How Do You Calculate the Initial Speed of a Bullet Fired into a Block?

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Homework Help Overview

The problem involves calculating the initial speed of a bullet that is fired into a block resting on a frictionless table. The bullet embeds itself in the block, which then falls a distance d after the impact. The context includes concepts from mechanics, specifically dealing with inelastic collisions and energy conservation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to relate potential energy and kinetic energy to find the speed of the bullet after the collision. Some participants question the role of the distance d in the calculations, indicating uncertainty about how to incorporate it into the solution.

Discussion Status

Participants are exploring different aspects of the problem, including the relationship between distance, time, and speed. Some guidance has been offered regarding the use of kinematic equations to find time, but no consensus has been reached on the overall approach.

Contextual Notes

There is a lack of clarity on how to effectively use the distance d in conjunction with the other variables, which may affect the direction of the discussion.

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1. Homework Statement

A bullet of mass m is fired into a block of mass M initially at rest on a frictionless table of height h. The bullet remains m in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

2. Homework Equations

I used (m +M)gh for the potential energy and set that equal to the kinetic energy ½(m + M)v₂² and found that v₂ = √(2gh).

And since it was a completely inelastic collision (right?) I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

Is this right?


3. The Attempt at a Solution

initial velocity = √(2gh)*((m + M)/m)
 
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What about the distance d?
 
rl.bhat said:
What about the distance d?

That's the thing, I din't know what do with it. That's why I'm here for help.
 
First solve for the speed of the bullet+block by finding time, (you already have distance):

[tex]d=v*t[/tex]

From kinematic equation [tex]h = 1/2 at^2[/tex], solve for t: [tex]t=\sqrt(2h/g)[/tex]

Now we know the speed of the bullet+block is just the distance "d" divided by time, so [tex]v_2 = d * \sqrt(g/2h)[/tex].

Using momentum equations, just as you did, [tex]v_1 = ((m+M)/m) * v_2 = ((m+M)/m) * [d * \sqrt(g/2h)][/tex]
 

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