impulse momentum theorem experiment

9.2 Impulse and Collisions

Learning objectives.

By the end of this section, you will be able to:

• Explain what an impulse is, physically
• Describe what an impulse does
• Relate impulses to collisions
• Apply the impulse-momentum theorem to solve problems

We have defined momentum to be the product of mass and velocity. Therefore, if an object’s velocity should change (due to the application of a force on the object), then necessarily, its momentum changes as well. This indicates a connection between momentum and force. The purpose of this section is to explore and describe that connection.

Suppose you apply a force on a free object for some amount of time. Clearly, the larger the force, the larger the object’s change of momentum will be. Alternatively, the more time you spend applying this force, again the larger the change of momentum will be, as depicted in Figure 9.5 . The amount by which the object’s motion changes is therefore proportional to the magnitude of the force, and also to the time interval over which the force is applied.

Mathematically, if a quantity is proportional to two (or more) things, then it is proportional to the product of those things. The product of a force and a time interval (over which that force acts) is called impulse , and is given the symbol J → . J → .

Let F → ( t ) F → ( t ) be the force applied to an object over some differential time interval dt ( Figure 9.6 ). The resulting impulse on the object is defined as

The total impulse over the interval t f − t i t f − t i is

Equation 9.2 and Equation 9.3 together say that when a force is applied for an infinitesimal time interval dt , it causes an infinitesimal impulse d J → d J → , and the total impulse given to the object is defined to be the sum (integral) of all these infinitesimal impulses.

To calculate the impulse using Equation 9.3 , we need to know the force function F ( t ), which we often don’t. However, a result from calculus is useful here: Recall that the average value of a function over some interval is calculated by

where Δ x = x f − x i Δ x = x f − x i . Applying this to the time-dependent force function, we obtain

Therefore, from Equation 9.3 ,

The idea here is that you can calculate the impulse on the object even if you don’t know the details of the force as a function of time; you only need the average force. In fact, though, the process is usually reversed: You determine the impulse (by measurement or calculation) and then calculate the average force that caused that impulse.

To calculate the impulse, a useful result follows from writing the force in Equation 9.3 as F → ( t ) = m a → ( t ) F → ( t ) = m a → ( t ) :

For a constant force F → ave = F → = m a → F → ave = F → = m a → , this simplifies to

Note that the integral form, Equation 9.3 , applies to constant forces as well; in that case, since the force is independent of time, it comes out of the integral, which can then be trivially evaluated.

Example 9.1

The arizona meteor crater.

Using the given data about the meteor, and making reasonable guesses about the shape of the meteor and impact time, we first calculate the impulse using Equation 9.6 . We then use the relationship between force and impulse Equation 9.5 to estimate the average force during impact. Next, we choose a reasonable force function for the impact event, calculate the average value of that function Equation 9.4 , and set the resulting expression equal to the calculated average force. This enables us to solve for the maximum force.

The average force during the impact is related to the impulse by

From Equation 9.6 , J → = m Δ v → J → = m Δ v → , so we have

The mass is equal to the product of the meteor’s density and its volume:

If we assume (guess) that the meteor was roughly spherical, we have

Thus we obtain

The problem says the velocity at impact was −1.28 × 10 4 m/s j ^ −1.28 × 10 4 m/s j ^ (the final velocity is zero); also, we guess that the primary impact lasted about t max = 2 s t max = 2 s . Substituting these values gives

This is the average force applied during the collision. Notice that this force vector points in the same direction as the change of velocity vector Δ v → Δ v → .

Next, we calculate the maximum force. The impulse is related to the force function by

We need to make a reasonable choice for the force as a function of time. We define t = 0 t = 0 to be the moment the meteor first touches the ground. Then we assume the force is a maximum at impact, and rapidly drops to zero. A function that does this is

(The parameter τ τ represents how rapidly the force decreases to zero.) The average force is

where Δ t = t max − 0 s Δ t = t max − 0 s . Since we already have a numeric value for F ave F ave , we can use the result of the integral to obtain F max F max .

