Speed of Bullet: Solving for Velocity in Wood Block Collision

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Homework Help Overview

The discussion revolves around a physics problem involving a bullet colliding with a wood block, where the goal is to determine the bullet's speed based on the subsequent motion of the block. The context includes concepts of kinetic energy, momentum, and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of conservation of energy and momentum principles. Questions arise regarding the interpretation of the equations provided, particularly the use of kinetic energy and the implications of friction on the system's energy.

Discussion Status

The discussion is active, with participants offering insights and questioning the original poster's approach. Some guidance has been provided regarding the relationship between kinetic energy and work done by friction, although there is no explicit consensus on the correct method to solve the problem.

Contextual Notes

There are indications of confusion regarding the equations used, particularly the potential misapplication of conservation of momentum in the context of an inelastic collision. The original poster expresses uncertainty about their calculations and the validity of their approach.

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A 12 g bullet is fired into a 8.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.

What was the speed of the bullet? Relevant equations
My professor told us that we would need these equations:
(1/2)mv2 = mgx

mvi = (m1+m2)v2The attempt at a solution
v2 = 2(9.81)(0.20)(0.05)
v = 0.443

0.012vi = (8.0+0.012)(0.443)2
vi = 131 m/s

This seemed like a logical process and the answer seemed fairly reasonable (bullets move fast) however, this is not the right answer and I don't know what I am doing wrong.
 
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Do you understand what the equations you use are telling you?

You are basically using conservation of energy. Before the bullet hits the block, the block is at rest and the bullet has a kinetic energy E = 1/2 m_\text{bullet} v_\text{bullet}^2.
After the bullet hits the block, the bullet and the block start moving together. So they get some kinetic energy. How much (expressed in E)?
There acts a frictional force on the block (with bullet) which will bring it to a stop. How much energy will the bullet+block have when it has come to rest? What happened to the energy that is missing? Can you write down a formula for the work done by the frictional force?
 
Wow, I've never heard of a professor just giving out equations like that. I think you might have copied his/her second equation wrong. That equation looks like conservation of momentum for an inelastic collision, in which case you should not have a v squared.
 
In other words: there is a kinetic energy imparted into the block when the bullet collides with it. That kinetic energy is the kinetic energy of the new block-bullet system. Next the block moves a distance d until it reaches a stop. This means that friction does work on it over that distance d. Work is equal to change in kinetic energy. The initial kinetic energy is just the kinetic energy imparted to the block by the bullet while the final is zero because it comes to a stop.
 
That's a nice explanation NovaKing, but why on Earth did you revive a thread that's been dead for almost a year?
 
My friend needed help on the question and when he googled it he got this. I thought I would give an alternative explanation for anyone else in his class who also googled it. heh just helping I suppose...if I was right that is.
 

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