Momentum Change and Impulse

You can easily move an empty shopping cart if you push it with consistent force for a set amount of time. However, you'll ultimately load the cart with things if you keep shopping, and it will get much heavier. When you try to move the cart with the same force as when it was empty, it becomes much more difficult to move. To move the cart faster, you will have to exert a greater force or impulse on it.

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What is the impulse?

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Impulse is a scalar quantity. 

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In Force-Time Graphs of an object, the area between the graph and the horizontal axis gives the impulse applied to the object. 

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In the Linear Momentum vs. Time Graph, the slope gives the impulse. 

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A force of \(20.0\,\mathrm{N}\) is applied to an object with a mass of \(1.0\,\mathrm{kg}\) in a frictionless plane and in a horizontal direction.As a result of this force, the magnitude of the velocity of the object increased from \(2.0\,\mathrm{m/s}\) to \(7.0\,\mathrm{m/s}\). Find the time it takes for the force to be applied to the object.

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A force of \(10.0\,\mathrm{N}\) is applied to the object with a mass of \(2.0\,\mathrm{kg}\) in a frictionless plane and in a horizontal direction.As a result of this force, the magnitude of the velocity of the object increased from \(3.0\,\mathrm{m/s}\) to \(8.0\,\mathrm{m/s}\). Find the time it takes for the force to be applied to the object.

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Forces of magnitude \(F\) are applied to bodies of mass \(m\) and \(4m\), which are at rest in a frictionless plane.For the bodies to have the same linear momentum change, what should be the ratio of \(\frac{t_1}{t_2}\) if the impact times are \(t_1\) and \(t_2\) respectively?

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Forces of magnitude \(F\) are applied to bodies of mass \(m\) and \(4m\), which are at rest in a horizontal plane.For bodies to have the same kinetic energy, what should be the ratio of \(\frac{t_1}{t_2}\) if the impact times are \(t_1\) and \(t_2\) respectively? 

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What is the impulse?

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Impulse is a scalar quantity. 

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In Force-Time Graphs of an object, the area between the graph and the horizontal axis gives the impulse applied to the object. 

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  • Mo

In the Linear Momentum vs. Time Graph, the slope gives the impulse. 

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  • + Add tag
  • Immunology
  • Cell Biology
  • Mo

A force of \(20.0\,\mathrm{N}\) is applied to an object with a mass of \(1.0\,\mathrm{kg}\) in a frictionless plane and in a horizontal direction.As a result of this force, the magnitude of the velocity of the object increased from \(2.0\,\mathrm{m/s}\) to \(7.0\,\mathrm{m/s}\). Find the time it takes for the force to be applied to the object.

Show Answer
  • + Add tag
  • Immunology
  • Cell Biology
  • Mo

A force of \(10.0\,\mathrm{N}\) is applied to the object with a mass of \(2.0\,\mathrm{kg}\) in a frictionless plane and in a horizontal direction.As a result of this force, the magnitude of the velocity of the object increased from \(3.0\,\mathrm{m/s}\) to \(8.0\,\mathrm{m/s}\). Find the time it takes for the force to be applied to the object.

Show Answer
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  • Immunology
  • Cell Biology
  • Mo

Forces of magnitude \(F\) are applied to bodies of mass \(m\) and \(4m\), which are at rest in a frictionless plane.For the bodies to have the same linear momentum change, what should be the ratio of \(\frac{t_1}{t_2}\) if the impact times are \(t_1\) and \(t_2\) respectively?

Show Answer
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  • Mo

Forces of magnitude \(F\) are applied to bodies of mass \(m\) and \(4m\), which are at rest in a horizontal plane.For bodies to have the same kinetic energy, what should be the ratio of \(\frac{t_1}{t_2}\) if the impact times are \(t_1\) and \(t_2\) respectively? 

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    You convey these impulses to the cart as linear changes of momentum. Heavier objects are harder to move and their momentum is harder to change, remember the formula \(p=mv\). This article will explain everything you need to know about momentum change and impulse: their formula, their relationship, and their differences. We will also explain the change in momentum and impulse theorem.

    Change in Momentum and Impulse An Image of Shopping Cart StudySmarterFig. 1 - It is more difficult to accelerate a full shopping cart than an empty one. To raise the speed of the full cart to the speed of the empty cart over the same period of time, it is necessary to apply a greater force

    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.

    Momentum, \( \vec{p} \) is a vector quantity equal to the product of an object's mass and its velocity

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

    Impulse, \( \vec{J} \) is a vector quantity that quantifies the change of momentum, \( \vec{p} \) in a system.

    The impulse vector has the same direction as the net force exerted on the system.

    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.

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

    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.

    Momentum Change and Impulse A Drawing of Force-Time Graph StudySmarter

    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 \(J=F_1 t-F_2 t\).

    Momentum Change and Impulse A Drawing of Linear Momentum-Time Graph StudySmarterFig. 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.

    MomentumImpulse
    Calculated as the product of mass and velocity.Calculated as the change of momentum.
    We use it to calculate the external force that acts over the system.We use it to measure the effect of external force on the system.
    Defines the instantaneous force that acts on the system.Takes into account the effects of both the force acting on the system and the duration of time for which it acts.
    In a Momentum vs. Time Graph, the slope of the curve at a given point represents the force.In a Force vs. Time Graph, the area under the curve at a given interval represents the impulse.

    Examples of Change in Momentum and Impulse

    Now let's solve some examples to put into practice what we have learned about the impulse-linear momentum theorem: \(\vec J=\Delta \vec p\).