Choosing τ = 1 e t max τ = 1 e t max (this is a common choice, as you will see in later chapters), and guessing that t max = 2 s t max = 2 s , this integral evaluates to

Thus, the maximum force has a magnitude of

The complete force function, including the direction, is

This is the force Earth applied to the meteor; by Newton’s third law, the force the meteor applied to Earth is

which is the answer to the original question.

Significance

Notice that the area under each plot has been filled in. For the plot of the (constant) force F ave F ave , the area is a rectangle, corresponding to F ave Δ t = J F ave Δ t = J . As for the plot of F ( t ), recall from calculus that the area under the plot of a function is numerically equal to the integral of that function, over the specified interval; so here, that is ∫ 0 t max F ( t ) d t = J ∫ 0 t max F ( t ) d t = J . Thus, the areas are equal, and both represent the impulse that the meteor applied to Earth during the two-second impact. The average force on Earth sounds like a huge force, and it is. Nevertheless, Earth barely noticed it. The acceleration Earth obtained was just

which is completely immeasurable. That said, the impact created seismic waves that nowadays could be detected by modern monitoring equipment.

Example 9.2

The benefits of impulse.

• What average force does the driver experience during the collision?
• Without the seatbelt and airbag, his collision time (with the steering wheel) would have been approximately 0.20 s. What force would he experience in this case?
• Define the + x -direction to be the direction the car is initially moving. We know J → = F → Δ t J → = F → Δ t and J → = m Δ v → . J → = m Δ v → . Since J is equal to both those things, they must be equal to each other: F → Δ t = m Δ v → . F → Δ t = m Δ v → . We need to convert this weight to the equivalent mass, expressed in SI units: 860 N 9.8 m/s 2 = 87.8 kg . 860 N 9.8 m/s 2 = 87.8 kg . Remembering that Δ v → = v → f − v → i Δ v → = v → f − v → i , and noting that the final velocity is zero, we solve for the force: F → = m 0 − v i i ^ Δ t = ( 87.8 kg ) ( − ( 27 m / s ) i ^ 2.5 s ) = − ( 948 N ) i ^ . F → = m 0 − v i i ^ Δ t = ( 87.8 kg ) ( − ( 27 m / s ) i ^ 2.5 s ) = − ( 948 N ) i ^ . The negative sign implies that the force slows him down. For perspective, this is about 1.1 times his own weight.
• Same calculation, just the different time interval: F → = ( 87.8 kg ) ( − ( 27 m / s ) i ^ 0.20 s ) = − ( 11,853 N ) i ^ F → = ( 87.8 kg ) ( − ( 27 m / s ) i ^ 0.20 s ) = − ( 11,853 N ) i ^ which is about 14 times his own weight. Big difference!

Effect of Impulse

Since an impulse is a force acting for some amount of time, it causes an object’s motion to change. Recall Equation 9.6 :

Because m v → m v → is the momentum of a system, m Δ v → m Δ v → is the change of momentum Δ p → Δ p → . This gives us the following relation, called the impulse-momentum theorem (or relation).

Impulse-Momentum Theorem

An impulse applied to a system changes the system’s momentum, and that change of momentum is exactly equal to the impulse that was applied:

The impulse-momentum theorem is depicted graphically in Figure 9.10 .

There are two crucial concepts in the impulse-momentum theorem:

• Impulse is a vector quantity; an impulse of, say, − ( 10 N · s ) i ^ − ( 10 N · s ) i ^ is very different from an impulse of + ( 10 N · s ) i ^ + ( 10 N · s ) i ^ ; they cause completely opposite changes of momentum.
• An impulse does not cause momentum; rather, it causes a change in the momentum of an object. Thus, you must subtract the initial momentum from the final momentum, and—since momentum is also a vector quantity—you must take careful account of the signs of the momentum vectors.

The most common questions asked in relation to impulse are to calculate the applied force, or the change of velocity that occurs as a result of applying an impulse. The general approach is the same.

Problem-Solving Strategy

• Express the impulse as force times the relevant time interval.
• Express the impulse as the change of momentum, usually m Δ v m Δ v .
• Equate these and solve for the desired quantity.

Example 9.3

Moving the enterprise.

