Trouble applying the concept of momentum conservation

In summary, the conversation discusses the concept of momentum conservation and how it applies to a specific problem. The participants consider the conservation of momentum in each dimension and how it relates to the final velocity of Puck B. They also discuss the possibility of solving for Puck B's final velocity by setting the initial and final momentum in each dimension equal to each other. The conclusion is that momentum is conserved, but some of it is transferred to the other puck.
  • #1
vineroon
12
0
v60yz4.jpg


I am having trouble applying the concept of momentum conservation to this problem. The particular problem I am having is in figuring out if I did part (c) correctly.

A look at Puck A:

It is evident that total the momentum is conserved before the collision because the external forces acting on the pucks (the normal force exerted by the ground on the puck and the weight of the puck) both add up to zero. Therefore, according to definition, the total momentum is conserved.

However, the specific text I am referencing for this material states that "conservation of momentum means conservation of its components." Yet, the momentum vector before collision is .400i + .300j and the momentum vector after collision is .300i + .400j. The components of these two vectors are not equal, so does that suggest that momentum ISN'T conserved?

Part (c):

Could you assume that if the total momentum is conserved, you can solve for the final velocity of Puck B by setting (m)(v_1x) = (m)(v_2x) and (m)(v_1y) = (m)(v_2y)? This would give you the exact x and y components of initial Puck B velocity, including the signs. However, the magnitude would still remain the same and agrees with the assumption.

Any help would be greatly appreciated.
 
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  • #2


vineroon said:
[...]Yet, the momentum vector before collision is .400i + .300j and the momentum vector after collision is .300i + .400j. The components of these two vectors are not equal, so does that suggest that momentum ISN'T conserved?

Momentum is conserved, it's just that some of the momentum has been passed off to the other puck. The net momentum in each dimension will always be the same (in this example).

vineroon said:
Part (c):

Could you assume that if the total momentum is conserved, you can solve for the final velocity of Puck B by setting (m)(v_1x) = (m)(v_2x) and (m)(v_1y) = (m)(v_2y)? This would give you the exact x and y components of initial Puck B velocity, including the signs. However, the magnitude would still remain the same and agrees with the assumption.

Any help would be greatly appreciated.

So sum up the total momentum in each plane, subtract final momentum of puck A from this, you get final momentum for puck B.
Take the magnitude of this for final momentum, divide by mass for final velocity.
 
  • #3


Rake-MC said:
Momentum is conserved, it's just that some of the momentum has been passed off to the other puck. The net momentum in each dimension will always be the same (in this example).
So sum up the total momentum in each plane, subtract final momentum of puck A from this, you get final momentum for puck B.
Take the magnitude of this for final momentum, divide by mass for final velocity.

Alright, that seems a lot more straightforward. Thanks a lot. :D
 
  • #4


Not necessarily.
Because both pucks have identical mass, it is infact possible to say that the final velocity of the system in each dimension is the same as the initial. But you cannot simply say the final velocity of each puck is the same as initial (in each dimension).
 

1. What is momentum conservation?

Momentum conservation is a fundamental principle in physics that states that the total momentum of a closed system remains constant in the absence of external forces.

2. How is momentum conserved?

Momentum is conserved because, according to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that when two objects interact, the total momentum before and after the interaction must be equal.

3. Why is it important to understand momentum conservation?

Understanding momentum conservation is crucial in many areas of science and engineering, including mechanics, fluid dynamics, and astrophysics. It allows us to accurately predict the behavior of objects and systems and has practical applications in fields such as transportation and sports.

4. Can momentum be lost or gained?

No, momentum cannot be lost or gained in a closed system. It can only be transferred from one object to another. This is known as the law of conservation of momentum.

5. How is momentum conservation applied in real-world situations?

Momentum conservation can be applied in various real-world situations, such as car crashes, rocket launches, and collisions between billiard balls. It allows us to analyze and understand the motion and interactions of objects in these scenarios.

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