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# Variable force motion watch

1. Attachment 395267I've been set a worksheet with a few questions on Variable forces as a prompt to learn the topic by doing questions. I haven't been given any notes or prompts on it, so I was hoping for some hints.

What I have so far, is, letting the mass of the particle be , is the force in the direction towards the origin, is the distance from the origin and is a positive constant, so that

The momentum of the particle at B and A respectively are and .

The forces on the particle at B and A respectively are and .

However, I don't really see how to go on from there to show that .Any hints/prompts?
2. (Original post by Zacken)
.Any hints/prompts?
You need to be useing F=ma, and looking at differential equations.

Also, note that the force is towards the origin, so F= - k/x
3. (Original post by ghostwalker)
You need to be useing F=ma, and looking at differential equations.

Also, note that the force is towards the origin, so F= - k/x
I took the direction towards the origin as positive, so , right?

?

4. I obviously did something wrong with the limits in the integral, because this gives .
5. (Original post by Zacken)
I obviously did something wrong with the limits in the integral, because this gives .
v1, v2 are specific instances of the velocity - they are constants. It's almost meaningless to talk about differentiating them.

You need the equation of motion at a general position "x" to start.
6. (Original post by ghostwalker)
v1, v2 are specific instances of the velocity - they are constants. It's almost meaningless to talk about differentiating them.

You need the equation of motion at a general position "x" to start.
How do you integrate

Given that x depends on t in someway, or can I just treat it as a constant?
7. (Original post by Zacken)
How do you integrate

Given that x depends on t in someway, or can I just treat it as a constant?
You don't want to integrate wrt t.

You should be familiar with this process from SHM work.

So you end up with a relationship between v and x.

PS: I would be inclined to use the origin and x values as given and have the force as negative. Otherwise your "x" and their "x" are going to mean different things.
8. (Original post by ghostwalker)
You don't want to integrate wrt t.

You should be familiar with this process from SHM work.

So you end up with a relationship between v and x.

PS: I would be inclined to use the origin and x values as given and have the force as negative. Otherwise your "x" and their "x" are going to mean different things.
Thanks! Yeah, I did end up using negative force!

I haven't really done SHM yet, which is why I keep forgetting that! :P
9. (Original post by ghostwalker)
You don't want to integrate wrt t.

You should be familiar with this process from SHM work.

So you end up with a relationship between v and x.

PS: I would be inclined to use the origin and x values as given and have the force as negative. Otherwise your "x" and their "x" are going to mean different things.
.

.

Doing the same for yields .

So, something has to be wrong?
10. (Original post by Zacken)
.

.

Doing the same for yields .

So, something has to be wrong?
You're missing a "v" on the LHS under the integral sign. Not checked the rest.
11. (Original post by ghostwalker)
You're missing a "v" on the LHS under the integral sign.
I am an idiot...
12. (Original post by ghostwalker)
You're missing a "v" on the LHS under the integral sign. Not checked the rest.
Thanks a lot, proved it!

By the way, how would I show that for a particle moving along the x-axis, it's acceleration can be expressed as

Where x is the displacement from the origin and v is the velocity.
13. (Original post by Zacken)
Thanks a lot, proved it!

By the way, how would I show that for a particle moving along the x-axis, it's acceleration can be expressed as

Where x is the displacement from the origin and v is the velocity.
I would leave it as v dv/dx.

As to the format you suggest, it looks as if it should be obvious, but I think I'm missing somthing subtle, so don't know.
14. (Original post by ghostwalker)
I would leave it as v dv/dx.

As to the format you suggest, it looks as if it should be obvious, but I think I'm missing somthing subtle, so don't know.
I would as well, but the next question asks me to show that, can't get away with leaving it as v dv/dx. xD
15. (Original post by Zacken)
I would as well, but the next question asks me to show that, can't get away with leaving it as v dv/dx. xD

Chain rule, to get to v (dv/dx)

Then reverse the process in the post where I originally introduced the idea.
16. (Original post by ghostwalker)

Chain rule, to get to v (dv/dx)

Then reverse the process in the post where I originally introduced the idea.
Thanks! PRSOM.

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