How High to Release a Granite Cube to Achieve Specific Speed in a Steel Cube?

[mrjoe2]

Homework Statement

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

Homework Equations

Ek=1/2mv^2
Eg=mgy
maybe Ptot=ptot'

The Attempt at a Solution

first i tried the problem with the conservation of energy where the energy from the first block is transferred to the second (the Ek of 1st block is transferred to Ek second block). that didnt work. then i tried the conservation of momentum but you can't use it because you don't know if it is a perfectly elastic collision of the cubes stick together. you cannot assume anything, so i think i need to utilize the conservation of momentum, but its not giving me the right answer! the answer should be 25.9cm and i keep getting 23cm. also, using the conservation of energy, the angle is irrelevant.

 

[natives]

At first use conservation of energy u=sqrroot(2gh) when the granite block slides down the ramp...Then use both conservation of momentum m1u1+m2u2=m1v1+m2v2 and conservation of energy m1sqr u1+m2sqr u2=m1sqr v1+m2sqr v2...If m1 is granite then u1 is initially zero..Sort out the two equations using the momentum equation to eliminate v2 in the energy equation...Yah,assume a perfectly elastic condition, they normally have a way of telling you if its not!

 

[mrjoe2]

this doesn't work. did you try and get the correct answer when you used your very long process?

 

[natives]

Yes,it did work and its not a very long process..You'll get used to it once you attend to more questions!Show me what you have done!?

 

[natives]

Yes,it did work and its not a very long process..You'll get used to it once you attend to more questions and be careful what value of g you use,refer to the book's tables!Show me what you have done!?

 

[mrjoe2]

Yes,it did work and its not a very long process..You'll get used to it once you attend to more questions!Show me what you have done!?

i set the total energy before the collision (mgh) equal to the kinetic energy of the initial cube and the now moving cube after collision. so

(.1)gh = .5(.1)v1'^2 + .5(.2)(1.5)^2

now i have two variables i need to solve for... first of all, this method doesn't work! if i put in conservation of momentum, then

.1(v1) = .1(v1') + .2(v2') now what!

 

[natives]

Remember v1 is the velocity of the first cube whose initial kinetic energy .5m1v1^2 u substitute with m1gh...So even though you didnt know but u had let .5m1v1^2=m1gh..cant you make v1 a subject and substitute in your momentum equation?then u'd 2 simultaneous equations which am sure you can solve easily!

 

[natives]

Remember v1 is the velocity of the first cube whose initial kinetic energy .5m1v1^2 you had substitute with m1gh...So even though you didnt know it but u had let .5m1v1^2=m1gh..cant you make v1 a subject out of this and substitute it in your momentum equation?then u'd 2 simultaneous equations which am sure you can solve easily!Also remember v2' was given as 150!

 

[natives]

Remember v1 is the velocity of the first cube whose initial kinetic energy .5m1v1^2 you had substitute with m1gh...So even though you didnt know it but u had let .5m1v1^2=m1gh..cant you make v1 a subject out of this and substitute it in your momentum equation?then u'd 2 simultaneous equations which am sure you can solve easily!Also remember v2' was given as 150 cm/s!