How Should Momentum Conservation Be Applied in a Swing Ball Lab?

In summary: So, to summarize, we need to find the weight of each object (mg) and multiply it by the difference in time between A and B and the difference in time between C and D. Then, we need to subtract these values from the total change in momentum (PDt) to find the change in momentum due to gravity.I hope this explanation helps you better understand the equation and the values you need to use for it. Please let me know if you have any further questions or need any clarification. Good luck with your lab!In summary, the equation given by your teacher is used to calculate the change in momentum due to the work done by gravity in a specific lab scenario. To
  • #1
DalekSec
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Homework Statement


The problem I am having, is my physics teacher will not explain a solution that he is having us use for a lab, and so I don't know what values to use for the equation.

The equation is to find the increase in momentum due to the work done by gravity. This is for a lab in which one metal ball berring is released from rest at a height 1.215 m above the ground. It swings downward on two strings, equal tension assumed, and strikes another ball at height .675 m above the ground, both ball are of equal mass. This ball then flies off the table and strikes the ground. The four reference frames are A: Both balls at rest, B: Ball on string was released and is about to hit the second ball at 3.255 m/s, C: second ball flies through the air and has an assumed air time of .371 sec., D: second ball is about to strike the ground. The problem is he wants us to use this equation, which I have no idea what to do with, and he won't tell us how to get these values.

I am a high school senior in AP Mechanics, and this is only a small portion of the lab. I just need to know what I am supposed to use for these values.


Homework Equations


%(delta)P= [PDt-(impulse due to gravity)(from time A to B)(in the vertical direction) -(impulse due to gravity)(from time C to D)(in the vertical direction)]/[(impulse due to gravity)(from A to B)(total),{i think that is what the t means},+ (impulse due to gravity)(from Cto D)(total)]


The Attempt at a Solution


There is no real attempted solution on this, I just need to know what some of these values are supposed to be, because he writes one thing and says another over the phone, and has confused me to know end.
 
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  • #2


Dear student,

I apologize for the confusion you are experiencing with your physics lab. it is important for me to help you understand the concepts and equations being used in your lab so that you can successfully complete it.

Firstly, let's break down the equation that your teacher has provided for you. It is a bit complicated, but I will explain it step by step.

%(delta)P= [PDt-(impulse due to gravity)(from time A to B)(in the vertical direction) -(impulse due to gravity)(from time C to D)(in the vertical direction)]/[(impulse due to gravity)(from A to B)(total),{i think that is what the t means},+ (impulse due to gravity)(from Cto D)(total)]

The symbol %(delta)P represents the change in momentum. This is the quantity that we are trying to calculate in this lab.

The first part of the equation, [PDt], represents the total impulse due to gravity from time A to time D. This means we are looking at the total change in momentum of the system from the initial time (when both balls are at rest) to the final time (when the second ball strikes the ground).

The next part of the equation, -(impulse due to gravity)(from time A to B)(in the vertical direction), represents the impulse due to gravity acting on the system from time A to time B. This is the time when the first ball is released and begins to swing downward on the strings.

Similarly, the next part, -(impulse due to gravity)(from time C to D)(in the vertical direction), represents the impulse due to gravity acting on the system from time C to time D. This is the time when the second ball is flying through the air and about to strike the ground.

Now, for the denominator, we have [(impulse due to gravity)(from A to B)(total),+ (impulse due to gravity)(from Cto D)(total)]. This represents the total impulse due to gravity acting on the system from time A to time B and from time C to time D.

Now, to find these values, we need to use the equation for impulse, which is given by FΔt = ΔP. In this case, since we are dealing with gravity, the force will be the weight of the objects (mg), and the time interval (Δt) will be
 
  • #3


As a scientist, it is important to have clear and thorough explanations in order to properly understand and conduct experiments. It is concerning that your physics teacher is not providing you with the necessary information to complete the lab and understand the equations being used. It is important to have a solid understanding of the concepts and equations being used in order to accurately analyze and interpret the results of the experiment. I would suggest discussing your concerns with your teacher and asking for a more thorough explanation of the equation and values being used. It is also important to ask for clarification on any discrepancies between written instructions and verbal instructions. Remember, as a scientist, it is important to have a clear understanding of the methods and data being used in order to draw accurate conclusions from experiments.
 

FAQ: How Should Momentum Conservation Be Applied in a Swing Ball Lab?

What is momentum conservation?

Momentum conservation refers to the principle in physics that states the total momentum of a closed system remains constant unless acted upon by an external force. This means that the total momentum of all objects involved in a collision or interaction will remain the same before and after the event.

How is momentum conserved?

Momentum is conserved through the transfer of momentum between objects in a system. In a closed system, the total momentum before and after an event will be the same, even if individual objects may experience changes in momentum. This is due to the principle of action and reaction, where every force has an equal and opposite reaction force.

Can momentum not be conserved?

In some cases, momentum may not be conserved. This can occur in open systems where external forces are present, such as air resistance or friction. In these cases, the total momentum of the system may change due to the transfer of energy between the system and its surroundings.

How does momentum conservation apply to real-life situations?

Momentum conservation is applicable to a wide range of real-life situations, such as collisions between objects, explosions, and rocket propulsion. It is also used in engineering to design systems that effectively transfer momentum, such as in cars and airplanes.

What are the implications of momentum conservation in the field of physics?

Momentum conservation is a fundamental principle in physics and has many implications in the study of mechanics, dynamics, and other branches of physics. It is a crucial concept in understanding and predicting the behavior of objects in motion and is often used in calculations and experiments to analyze physical systems.

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