What Height Must the Granite Cube Start From to Propel a Steel Cube to 150 cm/s?

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In summary, the problem involves a 100g granite cube sliding down a 40 degree frictionless ramp and colliding with a 200g steel cube at rest. The goal is to find the height at which the granite cube should be released in order to give the steel cube a final speed of 150cm/s. Two approaches can be used, either assuming an elastic collision and using conservation of momentum and energy, or breaking the problem into two parts and solving for the initial velocity of the granite cube and then using conservation of energy to find the height. Both methods should give the same answer.
  • #1
mrjoe2
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Homework Statement


no one gave correct help the first time i sent this out. if you can't do it, don't worry.

a 100g granite cube slides down a 40 degree frictionless ramp. at the bottom, just as it exits onto a horizontal table, it collides with a 200 g steel cube at rest, how high above the table should the granite cube be released to give the steel cube a speed of 150cm/s


Homework Equations


energy equations, momentum equations


The Attempt at a Solution


do not assume this is straight forward. i am a first year university student with a lot of physics behind me and i can't get the solution. i tried posting it before, but no one could get the correct answer. before you help, please post your answer.
i used regular equivalent momentums before and after then subbed in the remaining velocity into the energy equations to yield h. sounds simple huh? try it.
 
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  • #2
Let's assume the collision is elastic, so that all the kinetic energy is transferred to the steel cube.
We can use the definition of momentum to find the final momentum of the steel cube:

Psteel = mv

Just plug in the given constants to find the momentum.
In order for our granite cube to have that momentum, it must satisfy:

Pgranite = Psteel
mgranitevgranite = Psteel
vgranite = Psteel / mgranite

Since mgranite is given to us, we can find vgranite.

Now, that's just the setup. Did you get that far?
The tricky part is finding the needed table height.
How might you go about finding the height that would result in a final velocity found above?
I'll give you a hint: Take a look at Newton's equations. Which one is the most useful here?
 
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  • #3
ok so, that was one of my many numerous methods. try getting the answer, it will be wrong. you will probably get 40 something. also, you can't assume the collision is inelastic. in fact, my first post of this question had numerous people say i have eto assume it is elastic!

gendou2 said:
Let's assume the collision is inelastic, so that all the kinetic energy is transferred to the steel cube.
We can use the definition of momentum to find the final momentum of the steel cube:

Psteel = mv

Just plug in the given constants to find the momentum.
In order for our granite cube to have that momentum, it must satisfy:

Pgranite = Psteel
mgranitevgranite = Psteel
vgranite = Psteel / mgranite

Since mgranite is given to us, we can find vgranite.

Now, that's just the setup. Did you get that far?


The tricky part is finding the needed table height.
How might you go about finding the height that would result in a final velocity found above?
I'll give you a hint: Take a look at Newton's equations. Which one is the most useful here?
 
  • #4
I am quite sure this can be considered an elastic collision. I got 25.8 cm. But I did it pretty quickly too. Do you know the answer?
You were doing correctly in your other post. I do not know why you stopped!
 
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  • #5
can you give me your steps please? general steps nothing perfuse. thanks! you got the correct answer!

Saladsamurai said:
I am quite sure this can be considered an elastic collision. I got 25.8 cm. But I did it pretty quickly too. Do you know the answer?
You were doing correctly in your other post. I do not know why you stopped!
 
  • #6
mrjoe2 said:
...you can't assume the collision is inelastic...
You're quite right. I meant to type elastic rather than inelastic. A typo. Sorry about that.
 
  • #7
Yes. Break the problem into two parts. That may be easier than trying to do it all at once like in your other post.

Part 1 should be to determine at what speed does the granite clock need to be traveling at when it makes contact with the steel block. This can be done by noting that both linear momentum and kinetic energy are conserved assuming an elastic collision.

That gives you two equations/two unknowns-->solve for the initial velocity of the granite block.

Part 2) Now using that velocity as the final velocity of the granite block in a conservation of mechanical energy scenario involving only the granite block, find out what height the granite block must be released from to achieve that velocity.

After doing this, perhaps take a good look at the problem and note how those two parts could have been merged into one.
 

Related to What Height Must the Granite Cube Start From to Propel a Steel Cube to 150 cm/s?

1. What is the purpose of trying the blocks again?

The purpose of trying the blocks again is to gather more data and information from the experiment in order to improve our understanding and results.

2. How many times should we try the blocks again?

It depends on the specific experiment and what information we are trying to gather. Generally, multiple trials are recommended to ensure accuracy and validity.

3. How long should we wait before trying the blocks again?

Again, this depends on the experiment and what we are trying to achieve. Some experiments may require immediate repetition while others may require a longer waiting period to allow for potential changes.

4. How do we ensure consistency when trying the blocks again?

To ensure consistency, it is important to follow the same procedures and methods as the initial experiment. This includes using the same materials, techniques, and controls.

5. What do we do if the results are different when trying the blocks again?

If the results are significantly different when trying the blocks again, it is important to analyze and evaluate the potential factors that may have caused the variation. This may include checking for errors or reevaluating the experimental design.

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