Momentum Change and Impulse

Momentum Change and Impulse Introduction

An unbalanced force always accelerates an object. When a force acts in the opposite direction of an object's motion, it slows down. Conversely, the object accelerates positively when a force acts in the same direction as an object's motion. Therefore, an external force will always change the velocity of an object. And, if the object's velocity changes, the object's momentum changes as well.

This change in momentum resulting from the applied external force is called impulse.

Momentum Change and Impulse Relationship

Let's begin by reviewing an important result known as the impulse-momentum theorem it says the following:

"When an object in a collision, is affected by a force for a specified period of time, this results in a change in its momentum. This change in momentum is called impulse and it equals the product of the average force times the period of time during which it acted."

Though that may sound like a lot of physics jargon. In essence, impulse quantifies how much momentum changes and tells us how the force acting on a system for a certain amount of time affects its motion.

We can calculate impulse using the following formula.

$$\boxed{\vec J = \int_{t_0}^t F(t) \mathrm{d}t}$$

We will later discuss how we can arrive at this result but it is worth noticing that, due to this relationship, we can tell that the units for impulse in the SI system are \( \mathrm{\frac{N}{s}} \), this is, units of force times units of time.

Momentum Change and Impulse Formula

Now, we will see why the external force acting on the system is related to the impulse and how we can obtain the impulse-momentum theorem equation.

According to Newton's Second Law of Motion, objects under the influence of a net force move with acceleration:

$$\vec F = m\vec a.$$

If we rearrange this expression, we can find the relationship between impulse and linear momentum. Since acceleration is the rate of change of the velocity. Therefore we can rewrite our equation as

$$\vec F = m\frac{\Delta\vec v}{\Delta t}\mathrm{,}$$

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.

$$\vec F_{net} =\frac{\mathrm{d} \vec p}{\mathrm{d} t} = \frac{\mathrm{d} m}{\mathrm{d} t}\vec{v} $$

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.

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.

If the magnitudes of the \(F_1\) and \(F_2\) forces are equal, the Linear Momentum-Time Graph of the object will be as shown in Fig. 3. In the Linear Momentum vs. Time Graph, the slope gives the force. Note that the slope between \(t=0\) to \(t\) is \(\frac{\Delta p}{\Delta t}=\frac{F\Delta t}{\Delta t} = F\) but from \(t\) to \(2t\) is \( -F. \)

Momentum Change and Impulse Difference

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.

Examples of Change in Momentum and Impulse

If you haven't had enough of Force vs. Time Graphs, you will like this next example.

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.

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.

Your shopping cart is probably pretty full of physics concepts by now. So after you register all this information in your brain bank, go do something totally mind-numbing. But first, store these concepts to hold for later because they're pretty important.