Bullet shot into a block #2

  • Thread starter oadeyemi
  • Start date
  • #1
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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)
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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What about the distance d?
 
  • #3
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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.
 
  • #4
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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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