Conservation of linear momentum problem

In summary, the problem the person is having is that they are trying to solve a conservation of linear momentum problem without taking into account the horizontal component of the velocity. They suggest using a shorter approach that calculates the resultant velocity of the system, which is 9.6 meters per second.
  • #1
133
0
Hi friends,
I have an issue in solving a Conservation of linear momentum problem.
Please Help me in solving this.
Thank you all in advance.

The problem is as:

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1379404_1432382880322151_1349912335_n.jpg


Attempt:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn2/q80/s720x720/1234574_1432380606989045_407245962_n.jpg
https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash4/q84/s720x720/1391595_1432380693655703_759644108_n.jpg

Friends I am not sure of that, what I am doing is absolutely correct. Please try to help me.
I will appreciate the help.
 
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  • #2
coldblood said:
Hi friends,
I have an issue in solving a Conservation of linear momentum problem.
Please Help me in solving this.
Thank you all in advance.

The problem is as:

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1379404_1432382880322151_1349912335_n.jpg


Attempt:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn2/q80/s720x720/1234574_1432380606989045_407245962_n.jpg
https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash4/q84/s720x720/1391595_1432380693655703_759644108_n.jpg

Friends I am not sure of that, what I am doing is absolutely correct. Please try to help me.
I will appreciate the help.

The thing you are doing wrong is that you found a force acting on the particle due horizontal. No. There is no net external force acting on the horizontal. So the linear momentum of particle+wedge system remains conserved. Make use of that knowledge. Initial linear momentum of the system=0.
 
  • #3
sankalpmittal said:
The thing you are doing wrong is that you found a force acting on the particle due horizontal. No. There is no net external force acting on the horizontal. So the linear momentum of particle+wedge system remains conserved. Make use of that knowledge. Initial linear momentum of the system=0.

Well my friend sankalpmittal,

I am taking forces for the isolated bodies.Not for the system Hence I think for can be taken in the horizontal direction.

Well for the system conserving momentum in the horizontal direction,

momentum before = momentum after

0 = 0.2 x 2.4 - v x 0.1
that gives, v = 4.8 m/s

But this velocity is only for the horizontal direction. For the vertical direction we have to calculate the time or vertical height fall.
Is in this way Can the answer be calculated in the short manner and how?
 
  • #4
coldblood said:
Well my friend sankalpmittal,

I am taking forces for the isolated bodies.Not for the system Hence I think for can be taken in the horizontal direction.

Well for the system conserving momentum in the horizontal direction,

momentum before = momentum after

0 = 0.2 x 2.4 - v x 0.1
that gives, v = 4.8 m/s

But this velocity is only for the horizontal direction. For the vertical direction we have to calculate the time or vertical height fall.
Is in this way Can the answer be calculated in the short manner and how?

Firstly there was a vertical force on the system downwards which we call gravitation force. That was the only net force on system. If you resolve it, how can you get its effect in the direction perpendicular to it ? In fact, effect of a net force in a direction perpendicular to it is 0.

Finding component of a component which you did is a meaningless operation.

And again what you did not wrong but lengthy approach, that I have marked red. You conserve linear momentum in horizontal. So you have to account for the horizontal component of the velocity.

0 = 0.2 x 2.4 - vcos(60) x 0.1

Here v is the resultant velocity of particle indeed. This is a shorter approach? Ain't it ? :p

(Note: The question does not end here.)
 
  • #5
sankalpmittal said:
Firstly there was a vertical force on the system downwards which we call gravitation force. That was the only net force on system. If you resolve it, how can you get its effect in the direction perpendicular to it ? In fact, effect of a net force in a direction perpendicular to it is 0.

Finding component of a component which you did is a meaningless operation.

And again what you did not wrong but lengthy approach, that I have marked red. You conserve linear momentum in horizontal. So you have to account for the horizontal component of the velocity.

0 = 0.2 x 2.4 - vcos(60) x 0.1

Here v is the resultant velocity of particle indeed. This is a shorter approach? Ain't it ? :p

Really sankalp this is shorter approach. I got the horizontal component of velocity relative to wedge as 9.6m/s second.
But there will be vertical component also. I think for to find this I have to find time duration or something else, isn't it?
Please resolve, I appreciate the help.
 
  • #6
coldblood said:
Really sankalp this is shorter approach. I got the horizontal component of velocity relative to wedge as 9.6m/s second.
But there will be vertical component also. I think for to find this I have to find time duration or something else, isn't it?
Please resolve, I appreciate the help.

Also from my approach the resultant is coming 9.6m/s..
 
  • #8
coldblood said:
Also from my approach the resultant is coming 9.6m/s..

Answer does not matter! What matters is the approach ! And see my reasoning again. And please write the solution with the procedure I suggested you. If not, then find the correct approach yourself.

No you do not need time. The v you calculated is speed absolute.
The relative horizontal speed = vcos(60)+v of wedge
Then you can calculate vertical velocity relative as vsin(60).
Then net relative speed can be obtained.

And the another thread which you have directed me to, seems to be question of FIITJEE. The formula you applied there is not workable for oblique collision. Please refer theory for oblique collision.
 

1. What is the conservation of linear momentum problem?

The conservation of linear momentum problem is a fundamental principle in physics that states that the total linear momentum of a closed system remains constant over time, regardless of any internal forces or external forces acting on the system. This means that the total momentum of all objects involved in a collision or interaction will remain the same before and after the event.

2. Why is conservation of linear momentum important?

The conservation of linear momentum is important because it allows us to predict the motion of objects in a system without having to consider all of the individual forces acting on each object. It is a fundamental law of physics and is essential in understanding and describing many physical phenomena, such as collisions, explosions, and other interactions between objects.

3. How is conservation of linear momentum applied in real-world scenarios?

The conservation of linear momentum can be applied in a variety of real-world scenarios, such as in car accidents, sports collisions, and rocket launches. In these situations, the total momentum of the system is conserved, allowing scientists and engineers to predict the outcomes and design safe and efficient systems.

4. What are the conditions for conservation of linear momentum to hold true?

The conservation of linear momentum holds true in a closed system where there are no external forces acting on the system. This means that the total momentum of all objects in the system remains constant, and any changes in momentum are due to internal forces between the objects.

5. How does the principle of conservation of linear momentum relate to other conservation laws?

The principle of conservation of linear momentum is closely related to other conservation laws, such as the conservation of energy and the conservation of angular momentum. These laws all stem from the fundamental principle of conservation of mass, which states that the total amount of matter in a closed system remains constant over time.

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