When Captain Picard commands, “Take us out,” the starship Enterprise ( Figure 9.11 ) starts from rest to a final speed of v f = 7.5 × 10 7 m/s v f = 7.5 × 10 7 m/s . Assuming this maneuver is completed in 60 s, what average force did the impulse engines apply to the ship?

Equating these expressions gives

Solving for the magnitude of the force and inserting the given values leads to

Check Your Understanding 9.1

The U.S. Air Force uses “10 g s” (an acceleration equal to 10 × 9.8 m/s 2 10 × 9.8 m/s 2 ) as the maximum acceleration a human can withstand (but only for several seconds) and survive. How much time must the Enterprise spend accelerating if the humans on board are to experience an average of at most 10 g s of acceleration? (Assume the inertial dampeners are offline.)

Example 9.4

The iphone drop.

We need to make a couple of reasonable estimates, as well as find technical data on the phone itself. First, let’s suppose that the phone is most often dropped from about chest height on an average-height person. Second, assume that it is dropped from rest, that is, with an initial vertical velocity of zero. Finally, we assume that the phone bounces very little—the height of its bounce is assumed to be negligible.

The impulse J → J → equals the change in momentum,

Next, the change of momentum is

We need to be careful with the velocities here; this is the change of velocity due to the collision with the floor. But the phone also has an initial drop velocity [ v → i = ( 0 m/s ) j ^ v → i = ( 0 m/s ) j ^ ], so we label our velocities. Let:

• v → i = v → i = the initial velocity with which the phone was dropped (zero, in this example)
• v → 1 = v → 1 = the velocity the phone had the instant just before it hit the floor
• v → 2 = v → 2 = the final velocity of the phone as a result of hitting the floor

Figure 9.12 shows the velocities at each of these points in the phone’s trajectory.

With these definitions, the change of momentum of the phone during the collision with the floor is

Since we assume the phone doesn’t bounce at all when it hits the floor (or at least, the bounce height is negligible), then v → 2 v → 2 is zero, so

We can get the speed of the phone just before it hits the floor using either kinematics or conservation of energy. We’ll use conservation of energy here; you should re-do this part of the problem using kinematics and prove that you get the same answer.

First, define the zero of potential energy to be located at the floor. Conservation of energy then gives us:

Defining h floor = 0 h floor = 0 and using v → i = ( 0 m/s ) j ^ v → i = ( 0 m/s ) j ^ gives

Because v 1 v 1 is a vector magnitude, it must be positive. Thus, m Δ v = m v 1 = m 2 g h drop m Δ v = m v 1 = m 2 g h drop . Inserting this result into the expression for force gives

Finally, we need to estimate the collision time. One common way to estimate a collision time is to calculate how long the object would take to travel its own length. The phone is moving at 5.4 m/s just before it hits the floor, and it is 0.14 m long, giving an estimated collision time of 0.026 s. Inserting the given numbers, we obtain

Check Your Understanding 9.2

What if we had assumed the phone did bounce on impact? Would this have increased the force on the iPhone, decreased it, or made no difference?

Momentum and Force

In Example 9.3 , we obtained an important relationship:

In words, the average force applied to an object is equal to the change of the momentum that the force causes, divided by the time interval over which this change of momentum occurs. This relationship is very useful in situations where the collision time Δ t Δ t is small, but measureable; typical values would be 1/10th of a second, or even one thousandth of a second. Car crashes, punting a football, or collisions of subatomic particles would meet this criterion.

For a continuously changing momentum—due to a continuously changing force—this becomes a powerful conceptual tool. In the limit Δ t → d t Δ t → d t , Equation 9.2 becomes

This says that the rate of change of the system’s momentum (implying that momentum is a function of time) is exactly equal to the net applied force (also, in general, a function of time). This is, in fact, Newton’s second law, written in terms of momentum rather than acceleration. This is the relationship Newton himself presented in his Principia Mathematica (although he called it “quantity of motion” rather than “momentum”).

If the mass of the system remains constant, Equation 9.3 reduces to the more familiar form of Newton’s second law. We can see this by substituting the definition of momentum:

The assumption of constant mass allowed us to pull m out of the derivative. If the mass is not constant, we cannot use this form of the second law, but instead must start from Equation 9.3 . Thus, one advantage to expressing force in terms of changing momentum is that it allows for the mass of the system to change, as well as the velocity; this is a concept we’ll explore when we study the motion of rockets.

