What is the final speed of a box sliding down a ramp onto a horizontal floor?

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The discussion focuses on calculating the final speed of a box sliding down a frictionless ramp onto a horizontal floor. The user attempts to apply energy conservation principles but encounters confusion regarding the velocities involved. Another participant suggests using conservation of momentum, clarifying that momentum is conserved in any direction without external forces. The conversation also touches on the use of forum tags for mathematical notation. The key takeaway is the importance of applying both energy conservation and momentum conservation principles in solving the problem.
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Homework Statement


http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam2/fa08/fig5a.gif

A frictionless ramp of mass 3m is initially at rest on a horizontal frictionless floor. A small box of mass m is placed at the top of the ramp and then released from rest. After the box is released, it slides down the ramp and onto the horizontal floor, where it is measured to have a speed v, having fallen a total distance h.
What is the speed v of the box after it has left the ramp?
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam2/fa08/fig5b.gif

Homework Equations



d(K.E.) = -d(P.E.)


The Attempt at a Solution



I don't quite get the situation here..(maybe poorly worded)

What's the meaning of the velocity of "pointing right" in the figure?

Anyway, when interpreting that as the same variable as velocity of the block,
i tried to do like the following:

Using the above equation,(Energy conservation)

0.5*m*v^2 + 0.5*(3m)*v^2 = mgh

Then the answer coming from this is not in the options...

What's wrong in my attempt?

Could someone help me out here, please?
 
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Hi nahanksh! :smile:

(try using the X2 tag just above the Reply box :wink:)
nahanksh said:
0.5*m*v^2 + 0.5*(3m)*v^2 = mgh

You mean 0.5*m*v^2 + 0.5*(3m)*V^2.

Now use conservation of momentum :smile:
 
I never thought i could use the conservation of momentum !
(I thought it could be used only when there is a collision)

Could you briefly tell me the condition of eligibility for using "conservation of momentum" ?

Thanks a lot ,tiny-tim.

Oh, BTW, x^2 using Tags?
My Tags section shows me "NONE", how did you do it? lol
 
nahanksh said:
Could you briefly tell me the condition of eligibility for using "conservation of momentum" ?

Always always always

momentum (unlike mechanical energy) is always conserved in any direction in which there are no external forces. :smile:
Oh, BTW, x^2 using Tags?
My Tags section shows me "NONE", how did you do it? lol

When you click the QUOTE button, you should get the Reply page, and there are lots of tags just above the Reply box.

[noparse](or just type "x2"[/noparse]: x2 :wink:)
 
Thanks a lot !
God bless you !
 
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