$$\vec F = m\frac{\Delta\vec v}{\Delta t}\mathrm{,}$$
and then rearrange our variables to get
$$\vec F\Delta t=m\Delta\vec v\mathrm{.}$$
Since \(\Delta\vec v=\vec{v_f}-\vec{v_i}\), we can also express it as
$$\vec F\Delta t = m(\vec{v}_f-\vec{v}_i),$$
and then use the distributive property of multiplication to yield the equation
$$\vec F\Delta t=m\vec{v}_f-m\vec{v}_i\mathrm{.}$$
Also, because \(\vec p=m\vec v\), we can now write our equation more concisely as
$$\vec F\Delta t=\vec{p}_f-\vec{p}_i,$$
and then rearrange terms to get
$$\vec F\Delta t=\Delta\vec p\mathrm{.}$$
According to this expression, the impulse applied to an object is equal to the change in the linear momentum of the object. This formula works if we use the average force acting during the time interval.
$$\boxed{\vec F_{\text{avg}}\Delta t=\Delta\vec p}.$$
Notice also that dividing by the \(\Delta t\) proves that the change in a system's momentum with respect to time equals the net force exerted on the system:
$$\vec F_\text{net}\Delta t = \Delta\vec p$$
$$\vec F_\text{net} = \frac{\Delta \vec p}{\Delta t}\\\mathrm{.}$$
In calculus terms, we describe a rate of change using derivatives. Therefore, we can rewrite our equation as
$$\vec F_\text{net} = \frac{\mathrm{d} \vec p}{\mathrm{d} t}\\\mathrm{.}$$
Se operating variables and integrating both sides of the above equation we find the integral expression for the impulse-momentum theorem
\begin{aligned}\vec F_\text{net} &= \frac{\mathrm{d} \vec p}{\mathrm{d} t}\\[8pt] \mathrm{d} \vec p & = \vec F_\text{net} dt\\[8pt] \int \mathrm{d} \vec p & = \int \vec F_\text{net} dt\\[8pt] \Delta \vec{p} &= \int_{t_0}^t \vec F(t) \mathrm{d}t\\[8pt]\vec J &= \int_{t_0}^t \vec F(t) \mathrm{d}t\end{aligned}
And finally, we find the expected result: the impulse equals the integral of the net force exerted on our system over a specific period of time.
Rate of Change of Momentum
Now, let's prove how the rate of change of momentum is equivalent to the net force acting on the object or system.
We've all heard that Newton's second law is \(F = ma\); however, when Newton was first writing the law, he had in mind the idea of linear momentum. Therefore, let's see if we can write Newton's second law a little differently. Starting with
\[\vec{F}_{net} = m\vec{a}\]
allows us to see a correlation between Newton's second law and linear momentum. Recall that acceleration is the derivative of velocity. Therefore, we can write our new force formula as
\[\vec{F}_{net} = m\frac{d\vec{v}}{dt}\]
It is essential to note the change that was made. Acceleration is just the rate of change in velocity, so to replace it with \(\frac{d\vec{v}}{dt}\) is valid. As the mass m stays constant, we see that the net force is equal to the rate of change of momentum:
\[\vec{F_{net}} = \frac{d(m\vec{v})}{dt} = \frac{d\vec{p}}{dt}\]
We can rearrange this to get
\[d\vec{p} = \vec{F}_{net} dt\]
With this new outlook on Newton's second law, we see that the change of momentum, or impulse, can be written as follows:
\[\vec{J} = \Delta \vec{p} = \int{d\vec{p}} = \int{\vec{F}_{net}dt}\]
- The change of momentum, or impulse (represented by the capital letter \(\vec{J}\)) , is the difference between a system's initial and final momentum. Therefore, it is equal to the mass times the change in velocity.
- Newton's second law is a direct result of the impulse-momentum theorem when mass is constant! The impulse-momentum theorem relates the change of momentum to the net force exerted: \(\vec{F}_{net} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt} = m \vec{a}\)
- As a result, the impulse is given by: \(\vec{J} = \int{\vec{F}_{net} dt}\)
Momentum Change and Impulse Theorem
As we already mentioned, can express the Impulse-momentum theorem mathematically as follows:
$$\boxed{\vec J = \int_{t_0}^t F(t) \mathrm{d}t = \Delta \vec p.}$$
But did you notice that Newton's Second Law is just a consequence or implication of the impulse-linear momentum theorem when mass is constant?
$$F_{net} = m\vec a =m\frac{\mathrm{d} \vec v}{\mathrm{d} t} =\frac{\mathrm{d} (m\vec v)}{\mathrm{d} t} = \frac{\mathrm{d} \vec p}{\mathrm{d} t}. $$
With it, we can derive the relationship between impulse and the change in momentum through the use of calculus and the definition of impulse! Substituting this result in the definition of impulse, then simplifying and integrating leads to the expected result.