    Momentum Change and Impulse A Drawing of Force-Time Graph for Example 1 StudySmarter

    Fig. 4 - This graph shows the Force-Time Graph in a frictionless plane.

    Above, you can see the graph of the change of the force applied to an object in time in a frictionless plane. Find the magnitude of the impulse exerted on the object.

    In a Force-Time Graph, the algebraic sum of the areas between the graph and the horizontal axis gives the magnitude of the impulse applied to the object:

    $$\begin{align*} \mathrm{Area} &= \left( \frac{10\times 2}2\right)+(10\times 2)-(15\times 2) \\ \mathrm{Area} &= 10+20-30 \\ \mathrm{Area} &= 0\mathrm{.} \\ \end{align*}$$

    Since the algebraic sum of the area under the graph is \(0\), the impulse is \(0\). This means that there is no change in momentum, thus momentum is conserved.

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

    Momentum Change and Impulse A Drawing of Force-Time Graph for Example 2 StudySmarterFig. 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.

    Momentum Change and Impulse A Drawing of Linear Momentum-Time Graph for Example 2 StudySmarter

    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.

    Momentum Change and Impulse A Drawing of Linear Momentum-Time Graph for Example 3 StudySmarter

    Fig. 7 - The image shows a Linear Momentum-Time Graph for objects \(X, Y,\) and \(Z\) that are under the influence of different forces.

    Above, we have the Linear Momentum-Time Graphs of objects \(X\), \(Y\), and \(Z\) under the influence of forces \(F_{X}\), \(F_{Y}\) and \(F_{Z}\), respectively.

    Order the magnitudes of the forces exerted on the objects.

    Solution

    Recall that the net force is the rate of change of momentum.

    $$\vec{F}_net = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}\$$

    Since all the functions all linear, their slope (and thus the force) are constant, and we can order the forces by ordering the lines according to their steepness.

    \(F_X > F_Y > F_Z\)

    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.

    Change in Momentum and Impulse - Key takeaways

    • Impulse is a vector that quantifies the effect of an external force acting over time. It is defined as integral of the force exerted on our system over a period of time.$$\vec J = \int_{t_0}^t \vec F(t) \mathrm{d}t$$
    • The change in momentum, the difference of the initial and final momentum, can be represented with the notation \(\Delta \vec p = \vec p_f - \vec p_i \).

    • Impulse can also be defined as the change of momentum of the system.$$\vec J = \int_{t_0}^t F(t) \mathrm{d}t = \Delta \vec p\mathrm{.}$$

    • Newton's Second Law is a consequence of the impulse-momentum theorem when mass is constant:$$F_{net} = m\vec a =m\frac{\mathrm{d} \vec v}{\mathrm{d} t} =\frac{\mathrm{d} (m\vec v)}{\mathrm{d} t} = \frac{\mathrm{d} \vec p}{\mathrm{d} t}. $$

    • We can also write an equation for a variable mass using the impulse-momentum theorem.$$\vec F_{net} =\frac{\mathrm{d} \vec p}{\mathrm{d} t}\\ = \frac{\mathrm{d} m}{\mathrm{d} t}\\\vec v$$

    • According to Newton's Second Law of Motion, objects under the influence of force move with acceleration. If we work this relation, we can find the relationship between impulse and linear momentum.

    • We can also calculate the impulse by multiplying the average force and the time period during which it acted. $$\vec F_\text{avg}\Delta t = \Delta\vec p\mathrm{.}$$

    • In Force-Time Graphs of an object, the area between the graph and the horizontal axis gives the impulse applied to the object. The impulse is positive if the area between the graph and the horizontal axis is above the time axis, and negative otherwise.

    • The change in a system's momentum with respect to time equals the net force exerted on the system.$$\vec F_{net} = \frac{\mathrm{d} \vec p}{\mathrm{d} t}\\\mathrm{.}$$

    • The slope in the Linear Momentum-Time Graph gives the net force acting over the system.

    References

    1. Fig. 1 - Shopping Cart (https://www.flickr.com/photos/uacescomm/21133249788/) by uacescomm (https://www.flickr.com/photos/uacescomm/) is licensed under Public Domain (https://creativecommons.org/publicdomain/zero/1.0/)
    2. Fig. 2 - Force-Time Graph: Positive and Negative Impulse, StudySmarter Originals
    3. Fig. 3 - Linear Momentum vs. Time Graph Example 1, StudySmarter Originals
    4. Fig. 4 - Force vs. Time Graph Example 1, StudySmarter Originals
    5. Fig. 5 - Force vs. Time Graph Example 2, StudySmarter Originals
    6. Fig. 6 - Linear Momentum vs. Time Graph Example 2, StudySmarter Originals
    7. Fig. 7 - Linear Momentum vs. Time Graph: Analyzing the Slope, StudySmarter Originals
    Frequently Asked Questions about Momentum Change and Impulse

    What is the difference between change in momentum and impulse?

    There is no difference between them. Impulse is equal to the change in momentum.

    Why does impulse not change momentum?

    Impulse doesn't "cause" a change in momentum, force causes that change.

    What is the impulse-momentum theorem in simple terms?

    The impulse-momentum theorem states that the impulse applied to an object will be equal to the change in its momentum. 

    What is the relationship between impulse and momentum?

    The impulse applied to an object by a force exerted over a period of time will be equal to the change in its momentum. 

    What is change in momentum equal to? 

    The change in momentum (final momentum minus the initial momentum) is equal to the impulse. 

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    What is the impulse?

    Impulse is a scalar quantity. 

    In Force-Time Graphs of an object, the area between the graph and the horizontal axis gives the impulse applied to the object. 

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