Newton’s Second Law of Motion in Terms of Momentum

The net external force on a system is equal to the rate of change of the momentum of that system caused by the force:

Although Equation 9.3 allows for changing mass, as we will see in Rocket Propulsion , the relationship between momentum and force remains useful when the mass of the system is constant, as in the following example.

Example 9.5

Calculating force: venus williams’ tennis serve.

As noted above, when mass is constant, the change in momentum is given by

where we have used scalars because this problem involves only one dimension. In this example, the velocity just after impact and the time interval are given; thus, once Δ p Δ p is calculated, we can use F = Δ p Δ t F = Δ p Δ t to find the force.

Now the magnitude of the net external force can be determined by using

where we have retained only two significant figures in the final step.

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• Section URL: https://openstax.org/books/university-physics-volume-1/pages/9-2-impulse-and-collisions

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6.2: Activities

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• University of California, Davis

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• carts & track
• Pasco sensor devices
• cart linking accessories
• electronic weight scale

The General Idea

We have seen in the textbook that the impulse-momentum theorem is a repackaging of Newton's 2nd law:

\[\vec F = m\vec a = m\frac{d\vec v}{dt} = \frac{d\left(m\vec v\right)}{dt} = \frac{d\vec p}{dt}\;\;\;\Rightarrow\;\;\; \int\limits_a^b\vec F dt = \vec p_b-\vec p_a\]

We will use fancy sensing equipment to confirm this (as well as the 3rd law) by directly measuring all the physical quantities involved in different collisions.  Specifically, you will make these impulse/momentum and 3rd law tests for three separate scenarios, all of them involving a stationary target cart :

• magnetic repulsion collision with carts of equal masses
• magnetic repulsion collision with carts of unequal masses
• link-up collision with carts of whatever masses you choose

In addition to your analysis of impulse and momentum, you should also say a few words (backed-up with data!) about the conservation or non-conservation of kinetic energy in each case.

Some Things to Think About

The Pasco equipment and laptop can be finicky.  If you think it isn't working properly, don't waste time trying to debug it – call your TA over to assist you! Here are some things that should help, most of them related to proper use of the fancy equipment..

Here's how you get started with the equipment :

• If the carts are plugged into the laptop (they are charging), unplug them and press their power buttons.  A red light should come on near the battery icon, and then the red light near the Bluetooth icon should start flashing.
• Go to the "Student" login in the laptop (there is no password), and open the 9A folder.  Open the file "Momentum and Impulse."
• Run the application, and you will see what is shown below. Click the Bluetooth button at the top of the "Hardware Setup" window, and then scroll this window to the bottom.

• You should find carts available for Bluetooth pairing at the bottom of the "Hardware Setup" window. Check to make sure that the carts listed are the same ones you will be using by comparing their numbers, then click on those carts to pair them to the computer. They should then be listed higher in the window.

• When the time comes to do the link-up collision, you need to remove the magnets from the carts and replace them with the bottle-brush/tube links. Make sure that the brush that comes from one of the tubes pairs well with the other tube. If it does not, you can bend it slightly,  but remove the brush from both tubes completely before doing this – don't torque the brush/tube while connected to the cart, or you may break the cart .

Some pointers while taking data:

• During the magnetic repulsion runs, do not push the incoming cart so hard that it contacts the other cart – the magnets should repel without an audible "clack."  This will not only produce the best results, but will also avoid breaking the magnets.
• It is a good idea to set up your track so that the cart doesn't slow down or speed up very much when rolling on its own. There is going to be some effect due to friction in the wheels or from air resistance, but you should use the bubble level to make sure there is not a slant in the track affecting your results. Also, centering the carts in the track so that the sides don't drag is essential.

