A conservation of mechanical energy and momentum problem

In summary, the problem involves a 6-kg bullet fired into a 2-kg block initially at rest on a table of height 1 m. The bullet remains in the block, causing the block to land 2 m from the bottom of the table. The goal is to determine the initial speed of the bullet. Using the equations for conservation of momentum and conservation of mechanical energy, it is possible to solve for the initial velocity of the bullet. By considering the time it takes for the block to fall 1 m and the distance it travels horizontally, the necessary horizontal speed can be determined. It is important to pay attention to units and use the correct formula when solving for the unknown variable.
  • #1
BrainMan
279
2

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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  • #2
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 [itex]V_{f}[/itex] has to be.

There's only one [itex]V_{f}[/itex] 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 [itex]V_{f}[/itex])
 
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  • #3
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 [itex]V_{f}[/itex] has to be.

There's only one [itex]V_{f}[/itex] 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 [itex]V_{f}[/itex])

How do I find [itex]V_{f}[/itex] 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.
 
  • #4
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.
 
  • #5
CWatters said:
The time taken to do both is the same.

How does that knowledge help me solve this problem?
 
  • #6
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?
 
  • #7
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?
 
  • #8
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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  • #9
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!
 

FAQ: A conservation of mechanical energy and momentum problem

What is conservation of mechanical energy and momentum?

The conservation of mechanical energy and momentum is a fundamental physical principle that states that the total amount of energy and momentum in a closed system remains constant over time, regardless of any internal changes or interactions within the system.

How does conservation of mechanical energy and momentum apply to real-world situations?

In real-world situations, conservation of mechanical energy and momentum can be seen in many phenomena, such as the motion of objects on a frictionless surface, collisions between billiard balls, and the movement of planets in our solar system.

What are the equations used to calculate mechanical energy and momentum?

The equation for mechanical energy is E = K + U, where E is the total mechanical energy, K is the kinetic energy, and U is the potential energy. The equation for momentum is p = m * v, where p is the momentum, m is the mass, and v is the velocity.

How is conservation of mechanical energy and momentum related to the laws of motion?

Conservation of mechanical energy and momentum is closely related to Newton's laws of motion. The first law states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. The second law relates force, mass, and acceleration, while the third law states that for every action, there is an equal and opposite reaction. These laws help to explain the conservation of energy and momentum in different scenarios.

What happens when mechanical energy and momentum are not conserved?

If mechanical energy and momentum are not conserved in a closed system, it means that there must be external forces or interactions acting on the system. This can result in changes in the system's total energy and momentum, leading to different outcomes than what would occur if the conservation laws were followed.

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