$$ \begin{aligned}\vec J &= \int_{t_i}^{t_f} {F(t)}\mathrm{d}t\\[8pt] \vec J &= \int_{t_i}^{t_f} {\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} }\mathrm{d}t\\[8pt] J &= \int_{t_i}^{t_f} \mathrm{d}\vec{p}\\[8pt] J &= \vec p_f -\vec p_i\\[8pt] J &= \Delta \vec p. \end{aligned}$$
We can also write a version of Newton's Second Law for a system moving with constant velocity but variable mass by using the impulse-momentum theorem.
$$\vec F_{net} =\frac{\mathrm{d} \vec p}{\mathrm{d} t} = \frac{\mathrm{d} m}{\mathrm{d} t}\vec{v} $$
If both mass and velocity change with time, we would need to use this formula instead:
$$\vec F_{net} =\frac{\mathrm{d} \vec p}{\mathrm{d} t}\\ =\frac{\mathrm{d} m}{\mathrm{d} t}\vec{v} + m\frac{\mathrm{d} \vec{v}}{\mathrm{d} t}$$
Knowing this integral relationship has other advantages. In a Force-Time Graph for an object, the area between the graph and the horizontal axis gives the impulse applied to the object.
Since impulse and linear momentum are vector quantities, we need to pay attention to their signs when using the areas below the graph. For example, the impulse is positive if the area between the graph and the horizontal axis is above the time axis. Conversely, the impulse is negative if it is below the time axis.
Fig. 2 - In the Force-Time Graph, the area gives the impulse, which is the change of momentum. The impulse is positive if the area between the graph and the horizontal axis is above the time axis.
To find the area of the graph, we need to multiply the force and the corresponding time interval. The sum of the areas will be equal to the impulse. In this case, the impulse will be equal to \(J=F_1 t-F_2 t\).
Fig. 3 - In a Linear-Momentum-Time Graph, the slope gives the force exerted on an object. This force can be positive or negative depending on the slope.
Impulse equals the change in momentum, therefore, the momentum and the impulse are not the same. Since they are different but related concepts, it can be a bit confusing, so let's look at them in detail.
Fig. 5 - A Force-Time Graph in a frictionless plane.
The time-dependent graph above shows the force applied to an object standing in a frictionless plane. Draw a Linear Momentum-Time Graph of the object.
In a Force-Time Graph, the area gives the impulse, the change in momentum.
Between \(0-1\) seconds, the area is \(20\, \text{N}\,\text{s}\).
Between \(1-2\) seconds, the area is \(-10\, \text{N}\,\text{s}\).
Between \(2-3\) seconds, the area is \(-10\, \text{N}\,\text{s}\).
Since the impulse is positive between \(0-1\) seconds, the linear momentum should increase in the Linear Momentum-Time Graph. However, it then starts decreasing, resulting in zero change in momentum.
Fig. 6 - The image shows a Linear Momentum-Time Graph. The change in momentum at the end of the \(3\) seconds is zero.
Since \(F=\frac{\mathrm{d}p}{\mathrm{d}t} \) the slope of a Momentum-Time Graph is equivalent to the value of the force!
I know we've been pounding this into your brain, but just one last area-related example, I promise! There won't be a graph for this one though, so you can take a quick breather.
The work done by a force on an object as a function of time is given by the equation
$$F(t)=at^2+bt+1,$$
where \(a\) and \(b\) are both constants. What is the momentum of the object at time \(t\) if the initial momentum is \(0\) at \(t=0\)?
Recall that the impulse equals the integral of the force as a function of time:
$$\vec J = \int_{t_0}^t F(t) \mathrm{d}t.$$
Therefore, by integrating with respect to \(t\), we can find the momentum of the object at time \(t\) by making our bounds of integration \(0\) and \(t\):
$$\vec J = \int_{0}^t F(t) \mathrm{d}t.$$
This gives us the change in the object's momentum. However, since its initial momentum is \(0\), the total change in the object's momentum over a time \(t\) will be its regular momentum at that time. Therefore, integrating
$$\vec J = \int_0^t at^2 +bt +1\ \mathrm{d}t.$$
yields the answer of
$$\vec J = \frac{1}{3} at^3 +\frac{1}{2} bt^2 + t.$$
Remember that impulse equals the change in momentum. But since the initial momentum is zero we can find the final momentum.
$$\begin{align*} p_f - p_i&=\frac{1}{3} at^3 + \frac{1}{2} bt^2+t \\ p_f- 0&=\frac{1}{3} at^3 + \frac{1}{2} bt^2+t \\ pf&=\frac{1}{3} at^3 + \frac{1}{2} bt^2+t \\ \end{align*}$$
which is the value of the momentum at time \(t\).
Okay, okay, I hear you. So we'll switch things up. Instead of a Force vs. Time Graph, we'll do an example with a Linear Momentum vs. Time Graph.
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