• Important! You will have 4 graphs on your computer screen, and you will want to use tools on each one, but the software has a maddening "feature."  The graph into which the tool appears is the last one that was selected.  This is as it should be, but the software selects the graph by mouse-over, rather than by clicking.  So if you want a tool to appear in the bottom graph, you need to make sure that was the last graph that your cursor was over before clicking the tool, which requires you to navigate the cursor  around  the other graphs on its way to clicking the tool.
• This lab doesn't include data tables, so you need some way to display your raw data. You should do this by taking screen captures of the graphs generated (after manipulating/magnifying them so they are easy to read).  You can do this with command-shift-4, which gives you a cursor that you can drag into a box around the portion of the screen you wish to capture (the file will appear on the desktop).  Getting this file from the desktop to where you can add it to your lab report is up to you.  In order to do anything over a network (such as emailing the file to yourself or using Airdrop), you will need to log the lab laptop into eduroam under your login. You might also be able to take a photo of the computer screen.
• We will not be doing rigorous uncertainty analysis here (the uncertainties inherent to our black-box-measuring-apparatus are not immediately apparent), but some reckoning of how far off the results are from what you predict (oh, btw, you should be hypothesizing results in each case!) is expected.  As with most experiments we do, if you are off by more than 10%, you should probably assume you made an error somewhere.  Either you made an incorrect hypothesis, or something went wrong in your analysis of the data.
• When you are finished with the equipment, please power-down the carts and plug them into the laptops with their USB cords, so that they are charged for the next lab.

Lab Report 

Craft a lab report for these activities and analysis, making sure to include every contributing group member's name on the front page.  You are strongly encouraged  to refer back to the Read Me as you do this, to make sure that you are not leaving out anything important.  You should also feel free to get feedback from your lab TA whenever you find that your group requires clarification or is at an impasse.

Every member of the group must upload a separate digital copy of the report to their lab assignment in Canvas  prior to leaving the lab classroom .  These reports are not to be written outside the lab setting.

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Momentum Change and Impulse Connection

• Momentum Change and Impulse
• Applications of Impulse-Momentum Change Theorem

A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down . If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.  

F = m • a or F = m • ∆v / t  

If both sides of the above equation are multiplied by the quantity t, a new equation results.

F • t = m • ∆v

This equation represents one of two primary principles to be used in the analysis of collisions during this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse . And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum . The equation really says that the

In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v.

In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed and thus his momentum. If the motion was represented by a ticker tape diagram , it might appear as follows:

At approximately the tenth dot on the diagram, the collision occurs and lasts for a certain amount of time; in terms of dots, the collision lasts for a time equivalent to approximately nine dots . In the halfback-defensive back collision, the halfback experiences a force that lasts for a certain amount of time to change his momentum. Since the collision causes the rightward-moving halfback to slow down, the force on the halfback must have been directed leftward. If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N•s. This impulse would cause a momentum change of 720 kg•m/s. In a collision, the impulse experienced by an object is always equal to the momentum change.

Representing a  Rebounding Collision

Now consider a collision of a tennis ball with a wall. Depending on the physical properties of the ball and wall, the speed at which the ball rebounds from the wall upon colliding with it will vary. The diagrams below depict the changes in velocity of the same ball. For each representation (vector diagram, velocity-time graph, and ticker tape pattern), indicate which case (A or B) has the greatest change in velocity, greatest acceleration , greatest momentum change, and greatest impulse . Support each answer. Click the button to check your answer.

a. The velocity change is greatest in case B . The velocity changes from +30 m/s to -28 m/s. This is a change of 58 m/s (-) and is greater than in case A (-15 m/s).

b. The acceleration is greatest in case B . Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above)

c. The momentum change is greatest in case B . Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above).

d. The impulse is greatest in case B . Impulse equals momentum change and the momentum change is greatest in case B (as stated above).

a. The velocity change is greatest in case A . The v changes from +5 m/s to -3 m/s. This is a change of 8 m/s (-) and is greater than in case B (-4 m/s).

b. The acceleration is greatest in case A . Acceleration depends on velocity change and the velocity change is greatest in case A (as stated above).

c. The momentum change is greatest in case A . Momentum change depends on velocity change and the velocity change is greatest in case A (as stated above).

d. The impulse is greatest in case A . Impulse equals momentum change and the momentum change is greatest in case A (as stated above).

a. The velocity change is greatest in case B . In each case the initial velocity is the same. In case B, the object rebounds in the opposite direction with a greater speed than in case A. This is equivalent to a change from +10 m/s to -5 m/s; whereas, case A has a change from +10 m/s to -2 m/s.

d. The impulse is greatest in case B . Impulse equals momentum change and the momentum change is greatest in case B (as stated above)

• The impulse experienced by an object is the force•time.
• The momentum change of an object is the mass•velocity change.
• The impulse equals the momentum change.

