very random physics qn


Sep 17, 2008
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#1
this kinda sucked damn big time cause my differentiation is atrocious.
and so i need help:cry:

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

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
 

kokfann

Deregistered
Aug 2, 2009
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#3
yeah..f = m.dv/dt + v.dm/dt
 

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Sep 17, 2008
3,656
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#4
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:thumbsup:

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
 

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Limsgp

New Member
Dec 16, 2005
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#7
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.


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

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Sep 17, 2008
3,656
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0
#8
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:thumbsup:

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

koplady11

New Member
Dec 2, 2009
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East forever and ever
#9
yea. until u realsie this is damn basic:rolleyes:

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. :bsmilie:

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

#10
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:thumbsup:

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.
 

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Sep 17, 2008
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#11
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:sweat: not in a lvl sylabus as far as i know. but i feel pissed if i dun understand the whole logic. its an obsession:rolleyes:
 

Sep 17, 2008
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#13
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:rolleyes:

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:rolleyes:)
 

Sep 17, 2008
3,656
0
0
#15
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:nono:
thou shalt not disturb poor me who cant solve mdv and vdm.
 

#16
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.
 

Sep 17, 2008
3,656
0
0
#17
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.:thumbsup:

thanks loads!
 

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wildcat

Senior Member
Sep 8, 2004
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Bedok
#18
I just realize I can't teach basic Physics anymore for tuition :cry:
 

Sep 17, 2008
3,656
0
0
#19
I just realize I can't teach basic Physics anymore for tuition :cry:
huh? nono. to me its basic. something i gotta know. its relative!(aww einstein <3)

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

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