Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    3
    ReputationRep:
    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 m, F is the force in the direction towards the origin, x is the distance from the origin and k is a positive constant, so that

    \displaystyle F = \frac{k}{x}

    The momentum of the particle at B and A respectively are mv_1 and mv_2.

    The forces on the particle at B and A respectively are \frac{k}{2b} and \frac{k}{b}.

    However, I don't really see how to go on from there to show that \displaystye v_2^2 = 2v_1^2.Any hints/prompts?
    • Study Helper
    Offline

    3
    (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
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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 F=\frac{k}{x}, right?

    \displaystyle \frac{\mathrm{d}v_1}{\mathrm{d}t  } = a_B = \frac{k}{2mb}

    \displaystyle \frac{\mathrm{d}v_2}{\mathrm{d}t  } = a_A = \frac{k}{mb}

    ?
    • Thread Starter
    Offline

    3
    ReputationRep:
    \displaystyle \frac{\mathrm{d}v_1}{\mathrm{d}t  } = a_B = \frac{k}{2mb}

    \displaystyle \frac{\mathrm{d}v_2}{\mathrm{d}t  } = a_A = \frac{k}{mb}


    \displaystyle v_1 = \int_0^t \frac{-k}{2bm} \, \mathrm{d}t \Rightarrow v_1^2 = \frac{k^2 t^2}{4b^2m^2}


    \displaystyle v_2 = \int_0^t \frac{-k}{mb} \, \mathrm{d}t \Rightarrow v_2^2 = \frac{k^2 t^2}{m^2b^2}


    I obviously did something wrong with the limits in the integral, because this gives v_2^2 = 4v_1^2.
    • Study Helper
    Offline

    3
    (Original post by Zacken)
    I obviously did something wrong with the limits in the integral, because this gives v_2^2 = 4v_1^2.
    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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 \displaystyle \int \frac{-k}{mx} \, \mathrm{d}t

    Given that x depends on t in someway, or can I just treat it as a constant?
    • Study Helper
    Offline

    3
    (Original post by Zacken)
    How do you integrate \displaystyle \int \frac{-k}{mx} \, \mathrm{d}t

    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.

    \displaystyle \text{acceleration }=\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}=v   \frac{dv}{dx}

    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    You don't want to integrate wrt t.

    \displaystyle \text{acceleration }=\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}=v   \frac{dv}{dx}

    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
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    You don't want to integrate wrt t.

    \displaystyle \text{acceleration }=\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}=v   \frac{dv}{dx}

    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.
    \displaystyle \int \mathrm{d}v = \frac{-k}{m} \int \frac{1}{x} \, \mathrm{d}x.

    \displaystyle v_1:

    \displaystyle \int_0^{v_1} \mathrm{d}v = \frac{-k}{m}\int_{4b}^{2b} \frac{1}{x}.

    \displaystyle v_1 = \frac{k}{m} \ln 2

    Doing the same for \displaystyle v_2 yields \displaystyle v_2 = \frac{2k}{m} \ln 2.

    So, something has to be wrong?
    • Study Helper
    Offline

    3
    (Original post by Zacken)
    \displaystyle \int \mathrm{d}v = \frac{-k}{m} \int \frac{1}{x} \, \mathrm{d}x.

    \displaystyle v_1:

    \displaystyle \int_0^{v_1} \mathrm{d}v = \frac{-k}{m}\int_{4b}^{2b} \frac{1}{x}.

    \displaystyle v_1 = \frac{k}{m} \ln 2

    Doing the same for \displaystyle v_2 yields \displaystyle v_2 = \frac{2k}{m} \ln 2.

    So, something has to be wrong?
    You're missing a "v" on the LHS under the integral sign. Not checked the rest.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    You're missing a "v" on the LHS under the integral sign.
    I am an idiot...
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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 \displaystyle a = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{v^2}{2}\right)

    Where x is the displacement from the origin and v is the velocity.
    • Study Helper
    Offline

    3
    (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 \displaystyle a = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{v^2}{2}\right)

    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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
    • Study Helper
    Offline

    3
    (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
    OK, just occurred to me. Start with the

    \frac{d}{dx}\left(\frac{v^2}{2} \right)

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

    Then reverse the process in the post where I originally introduced the idea.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    OK, just occurred to me. Start with the

    \frac{d}{dx}\left(\frac{v^2}{2} \right)

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

    Then reverse the process in the post where I originally introduced the idea.
    Thanks! PRSOM.
 
 
 
Poll
Would you rather
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.