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Thread: very random physics qn

  1. #1
    Deregistered allenleonhart's Avatar
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    Default very random physics qn

    this kinda sucked damn big time cause my differentiation is atrocious.
    and so i need help

    newton's second law states that force on an object is given by the rate of change in its momentum correct?

    so f=dp/dt

    if i expand dp, i get dmv.

    f=dmv/dt

    then now, i got to break up dmv. i'll let dt go over to F part.

    fdt=mdv+vdm

    now here is the part that i screw up. wads m and wads v?

    i'll throw in some numbers so u all can punch. i got weird values.

    assuming a car is travelling initially at 10ms-1, with the mass of the car+fuel being 100kg.

    at time 1s later, the car is travelling at 100ms-1, with the mass of car + fuel being 10kg.

    what is the force acting on the car during this 1s?

    if i take it as Fdt= change of momentum, then i punch in numbers...

    i shld get 0 resultant force as the momentum m1v1=m2v2 and hence no force.

    yet if u use mdv+vdm.....

    100kg (100-10)+ 10ms-1 (10-100)
    u get a weird value.

    i suspect i punched in wrong numbers for V and M. or my DV and DM is wrong. anyone can advise?

  2. #2

    Default Re: very random physics qn

    Quote Originally Posted by allenleonhart View Post
    this kinda sucked damn big time cause my differentiation is atrocious.
    and so i need help

    newton's second law states that force on an object is given by the rate of change in its momentum correct?

    so f=dp/dt

    if i expand dp, i get dmv.

    f=dmv/dt

    then now, i got to break up dmv. i'll let dt go over to F part.

    fdt=mdv+vdm

    now here is the part that i screw up. wads m and wads v?

    i'll throw in some numbers so u all can punch. i got weird values.

    assuming a car is travelling initially at 10ms-1, with the mass of the car+fuel being 100kg.

    at time 1s later, the car is travelling at 100ms-1, with the mass of car + fuel being 10kg.

    what is the force acting on the car during this 1s?

    if i take it as Fdt= change of momentum, then i punch in numbers...

    i shld get 0 resultant force as the momentum m1v1=m2v2 and hence no force.

    yet if u use mdv+vdm.....

    100kg (100-10)+ 10ms-1 (10-100)
    u get a weird value.

    i suspect i punched in wrong numbers for V and M. or my DV and DM is wrong. anyone can advise?
    You cannot just bring dt over to f. Differentiation is not like fraction.

    f= d(mv)/dt

    In this case, u apply product rule.

    f= m (dv/dt) + v (dm/dt)

    1) By keeping mass of the car constant with varying velocities => m(dv/dt)

    2) By keeping the velocity of the car constant with varying mass => v(dm/dt)

    For 1), in 1 second, m(dv/dt) = (100)((100-10)/1) = 9000kgm/s^2

    For 2), in 1 second, v(dm/dt) = (100)((10-100)/1) = -9000kgm/s^2
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  3. #3

    Default Re: very random physics qn

    yeah..f = m.dv/dt + v.dm/dt
    Last edited by kokfann; 21st June 2010 at 01:11 PM.

  4. #4
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    Default Re: very random physics qn

    Quote Originally Posted by chikubang View Post
    You cannot just bring dt over to f. Differentiation is not like fraction.


    1) By keeping mass of the car constant with varying velocities => m(dv/dt)

    2) By keeping the velocity of the car constant with varying mass => v(dm/dt)

    For 1), in 1 second, m(dv/dt) = (100)((100-10)/1) = 9000kgm/s^2

    For 2), in 1 second, v(dm/dt) = (100)((10-100)/1) = -9000kgm/s^2


    For 1), in 1 second, m(dv/dt) = (100)((100-10)/1) = 9000kgm/s^2

    For 2), in 1 second, v(dm/dt) = (100)((10-100)/1) = -9000kgm/s^2

    this part is the one i need clarification. why is the constant mass 100kg rather than 10kg, or why is the constant velocity 100 rather than 10ms-1? this is the part i dun get it

    and it shld be delta t rather than dt. dt is instantaneous. delta is for small changes, so u can still bring delta t over actually. i think its my fault here
    Last edited by allenleonhart; 21st June 2010 at 02:20 PM.

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    Member koplady11's Avatar
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    Default Re: very random physics qn

    Glad to see there are physics/maths geeks in the forum. Lol!
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  6. #6
    Deregistered allenleonhart's Avatar
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    Default Re: very random physics qn

    Quote Originally Posted by koplady11 View Post
    Glad to see there are physics/maths geeks in the forum. Lol!
    yea. until u realsie this is damn basic

    ARGH.

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    Default Re: very random physics qn

    F = dmv/dt

    F = m.dv/dt + v.dm/dt

    You'll get the Force acting on the system at the particular "instant". So at the particular instant, there'll be respective values of m and v to use. (Say, if dm/dt = constant, then, at t=0.5s, the m will be 55kg at the instant where t = 0.5s)

    If you want to find the average force, you'll need to integrate F over the interval of 1s, then divide by the interval 1s.


    Quote Originally Posted by allenleonhart View Post

    what is the force acting on the car during this 1s?
    Last edited by Limsgp; 21st June 2010 at 03:35 PM.

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    Default Re: very random physics qn

    Quote Originally Posted by Limsgp View Post
    F = dmv/dt

    F = m.dv/dt + v.dm/dt

    You'll get the Force acting on the system at the particular "instant". So at the particular instant, there'll be respective values of m and v to use. (Say, if dm/dt = constant, then, at t=0.5s, the m will be 55kg at the instant where t = 0.5s)

    If you want to find the average force, you'll need to integrate over the interval of 1s, then divide by the interval 1s.
    ok. let me digest a bit. i think this will work.

    so lets say i wanna find at time t0.5. so the respective M value assuming that the rate of fuel burnt is constant, so it will be 55kg. then velocity also. so its 55ms-1.

