion, conservation of momentum and energy, and understanding projectile motion. Additionally, knowledge of vectors and trigonometry is necessary to solve these types of problems.

## What are the steps to solve a Kinematics problem with a bullet in a block of wood?

The steps to solve a Kinematics problem with a bullet in a block of wood include: 1) identifying the given variables and information, 2) drawing a diagram to visualize the problem, 3) applying the relevant equations and concepts to solve for the unknowns, and 4) checking the solution for accuracy and reasonableness.

## What are some common mistakes made when solving a Kinematics problem with a bullet in a block of wood?

Some common mistakes made when solving a Kinematics problem with a bullet in a block of wood include: forgetting to account for air resistance, not considering the direction of motion (positive or negative), and using incorrect equations or neglecting certain variables. It is important to carefully read and understand the problem and double check all calculations.

## How can Kinematics problems with a bullet in a block of wood be applied in real life?

Kinematics problems with a bullet in a block of wood have practical applications in forensic science, ballistics, and engineering. These types of problems can also help us understand the physics behind gun recoil and how different materials may affect the motion of a projectile.

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