Year 12: Cambridge Physics Problems (2D kinetic theory)

In summary, the conversation is discussing the calculation of force on each side of a square frame due to the movement of 500 ants within it. Assuming perfectly elastic collisions and a random motion factor of 1/2, the force on each side is calculated to be 10-6N. The conversation also touches on the assumption that half of the ants are moving in the east-west direction and the other half in the north-south direction at any given time. The time interval during collisions is also mentioned as a necessary factor in calculating the force.
  • #1
johnconnor
62
0
Question:
A square frame of side 10cm rests on a table. Confined within the frame are 500 ants, each of mass 0.001g. The ants rush about randomly with constant speed 0.02m/s, colliding with each other and with the walls. Assuming the collisions are perfectly elastic, calculate for force on each side of the square due to the ants' movement.

Attempt:
Assuming a random motion factor ( I don't really know what's the proper name for it) of 1/2 instead of 1/3 as we are dealing with a 2-dimensional motion, and applying the usual kinetic theory formula, answer is 10-6N.

Thoughts:
What I don't entirely understand is this:
Assume that the word "random" implies that the motion is equivalent to two groups of 250 ants running at right angles to each other and parallel to the sides of the frame.

Are we trying to postulate that the net vector force acting in the region is zero? If that's the case then why the need to have the ants moving "parallel to the sides of the frame"? The resultant force will still be zero if they move at 45° to the planes AND perpendicular to each other.

Can anyone please provide a more elaborate explanation on the bold statement? Thank you!
 
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  • #2
Lets called the directions north,south east and west.
What it is saying is that at any instant approximately 50% of the ants are moving in the north-south direction and the other 50% in the east-west direction.
 
  • #3
johnconnor said:
Assuming a random motion factor ( I don't really know what's the proper name for it) of 1/2 instead of 1/3 as we are dealing with a 2-dimensional motion, and applying the usual kinetic theory formula, answer is 10-6N.

Fightfish said:
Lets called the directions north,south east and west.
What it is saying is that at any instant approximately 50% of the ants are moving in the north-south direction and the other 50% in the east-west direction.

Thanks! Could you please help me out with a proper explanation for my quote above? I'm not really sure whether my explanation is correct, or how could I improve it.
 
  • #4
johnconnor said:
Thanks! Could you please help me out with a proper explanation for my quote above? I'm not really sure whether my explanation is correct, or how could I improve it.

Yes, your answer is correct. 3 is the term that comes into describe how many possible dimensions in which the ant(or molecule, in terms of kinetic theory) could have gone in case of a three dimensional cube. But since you need the force only on one side, you divide it by 3. The same logic is applicable to a two dimensional square.

This is basically the assumption that at any given time half of the ants are moving east-west, and the other half are moving north-south.
 
  • #5
I am confused, don't we NEED the time of collision of the "ant" and the side of the square block to calculate the force ?

F=ma=(mv-mu)/t
we can calculate mv-mu=impulse
But to calculate the force we NEED the time interval during which the collision took place.
That wil give us the force exerted by ONE ant on one side of the container, then we multiply it by , say 250, (using the above argument) to find the force on ONE side.
 
  • #6
hms.tech said:
I am confused, don't we NEED the time of collision of the "ant" and the side of the square block to calculate the force ?

F=ma=(mv-mu)/t
we can calculate mv-mu=impulse
But to calculate the force we NEED the time interval during which the collision took place.
That wil give us the force exerted by ONE ant on one side of the container, then we multiply it by , say 250, (using the above argument) to find the force on ONE side.

Yes, you do need the time. But, if you carefully observe the problem, you will see it hidden in the given information :wink:
 
  • #7
hms.tech said:
I am confused, don't we NEED the time of collision of the "ant" and the side of the square block to calculate the force ?

F=ma=(mv-mu)/t
we can calculate mv-mu=impulse
But to calculate the force we NEED the time interval during which the collision took place.
That wil give us the force exerted by ONE ant on one side of the container, then we multiply it by , say 250, (using the above argument) to find the force on ONE side.

You're right, but in this case you obtain the force exerted on the sides by deriving the formula for Nm<c>^2/3 (which in this case the denominator is 2, not 3).
 

1. What is the purpose of studying Year 12: Cambridge Physics Problems (2D kinetic theory)?

The purpose of studying Year 12: Cambridge Physics Problems (2D kinetic theory) is to understand the behavior of gas particles in a two-dimensional space and how they interact with each other. This knowledge is essential in many areas of physics, including thermodynamics and fluid mechanics.

2. What are the main concepts covered in Year 12: Cambridge Physics Problems (2D kinetic theory)?

The main concepts covered in Year 12: Cambridge Physics Problems (2D kinetic theory) include the ideal gas law, kinetic theory of gases, and the Maxwell-Boltzmann distribution. Students will also learn about temperature, pressure, and the relationship between them.

3. How is Year 12: Cambridge Physics Problems (2D kinetic theory) different from other physics topics?

Year 12: Cambridge Physics Problems (2D kinetic theory) focuses specifically on the behavior of gas particles in a two-dimensional space, whereas other physics topics may cover a broader range of concepts and applications. It also involves a lot of mathematical calculations and problem-solving.

4. What are some real-world applications of Year 12: Cambridge Physics Problems (2D kinetic theory)?

Year 12: Cambridge Physics Problems (2D kinetic theory) has many real-world applications, such as understanding the behavior of gases in weather patterns, designing efficient engines and turbines, and studying the properties of materials at the atomic level. It also plays a crucial role in the development of new technologies and advancements in various industries.

5. Is it necessary to have a strong background in mathematics to study Year 12: Cambridge Physics Problems (2D kinetic theory)?

While a solid understanding of mathematics is beneficial, it is not necessary to have a strong background in mathematics to study Year 12: Cambridge Physics Problems (2D kinetic theory). Basic algebra and geometry skills are sufficient, and students will learn the necessary mathematical concepts as they progress through the course.

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