(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A .2kg block from rest slides down a frictionless ramp that is .5 meters long. The block lands on a flat surface and moves towards another block that is 1.4 meters away. The other block is not moving and it has a mass of .8kg. The Miu (sp?) of the flat surface is .1. Once the two blocks collide, what is the speed of the second block?

Here is a picture of the scenario:

http://i91.photobucket.com/albums/k285/happygolucky2442/abeapic.jpg

[tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

2. Relevant equations

P=MV

MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

[tex]f_k = \mu_k N[/tex]

[tex]\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/tex]

3. The attempt at a solution

[tex]v^2=2ad[/tex]

[tex]v^2=9.8(.5)[/tex]

[tex]v^2=4.9[/tex]

[tex]v=2.21m/s[/tex]

[tex]f_k = \mu_k N[/tex]

Ff= -.1(1.96)

Ff= -.196

A=f/m

a=-.1

[tex]v^2=2ad[/tex]

=4.9 - .28

=4.62

=2.15

MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

.2(2.15) +.8(0) = .2(0) + .8VB2

.43=.8VB2

VB2=.5375 m/s

Please check and see if my answer is correct. Thanks.

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# Conservation of Momentum- with ramp

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