# Question in the momemtum-impulse theorem derivation

• B
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}##?

2022 Award
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?

• PhDeezNutz
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}##?

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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} = ## ??

• just NOTHING
Gold Member
2022 Award
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
$$\dot{\vec{p}}=\vec{F}(\vec{x},t).$$

• PhDeezNutz
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