# Conservation of Momentum- with ramp

## Homework Statement

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

$$v^2 = v_0^2 + 2 a \Delta x$$

## Homework Equations

P=MV

MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

$$f_k = \mu_k N$$

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

## The Attempt at a Solution

$$v^2=2ad$$
$$v^2=9.8(.5)$$
$$v^2=4.9$$
$$v=2.21m/s$$

$$f_k = \mu_k N$$
Ff= -.1(1.96)
Ff= -.196

A=f/m

a=-.1

$$v^2=2ad$$
=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.

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

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

$$v^2 = v_0^2 + 2 a \Delta x$$

## Homework Equations

P=MV

MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

$$f_k = \mu_k N$$

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

## The Attempt at a Solution

$$v^2=2ad$$
$$v^2=9.8(.5)$$
$$v^2=4.9$$
$$v=2.21m/s$$

$$f_k = \mu_k N$$
Ff= -.1(1.96)
Ff= -.196

A=f/m

a=-.1 you've made an error here, instead of plugging in the mass to detrmine a, you plugged in the weight by mistake. This will affect your result for detrmining v below, which otherwise would be correct.
$$v^2=2ad$$
=4.9 - .28
=4.62
=2.15

MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

.2(2.15) +.8(0) = .2(0) + .8VB2, good formula, butwhy did you assume VA2 is 0? The problem is insufficiently worded to get a result without knowing whether the collision is elastic or inelastic, or without other given data. MA might actually bounce back in the opposite direction with a smaller speed, and MB move forward with not much speed, but more data is needed about the speed of MA before calculating the speed of MB

.43=.8VB2

VB2=.5375 m/s

Please check and see if my answer is correct. Thanks.
See my comments in red above.

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