What is the initial speed of a bullet shot into a block on a frictionless table?

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The discussion centers on calculating the initial speed of a bullet shot into a block on a frictionless table. The user applies conservation of energy principles, equating potential energy to kinetic energy, and derives a vertical speed after the block falls. However, there is confusion regarding the distinction between the vertical component of speed and the initial horizontal speed of the block and bullet combination after the collision. Clarifications indicate that the calculated speed is not the initial horizontal speed, which is necessary to determine how far the block lands from the table. The conversation emphasizes the need to correctly identify and separate the components of motion in this inelastic collision scenario.
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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.

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?

The Attempt at a Solution



initial velocity = √(2gh)*((m + M)/m)
 
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Do you have an attempt at the problem? If not, tell me your thought process for the problem.
 
satchmo05 said:
Do you have an attempt at the problem? If not, tell me your thought process for the problem.

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?
 
oadeyemi said:
I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

Something about this chunk doesn't seem right to me for some reason. Everything else is correct, from my understanding. Is this a formula you received from your text, or is this something that you derived yourself?

Check out the Wiki article, I think I see where you're going with this, but I see a slight error in your formula: http://en.wikipedia.org/wiki/Inelastic_collision
 
satchmo05 said:
Something about this chunk doesn't seem right to me for some reason. Everything else is correct, from my understanding. Is this a formula you received from your text, or is this something that you derived yourself?

Check out the Wiki article, I think I see where you're going with this, but I see a slight error in your formula: http://en.wikipedia.org/wiki/Inelastic_collision

It's from my textbook
 
I think the formula here is OK, wiki describes different situation (initial velocity of M not being zero) and uses different symbols, hence the confusion.

Problems are with the other part - v2 that you calculated from the energy conservation is a vertical component of the speed when block with a bullet hits the ground. That's not initial HORIZONTAL speed of the body.
 
Borek said:
I think the formula here is OK, wiki describes different situation (initial velocity of M not being zero) and uses different symbols, hence the confusion.

Problems are with the other part - v2 that you calculated from the energy conservation is a vertical component of the speed when block with a bullet hits the ground. That's not initial HORIZONTAL speed of the body.

Elaborate please
 
Once the bodies (m+M) left the table they behave just like something thrown norizontally - they maintain horizontal speed, but they start to fall down, till they hit the ground at distance d from the table border.
 
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