Click the button to view answers.  

There are a few observations that can be made in the above table that relate to the computational nature of the impulse-momentum change theorem. First, observe that the answers in the table above reveal that the third and fourth columns are always equal; that is, the impulse is always equal to the momentum change. Observe also that if any two of the first three columns are known, then the remaining column can be computed. This is true because the impulse=force • time. Knowing two of these three quantities allows us to compute the third quantity. And finally, observe that knowing any two of the last three columns allows us to compute the remaining column. This is true since momentum change = mass • velocity change.

We Would Like to Suggest ...

Check Your Understanding

Express your understanding of the impulse-momentum change theorem by answering the following questions. Click the button to view the answers.

1. A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds. Which cart (#1 or #2) has the greatest acceleration? Explain.

Cart #2 has the greatest acceleration . Recall that acceleration depends on force and mass. They each have the same mass, yet cart #2 has the greater force.

Which cart (#1 or #2) has the greatest impulse? Explain.

The impulse is the same for each cart . Impulse is force*time and can be calculated to be 1.0 N*s for each cart.

Which cart (#1 or #2) has the greatest change in momentum? Explain.

The momentum change is the same for each cart. Momentum change equals the impulse; if each cart has the same impulse, then it would follow that they have the same momentum change.

2. In a physics demonstration, two identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below.

Which balloon (A or B) has the greatest acceleration? Explain.

Balloon B has the greatest acceleration. The rate at which the velocity changes is greatest for Balloon B; this is shown by the fact that the speed (distance/time) changes most rapidly.

Which balloon (A or B) has the greatest final velocity? Explain.

Balloon B has the greatest final velocity. At the end of the diagram, the distance traveled in the last interval is greatest for Balloon B.

Which balloon (A or B) has the greatest momentum change? Explain.

Balloon B has the greatest momentum change. Since the final velocity is greatest for Balloon B, its velocity change is also the greatest. Momentum change depends on velocity change. The balloon with the greatest velocity change will have the greatest momentum change.

Which balloon (A or B) experiences the greatest impulse? Explain.

Balloon B has the greatest impulse. Impulse is equal to momentum change. If balloon B has the greatest momentum change, then it must also have the greatest impulse.

3. Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below.

At what approximate location on the diagram (in terms of dots) does each car begin to experience the impulse?

The collision occurs at approximately the ninth dot (plus or minus a dot). The diagram shows that it is at that location that the cars begin to slow down.

Which car (A or B) experiences the greatest acceleration? Explain.

Car A has the greatest acceleration. The velocity change of each car is the same. (They start with the same velocity and each finish with zero velocity.) Yet car A accomplishes this change in less time. Car A accelerates "most rapidly."

Which car (A or B) experiences the greatest change in momentum? Explain.

The momentum change is the same for each car. The velocity change of each car is the same (they start with the same velocity and each finish with zero velocity), and the mass of each car is the same. Thus, the momentum change is the same for each car.

Which car (A or B) experiences the greatest impulse? Explain.

The impulse is the same for each car. The impulse equals the momentum change. If the momentum change is the same for each car, then so must be the impulse.

a. In which case (A or B) is the change in velocity the greatest? Explain.

Case A has the greatest velocity change . The velocity change is -9 m/s in case A and only -5 m/s in case B.

b. In which case (A or B) is the change in momentum the greatest? Explain.

Case A has the greatest momentum change. The momentum change is dependent upon the velocity change; the object with the greatest velocity change has the greatest momentum change.

c. In which case (A or B) is the impulse the greatest? Explain.

The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for Car A, then so must be the impulse.

d. In which case (A or B) is the force that acts upon the car the greatest (assume contact times are the same in both cases)? Explain.

