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

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SUMMARY

The problem involves a 100g granite cube sliding down a 40-degree frictionless ramp, colliding with a 200g steel cube at rest, to achieve a final speed of 150 cm/s for the steel cube. The solution requires applying conservation of momentum and energy principles, specifically for elastic collisions. The granite cube must be released from a height of approximately 25.8 cm to impart the necessary velocity to the steel cube. The discussion emphasizes breaking the problem into two parts: calculating the velocity of the granite cube at impact and then determining the height needed for that velocity.

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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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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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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?
 
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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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!
 
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.
 
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.
 

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