    55kg 90/1 + 55 (-90/1) = 0.

    seems correct. thanks a lot

    so it is the respective value of M and V at the instant 0.5s.

  9. #9
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    Default Re: very random physics qn

    Quote Originally Posted by allenleonhart View Post
    yea. until u realsie this is damn basic

    ARGH.
    Haha. Chill. If I switched on the part of my brain that was once soaked in A levels Physics, I'm sure I could have helped you. But for now, my mind is tuned to H1 Maths coz I'm teaching a kid H1 maths. Initially everything looked so blur but if when I stepped back and relaxed a lil, the stuff pretty easy. No wonder can go thru maths then.

    That said, chill bro!! Sumtimes we just tend to overlook the simplest solutions to the worst problems.
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  10. #10

    Default Re: very random physics qn

    Quote Originally Posted by allenleonhart View Post
    ok. let me digest a bit. i think this will work.

    so lets say i wanna find at time t0.5. so the respective M value assuming that the rate of fuel burnt is constant, so it will be 55kg. then velocity also. so its 55ms-1.

    55kg 90/1 + 55 (-90/1) = 0.

    seems correct. thanks a lot

    so it is the respective value of M and V at the instant 0.5s.
    By using that formula, u only find the instantaneous force at a particular time. tt's why ur answer is weird.
    Last edited by chikubang; 21st June 2010 at 05:55 PM.
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    Default Re: very random physics qn

    Quote Originally Posted by chikubang View Post
    By using that formula, u only find the instantaneous force at a particular time. tt's why ur answer is weird.
    hmm. ok. cause i'm trying to play with forces related to varying force and mass at the same time not in a lvl sylabus as far as i know. but i feel pissed if i dun understand the whole logic. its an obsession

  12. #12

    Default Re: very random physics qn

    I also make a mistake. the mass should be the average mass and the velocity should be the average velocity.

    I also thought why u ask such physics qns that is not A lvl and u claimed that it's basic. LOL.
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    Default Re: very random physics qn

    Quote Originally Posted by chikubang View Post
    I also make a mistake. the mass should be the average mass and the velocity should be the average velocity.

    I also thought why u ask such physics qns that is not A lvl and u claimed that it's basic. LOL.
    as in Mdv the m is avg mass of start and final, Vdm the V is avg of initial and final? or is it at that moment in time?

    cause i'm doing physics olympiad competition. needa prep up on this

    well. for me the way i study is:
    its either 100 marks or 0. either u know or u dunno.

    everything is basic. if u cant derive anything, u dunno ur basics.

    (i'm a sadist. i like to self pressure a lot)

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    Default Re: very random physics qn

    Quote Originally Posted by chikubang View Post
    I also thought why u ask such physics qns that is not A lvl and u claimed that it's basic. LOL.
    he's liddat one. hi kp! study for your midyrs pls don't go so far T_T

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    Default Re: very random physics qn

    Quote Originally Posted by spitfire93 View Post
    he's liddat one. hi kp! study for your midyrs pls don't go so far T_T
    what the. why u suddenly pop up here! SHOO go back and mug your own stuff
    thou shalt not disturb poor me who cant solve mdv and vdm.

  16. #16

    Default Re: very random physics qn

    Quote Originally Posted by Limsgp View Post
    F = dmv/dt

    F = m.dv/dt + v.dm/dt

    You'll get the Force acting on the system at the particular "instant". So at the particular instant, there'll be respective values of m and v to use. (Say, if dm/dt = constant, then, at t=0.5s, the m will be 55kg at the instant where t = 0.5s)

    If you want to find the average force, you'll need to integrate F over the interval of 1s, then divide by the interval 1s.
    Like what Limsgp has said,

    Integrate F over interval of 1s and divide by the interval of 1s to find average force.

    What I did is to assume is that the force changes evenly, that's why I just took, average of the mass (100+10/2). But this may not be true, depending on the rate of change of F. Am I very confusing? LOL.
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    Default Re: very random physics qn

    Quote Originally Posted by chikubang View Post
    Like what Limsgp has said,

    Integrate F over interval of 1s and divide by the interval of 1s to find average force.

    What I did is to assume is that the force changes evenly, that's why I just took, average of the mass (100+10/2). But this may not be true, depending on the rate of change of F. Am I very confusing? LOL.
    ah crap. i forgot, rate of change of f also.
    and yes. for the numbers i gave u, the rate of change of F shld be 0. cause momentum of system isnt changing (while its going faster, its getting lighter), force is a constant 0.

    i think i understand where its weird liao. mine i had the rate of change of velocity = rate of change of mass.with starting mass and final mass of equal magnitude with velocity. if i were to put rate of change of velocity higher i think there will be a resultant force.

    thanks loads!
    Last edited by allenleonhart; 21st June 2010 at 07:04 PM.

  18. #18
    Senior Member wildcat's Avatar
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    Default Re: very random physics qn

    I just realize I can't teach basic Physics anymore for tuition

  19. #19
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    Default Re: very random physics qn

    Quote Originally Posted by wildcat View Post
    I just realize I can't teach basic Physics anymore for tuition
    huh? nono. to me its basic. something i gotta know. its relative!(aww einstein <3)

    dun worry wildcat. i'm sure ur a pro i got the idea for this qn from ICL first year paper so dun worry too much
    *zomg i realise i suck at my basics. sigh...*

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    Default Re: very random physics qn

    That is like S paper stuff.. normal A level (H2?) shouldn't have it lah..

    Quote Originally Posted by wildcat View Post
    I just realize I can't teach basic Physics anymore for tuition

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