The impulse is greatest for Car A. The force is related to the impulse (I=F*t). The bigger impulse for Car A is attributed to the greater force upon Car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds.

5. Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid hitting a deer crossing the road. She strikes the air bag, that brings her body to a stop in 0.500 s. What average force does the seat belt exert on her?

F = (mass * velocity change)/time

F = (50 * 35) / 0.500

If Jennifer had not been wearing her seat belt and not had an air bag, then the windshield would have stopped her head in 0.002 s. What average force would the windshield have exerted on her?

F = (50 * 35)/0.002

F = 875 000 N

Note that a 250-fold decrease in the time corresponds to a 250-fold increase in the force.

6. A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck.

Impulse = F*t = 80 N * 0.1 s

Impulse = 8 N*s

Note that not all the numbers are necessary for computing the impulse; don't "force" the value of mass into the computation.

7. If a 5-kg object experiences a 10-N force for a duration of 0.10-second, then what is the momentum change of the object?

Momentum Change = 1.0 kg*m/s

The momentum change = mass*velocity change. But since velocity change is not known another strategy must be used to find the momentum change. The strategy involves first finding the impulse (F*t = 1.0 N*s). Since impulse = momentum change, the answer is 1.0 N*s.

• The Law of Action-Reaction (Revisited)

Impulse and Momentum

• is a quantity that describes an object's resistance to stopping (a kind of "moving inertia").
• is represented by the symbol p (boldface).

p  =  m v

• is a vector quantity (since velocity is a vector and mass is a scalar).
• is a quantity that describes the effect of a net force acting on an object (a kind of "moving force").
• is represented by the symbol J (boldface).

J  =  F ∆ t

• is a vector quantity (since force is a vector and time is a scalar).

J  = ∆ p

F ∆ t  =  m ∆ v

F   dt  =  m d v  +  v   dm

• The impulse-momentum theorem is logically equivalent to Newton's second law of motion (the force law).
• The SI unit of impulse is the newton second .
• The SI unit of momentum is the kilogram meter per second .

[N s = kg m/s]

• is a measure of the efficacy of rocket propellants.
• is equal to exhaust velocity and has the SI unit of meter per second .

• is equal to exhaust velocity divided by acceleration due to gravity and has the SI unit of second .

Remember Me

Shop Experiment Impulse and Momentum Experiments​

Impulse and momentum.

Experiment #6 from Physics Explorations and Projects

Introduction

The goal of this activity is to relate impulse and momentum, and to determine that the impulse is equal to the change in momentum. The investigation is set up in two parts. First, students will evaluate how to quantify the event that causes a change in motion (i.e., impulse). The second is to develop a model for how impulse changes the velocity or momentum of an object.

In the Preliminary Observations, students observe a cart experiencing an impulse, using a hoop spring on a force sensor to change the momentum of a cart. Students address impulse in Part I of the investigation.

In Part II, students address the question of quantifying the change in the motion state of the cart. Students who investigate the relationship between impulse and change in velocity should find that the constant of proportionality is about equal to the mass of the cart. Students who investigate the relationship between impulse and change in momentum should find that the two values are nearly numerically equal.

• Identify variables, design and perform investigations, collect and analyze data, and draw a conclusion.
• Determine impulse and change in momentum based on measurements of force and velocity.
• Create a mathematical model of the relationship between impulse and the change in momentum.

Sensors and Equipment

This experiment features the following sensors and equipment. Additional equipment may be required.

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This experiment is #6 of Physics Explorations and Projects . The experiment in the book includes student instructions as well as instructor information for set up, helpful hints, and sample graphs and data.

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Table of contents.

