Conservation of Momentum- with ramp

[JerG90]

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 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.

 

[m2003]

I would like to commend you on your attempt to solve this problem using the conservation of momentum and the equations for motion. Your approach is correct and your calculations seem to be accurate. However, I would like to point out that in the equation MA1VA1+ MB1VB1=MA2VA2+ MB2VB2, the masses should be in kilograms and the velocities should be in meters per second. So, the correct equation would be 0.2(2.21) + 0.8(0) = 0.2(0) + 0.8VB2 which gives a final velocity of 0.4425 m/s for the second block. Also, when calculating the acceleration, it is important to consider the direction of the forces and velocities. In this case, the frictional force should be subtracted from the net force and the acceleration should be negative since it opposes the motion of the block. So, the correct equation would be a = (-0.196 - 0)/0.8 = -0.245 m/s^2. Overall, your approach and solution are good, but it is important to pay attention to units and direction when dealing with equations and calculations in physics.