Conservation of Momentum- with ramp

In summary, a .2kg block slides down a frictionless ramp and lands on a flat surface, moving towards another stationary block with mass .8kg. The coefficient of friction on the flat surface is .1. Using the equations v^2 = v_0^2 + 2 a \Delta x and MA1VA1+ MB1VB1=MA2VA2+ MB2VB2, we can calculate the speed of the second block to be 2.15 m/s, assuming an elastic collision. However, more data is needed to determine the exact speed of the second block.
  • #1
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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

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

Homework 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]


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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  • #2
JerG90 said:

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

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

Homework 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]


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 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.
[tex]v^2=2ad[/tex]
=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.
 
Last edited:
  • #3


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.
 

1. What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant. This means that in any interaction or collision between objects, the total momentum before and after the interaction is the same.

2. How does the conservation of momentum apply to a ramp?

In the context of a ramp, the conservation of momentum means that the total momentum of the object at the top of the ramp will be the same as the total momentum of the object at the bottom of the ramp. This is because the ramp is a closed system, meaning that no external forces are acting on the objects as they move down the ramp.

3. Does the mass of the objects on the ramp affect the conservation of momentum?

Yes, the mass of the objects on the ramp does affect the conservation of momentum. The heavier the objects, the greater their momentum will be, and the more difficult it will be to change their momentum. This is why larger objects tend to have more momentum and are harder to stop or change direction.

4. How does the angle of the ramp affect the conservation of momentum?

The angle of the ramp can affect the conservation of momentum in a few ways. If the ramp is steeper, the objects will gain more momentum as they move down the ramp. Additionally, the angle of the ramp can also affect the direction of the objects' momentum, as a steeper ramp will cause the objects to gain more vertical momentum compared to a shallower ramp.

5. Can the conservation of momentum be violated on a ramp?

No, the conservation of momentum cannot be violated on a ramp. As long as there are no external forces acting on the objects, the total momentum of the system will remain constant. If it appears that the conservation of momentum is violated, it is likely due to external forces acting on the objects, such as friction or air resistance.

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