A conservation of mechanical energy and momentum problem

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SUMMARY

The discussion centers on solving a conservation of mechanical energy and momentum problem involving a 6-kg bullet and a 2-kg block. The bullet embeds itself in the block, which then falls from a height of 1 meter and travels 2 meters horizontally. Participants clarify the use of the equations for conservation of momentum and mechanical energy, specifically m1v1i + m2v2i = (m1 + m2)vf and 1/2mvi2 + mgyi = 1/2mvf2 + mgyf. The correct approach involves calculating the time to fall and the necessary horizontal speed to solve for the bullet's initial velocity.

PREREQUISITES
  • Understanding of conservation of momentum
  • Familiarity with conservation of mechanical energy principles
  • Ability to calculate time of free fall from a height
  • Basic knowledge of projectile motion
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  • Learn how to apply conservation of momentum in inelastic collisions
  • Study the equations of motion for free-falling objects
  • Explore projectile motion concepts, including horizontal and vertical components
  • Practice solving ballistic problems involving energy conservation
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of conservation laws in action.

BrainMan
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Homework Statement


a 6-kg bullet is fired into a 2-kg block initially at rest at the edge of a table of height 1 m. The bullet remains in the block, and after the impact the block lands 2 m from the bottom of the table. Determine the initial speed of the bullet.


Homework Equations


m1v1i + m2v2i = (m1 +m2)bf

1/2mvi2 + mgyi = 1/2mvf2 +mgyf

The Attempt at a Solution


I attempted to do this problem by first putting the information into the equation for the conservation of momentum. So (.006)v1i +0 = (2.006)(vf). Unfortunately this equation has two unknowns. I know that in order to substitute for the unknowns I must use the formula for the conservation of mechanical energy. I don't know how to do this based on the information provided. Any help would be greatly appreciated.
 
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BrainMan said:
a 6-kg bullet.

I think that's a typo, or it's just a huge bullet haha

BrainMan said:
(.006)v1i +0 = (2.006)(vf). Unfortunately this equation has two unknowns.

Not a problem, because you can figure out what V_{f} has to be.

There's only one V_{f} that satisfies the constraints of the problem (moving 2 meters horizontally in the time it takes to fall 1 meter)

So the only unknown is the one you're solving for. (But first you need to write another equation and solve for V_{f})
 
Last edited:
Nathanael said:
I think that's a typo, or it's just a huge bullet haha



Not a problem, because you can figure out what V_{f} has to be.

There's only one V_{f} that satisfies the constraints of the problem (moving 2 meters horizontally in the time it takes to fall 1 meter)

So the only unknown is the one you're solving for. (But first you need to write another equation and solve for V_{f})

How do I find V_{f} based off the fact that it moves two meters horizontally in the time it takes to fall one meter? And its supposted to be grams my bad.
 
How do I find Vf based off the fact that it moves two meters horizontally in the time it takes to fall one meter?

The time taken to do both is the same.
 
CWatters said:
The time taken to do both is the same.

How does that knowledge help me solve this problem?
 
Until the block actually slides off the table, the vertical component of the velocity is 0. can you solve for the time it takes an object to hit the ground if it starts from rest at a height of 1m?
 
BiGyElLoWhAt said:
Until the block actually slides off the table, the vertical component of the velocity is 0. can you solve for the time it takes an object to hit the ground if it starts from rest at a height of 1m?

Yes I do according to my calculations it would take .45 seconds to hit the ground. Now what do I do?
 
BrainMan said:
Yes I do according to my calculations it would take .45 seconds to hit the ground. Now what do I do?

Find the horizontal speed necessary to move 2 meters in 0.45 seconds.

Does that speed play any special role in this problem?
 
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BrainMan said:
How does that knowledge help me solve this problem?

See replies from others.

Looks like you should get some practice with ballistic problems. You know the sort of thing... a projectile is thrown at an angle Θ to the ground with an initial velocity u. How far does it go? That sort of thing.
 
  • #10
Nathanael said:
Find the horizontal speed necessary to move 2 meters in 0.45 seconds.

Does that speed play any special role in this problem?

OK what I did now was realize that if it moved 2 meters in .45 seconds it must be traveling at a rate of .225 m/s. Then I plugged that into the original equation for the final velocity
(.006)v1f = (m1+m2)(.225)
I found the answer to be 75.225 which is clearly wrong. What did I do wrong?
 
  • #11
BrainMan said:
if it moved 2 meters in .45 seconds it must be traveling at a rate of .225 m/s.
...
What did I do wrong?

If it moves 0.225 meters in 1 second, (0.225 m/s) then how does it move 2 meters in 0.45 seconds?

(You divided 0.45 by 2 instead of 2 by 0.45)
 
  • #12
Nathanael said:
If it moves 0.225 meters in 1 second, (0.225 m/s) then how does it move 2 meters in 0.45 seconds?

(You divided 0.45 by 2 instead of 2 by 0.45)

I see my mistake. Whoops! Thanks I got it right now!
 

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