B Question in the momemtum-impulse theorem derivation

[just NOTHING]

when the net force is constant then
Q1. rate of change of momentum (dp/dt) is zero or constant
Q2. assuming dp/dt is constant we replaced it with ----> p2-p1(total change in momentum ) ? how?

 

[PhDeezNutz]

Can you clarify question 2.

As for question 1. Observe ( assume #m# is constant)

$F = ma = m \frac{dv}{dt} = \frac{d(mv)}{dt} = \frac{dp}{dt}$

We can bring $m$ inside the derivative since it is a constant.

If $F$ is constant what does that say about $\frac{dp}{dt}$?

 

[Ibix]

Q2. assuming dp/dt is constant we replaced it with ----> p2-p1(total change in momentum ) ? how?

You can't - the units don't match. Have you missed an integral sign somewhere?

 

[just NOTHING]

Can you clarify question 2.

As for question 1. Observe ( assume #m# is constant)

$F = ma = m \frac{dv}{dt} = \frac{d(mv)}{dt} = \frac{dp}{dt}$

We can bring $m$ inside the derivative since it is a constant.

If $F$ is constant what does that say about $\frac{dp}{dt}$?

 

[A.T.]

Q2. assuming dp/dt is constant we replaced it with ----> p2-p1(total change in momentum ) ? how?

That is not what your text says. Read the sentence to the end.

 

[just NOTHING]

That is not what your text says. Read the sentence to the end.

Sorry i got it wrong/
can you tell me what it is trying to say.

 

[just NOTHING]

is it about the derivation of impuse momentum theorem from 2nd law??

 

[PhDeezNutz]

It's not replacing $\frac{d \vec{p}}{dt}$ with $\vec{p}_2 - \vec{p}_1$. It's replacing the numerator of $\frac{d \vec{p}}{dt}$ with $\vec{p}_2 - \vec{p}_1$.

$\vec{J} = \vec{F}\left(t_2 - t_1 \right) \Delta t$

But $\vec{F}$ is constant so, which means the momentum changes at a constant rate (not necessarily that the momentum is always the same)

$\vec{J} = \vec{F} \Delta t = \frac{d\vec{p}}{dt} \Delta t$

Replace $d\vec{p}= \vec{p}_2 - \vec{p}_1$

$\vec{J} = \frac{\vec{p}_2 - \vec{p}_1}{dt} \Delta t$

$\Delta t$ and $dt$ cancel out leaving you with?

$\vec{J} =$ ??

 

[vanhees71]

It's just Newton's 2nd Law integrated
$$\int_{t_1}^{t_2} \mathrm{d}t \dot{\vec{p}}=\vec{p}(t_2)-\vec{p}(t_1) = \int_{t_1}^{t_2} \mathrm{d} t\vec{F}[\vec{x}(t),t],$$
where the trajectory $\vec{x}(t)$ is the solution of Newton's equation of motion
$$\dot{\vec{p}}=\vec{F}(\vec{x},t).$$

 

[just NOTHING]

It's not replacing $\frac{d \vec{p}}{dt}$ with $\vec{p}_2 - \vec{p}_1$. It's replacing the numerator of $\frac{d \vec{p}}{dt}$ with $\vec{p}_2 - \vec{p}_1$.

$\vec{J} = \vec{F}\left(t_2 - t_1 \right) \Delta t$

But $\vec{F}$ is constant so, which means the momentum changes at a constant rate (not necessarily that the momentum is always the same)

$\vec{J} = \vec{F} \Delta t = \frac{d\vec{p}}{dt} \Delta t$

Replace $d\vec{p}= \vec{p}_2 - \vec{p}_1$

$\vec{J} = \frac{\vec{p}_2 - \vec{p}_1}{dt} \Delta t$

$\Delta t$ and $dt$ cancel out leaving you with?

$\vec{J} =$ ??

thanks a lot i feel lighter