# Question in the momemtum-impulse theorem derivation

• B
• just NOTHING
In summary, when the net force is constant, the rate of change of momentum (dp/dt) is also constant. This can be derived from Newton's second law and the impulse-momentum theorem, where the numerator of dp/dt is replaced by the total change in momentum (p2-p1) and the time interval (dt) is cancelled out.
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?

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}##?

just NOTHING said:
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
PhDeezNutz said:
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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just NOTHING said:
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.

A.T. said:
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.

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

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
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
PhDeezNutz said:
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

## 1. What is the momentum-impulse theorem?

The momentum-impulse theorem states that the change in momentum of an object is equal to the impulse applied to the object. In other words, the force applied to an object over a certain period of time will result in a change in the object's momentum.

## 2. How is the momentum-impulse theorem derived?

The momentum-impulse theorem can be derived using the principles of Newton's second law of motion and the definition of impulse. It involves using calculus to integrate the force applied to an object over a specific time interval to find the change in momentum.

## 3. What is the equation for the momentum-impulse theorem?

The equation for the momentum-impulse theorem is p = F * t, where p is the change in momentum, F is the force applied, and t is the time interval over which the force is applied.

## 4. How is the momentum-impulse theorem used in real-world situations?

The momentum-impulse theorem is used in many real-world situations, such as calculating the force needed to stop a moving object or the force exerted by a collision. It is also used in sports, such as in a game of billiards where the force applied to the cue ball determines its momentum and direction.

## 5. Are there any limitations to the momentum-impulse theorem?

While the momentum-impulse theorem is a useful tool in physics, it does have some limitations. It assumes that the force applied to an object is constant, which may not always be the case in real-world situations. Additionally, it does not take into account other factors such as friction or air resistance, which can affect an object's momentum.

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