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Conservation of linear momentum problem

  • Thread starter coldblood
  • Start date
  • #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.
 

Answers and Replies

  • #2
785
15
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
133
0
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
785
15
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
133
0
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
133
0
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
785
15
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.
 

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