Use energy conservation to determine the forces

[Dick]

Yes. So it's a r-cubed force instead of an r-squared force like gravity.

 

[imy786]

is this correct

a (i) (ii)

 

[Dick]

You are correct.

 

[imy786]

r cubed force...is that a correct name for a force??

 

[Dick]

r cubed force...is that a correct name for a force??

No, it's not official. I just said it to point out that the force is related to a cube not a square (like gravity).

 

[Dick]

(iv) provide brief explanation in each case.

(a) Is the force law in (a-=r cubed force) attractive or repulsive?

repulsive as it has positive sign

(b)Does it satisfy Newton’s third law?

yes, as the object is two particles are opposing forces on each other
(c)Is it an inverse square law of force?

yes- it is inversely porptional
(d)Does it conserve linear momentum?

yes-momentum is alwayz converved in a colliosn

(e)Does it conserve angular momentum?

as momentum is alwayz conserved, so does angular momentum

[are these correct for these quesitons}

a) It's attractive. The potential was given as -k/r^2. Not k/r^2. Reverse your sign.

b) Give a clearer statement of Newtons third law.

c) WHY oh WHY would you say it is inverse square when you have just computed it to be inverse cube?

d) Linear momentum conservation is the same as Newtons third law.

e) Does your force exert any torques in the system?

 

[Dick]

so what is the r-cubed force called?

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n3rd law- "For every action, there is an equal and opposite reaction."

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(c)Is it an inverse square law of force
no its an inverse cube

(d)Does it conserve linear momentum?

yes-momentum is alwayz converved in a collioion and is same as Newtons 3rd law

(e)Does it conserve angular momentum?

no, as there is no torque not applied to the system.

(is this correct)

Inverse r cubed laws don't have any particular name. ARE YOU SURE THAT IN YOUR STATEMENT YOU DIDN'T MEAN TO SAY U=-k/r (then it would be gravity). To show Newton's third law you want to show that the force m1 exerts on m2 is equal and opposite to the force m2 exerts on m1. Then linear momentum is conserved. If there is no torque applied by the force then angular momentum must be conserved. How would you show there is no torque??

 

[imy786]

moment= force (torque * perpendicular distance)

so if there is no moment on the foice there would be no toruq

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quesitons does say

U= -k/ r^2

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can this be a mistake in the quesiton

 

[Dick]

moment= force (torque * perpendicular distance)

so if there is no moment on the foice there would be no toruq

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quesitons does say

U= -k/ r^2

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can this be a mistake in the quesiton

If that's what the question says, then the answer is our inverse cubed force. The forces are directed straight from one particle to the other - so there is no perpendicular component to cause a torque. Is that what you mean?

 

[imy786]

yeah ,

well are all the quesitons..all ok..now and correct..and any parts still needed more clarifcaiton??

 

[Dick]

If you are comfortable with the answers, then I am.

 

[Dick]

It's certainly true that if something is always conserved - then it is conserved. But that's not a very interesting answer. On the other hand you can SHOW N3LAW works. Why not cite that as evidence?