momentum and time

Impulse F.(Δt) can be positive or negative, depending on the direction of force.


2015-07-28 16:15:35 補充：
Your suppl question:
The rate of change of momentum is proportional to the applied force. This is Newton's Second Law.

2015-07-29 20:21:42 補充：
Your further question:
I don't know where you get this equation: momentum =f(time), but it is not correct.
I suppose the symbol f represents "force". Hence, force x time equals to "impulse". An impulse leads to a "CHANGE of momentum".

2015-07-29 20:24:01 補充：
(cont'd)...
The equation: impulse = change of momentum is just another way of expressing Newton's Second Law, which I have already given you in my ealier answer.

2015-07-29 20:27:22 補充：
(cont'd)...
Notice that the "time" in the equation refers to the "impact time",i.e. the duration in which the force acts. Hence, the longer the impact time, the more is the momentum change. This is just simple mathematics.

2015-07-29 22:56:54 補充：
Q:即係問momentum =f(time)的關係。
A: How the momentum of an object changes with time depends on how the applied force varies with time. Mathematically, it is
Momentum = integral { F(t).dt}
where F(t) is the function describing how the force F varies with time t.

2015-07-30 15:33:39 補充：
Q:If F(t)=m(d^2 x/dt^2),does momentum have negative relationship with time?
A: As said earlier, the equation F(t) = mass x acceleration, is a result of Newton's Second Law, force = rate of change of momentum. The change of momentum can be +ve, -ve (or even zero) depending on the direction of force.

2015-07-31 23:34:34 補充：
Q:if the time variable in the force function is doubled, and the "impact time" is cut in half, what will happen to the change of momentum?
A: The impulse and hence change of momentum will be the same as before.

2015-08-01 10:29:03 補充：
Q:WHY? The time variable is squared in the force function.
A: How come the time variable is squared?
Imppulse = force x time = F x t
if force F is doubled, and time t is halved, new impulse = (2F) x (t/2) = F x t

2015-08-02 19:56:50 補充：
已達字數限額，請看意見欄。

2015-08-02 20:10:18 補充：
Q:I mentioned " the time variable in the force function is doubled".
A:The time variable t indicates the "force interaction time". "The time varibale is doubled", I suppose you mean such interaction time is twice as before.

2015-08-02 20:13:05 補充：
(cont'd)...
In that case, the "impact time" would also be doubled.It cannot be halved.

New impulse = integral{F(t).dt} with limit of integration from t = 0 to t = 2(to) where to is the original impact time.

2015-08-03 15:48:51 補充：
Q:Is this correct that "force interaction time" is the time before and during impact, and "impact time" is the duration of the impact?

A: The "force interaction time" is just the time duration within which the force acts on the object of concern.

2015-08-03 15:49:21 補充：
(cont'd)...
This is just the same as "impact time", which usually used in the case of collision.

2015-08-03 15:52:46 補充：
Q:the force should be negatively related to the time variable in the force function, FORCE=f(TIME).

A: A force function can be of many many forms, depending on the nature of force and the situation applied. Not sure what your "-vely related to the time varibale" means.

2015-08-03 19:48:13 補充：
Q:I want to know the most is the relationship between momentum and time in positive direction, MOMENTUM=f(TIME).

A: The most simple answer, momentum = mass x velocity
Hence, how momentum varies with time depends on how the velocity of the object varies with time.

2015-08-03 19:51:39 補充：
Q:More time means more momentum, but more time also means less force under the assumption of "Δx and m are constant". But the time variable in the force function is 1/(t^2). Should there be negative relationship in MOMENTUM=f(TIME)?

2015-08-03 19:52:33 補充：
A: How do you get all these relationship? There is no rationale behind that such relationship is valid.

2015-08-03 23:15:44 補充：
Q:If the force function = m(Δx/Δt^2), the time variable is 1/(t^2) as "to"=0 was assumed.

2015-08-03 23:17:23 補充：
A: This is mathematically wrong.
Force = rate of change of momentum = (mΔv)/Δt = m[Δ(Δx/Δt)]/(Δt)
where (Δv) is the change of velocity of the object
and (Δx/Δt) is the instantaneous velocity of the object.

2015-08-03 23:19:18 補充：
(cont'd)...
In mathematics, the symbol Δ represents an infintesimal quantity.
You cannot multiply the two (Δt) together like an algebric quantity.

2015-08-04 16:12:40 補充：
Q: can we say Δx/Δx=Δ/Δ=1?
A: How come Δx/Δx=Δ/Δ ???
The symbol "Δ" is a mathematical operator, you cannot treat it like an algebric parameter.

2015-08-04 16:18:59 補充：
In fact, the first line of your question is already wrong.
Δp=FΔt=m(Δx/Δt^2)(Δt)=m(Δx/Δt) <---- this is wrong

F = Δ[m(Δx/Δt)]/Δt = m[Δ(Δx/Δt)]/Δt = m(Δv/Δt) = ma
where a = (Δv/Δt) is the acceleration

2015-08-04 23:16:30 補充：
Q:Should there be negative relationship in Δp=f(Δt)?
A: As said in the last post, Δp= F. Δt = ma(Δt)
Depending on the acceleration a is +ve (accelerating) or -ve (decelerating), Δp can be +ve (increase in momentum) or -ve (decrease in momentum).