• Experiment 1 - Heart Rate Meter
• Experiment 2 - Kinematics
• Experiment 3 - Newton's Second Law
• Experiment 4 - Conservation of Energy

Experiment 5 - Momentum and Impulse

• Experiment 6 - Biceps Muscle Model
• Experiment 7 - Rotation and Gyroscopic Precession

Click Here for Experiment 5 - Momentum and Impulse

Shown in the picture below:

• Glider with bumper and flag
• Force sensor

Not shown in the picture above:

• Computer and Pasco interface
• Set of masses in 50-g increments to 600 g
• Vernier calipers or meter stick to measure glider flag
• Scale and weight set

Newton's Second Law tells us that the net force acting on an object is equal to the object's mass multiplied by its acceleration: \(\textbf{F}_{\textrm{net}} = m\textbf{a}\). Using \(\textbf{a} = \textrm{d}\textbf{v}/\textrm{d}t\), where \(\textbf{v}\) is the object's velocity, we can rewrite this law as

\begin{eqnarray} \textbf{F}_{\textrm{net}} &=& m\,\textrm{d}\textbf{v}/\textrm{d}t. \end{eqnarray}

Newton himself believed that this relation should also account for the possibility that the mass is varying:

\begin{eqnarray} \textbf{F}_{\textrm{net}} &=& \textrm{d}(m\textbf{v})/\textrm{d}t. \end{eqnarray}

Examples of varying masses include rain falling into a rolling open box car and a rocket expelling gases. The above equation can be rewritten as

\begin{eqnarray} \textbf{F}_{\textrm{net}} &=& \textrm{d}\textbf{p}/\textrm{d}t, \label{eqn_1} \end{eqnarray}

where \(\textbf{p} = m\textbf{v}\) is the momentum of the object. Eq. \eqref{eqn_1} is the most general definition of force: the change of momentum with time. If we write it as a differential equation,

\begin{eqnarray} \textrm{d}\textbf{p} &=& \textbf{F}_{\textrm{net}} \textrm{d}t, \end{eqnarray}

and integrate with respect to time, then Eq. \eqref{eqn_1} becomes

\begin{eqnarray} \Delta \textbf{p} &=& \textbf{p}_2 - \textbf{p}_1 = \int \textbf{F}_{\textrm{net}}\,\textrm{d}t. \label{eqn_2} \end{eqnarray}

The right side of Eq. \eqref{eqn_2} is known as the impulse , and the left side is the change in momentum. The notion of impulse is often associated with a force that acts for a short period of time. Examples of such forces include a bat hitting a ball and the impact between two objects moving at relatively high speeds.

In this experiment, you will verify Eq. \eqref{eqn_2} by allowing a glider on an air track to pass through a photogate and strike a force sensor. The sensor allows you to measure the force on the glider as a function of time. This time interval is relatively short, so the impulse approximation is valid. The velocity of the glider is measured when it crosses the photogate, just before and just after the collision. These two velocity measurements, along with knowledge of the glider's mass, allow you to calculate the change in momentum (i.e., the left side of Eq. \eqref{eqn_2}). The sensor generates a force-versus-time curve on the computer, which can be integrated to obtain the impulse (i.e., the right side of Eq. \eqref{eqn_2}). The glider has a foam bumper, so its collision with the force sensor is inelastic . In other words, kinetic energy is not conserved during the collision, but the change in momentum is still equal to the impulse.

INITIAL SETUP: CALIBRATING THE FORCE SENSOR

Arrange the force sensor so you can hang masses from it. Set up the sensor in Data Studio, and plug it in. Double-click on the force sensor icon in the setup window, click the “Measurement” tab, and check the box for “Voltage”. The sensor will produce force readings, but we are going to ignore these and do our own calibration of the sensor, so we can convert its voltage readings to forces. We will hang masses from the sensor, type in the force, and let the computer read the corresponding voltage. Note that the sensor has two interchangeable end parts: a screw hook for hanging masses in the calibration phase of the experiment, and a rubber bumper for the impulse measurements in the actual experimental runs. When calibrating the force sensor, use the screw hook, and mount the sensor vertically downward on the horizontal bar of the ring stand.

Click “Sampling Options” in the setup window tool bar, check the box for “Keep data values only when commanded”, and type in “mass” for a name and “grams” for units. Drag a table to the force sensor, and drag the voltage data and the Keyboard 1 data (the mass values) over to the table, so that you get two columns showing these data. We will convert the masses to forces later in Excel.

Click “Start”. The “Start” button changes to a keep-and-stop button as shown below:

Also, you will see in your table the running voltage reading. With no mass hanging on the force sensor, push the “Tare” button on the side of the sensor to zero the reading, and click “Keep”. You will be prompted to enter the mass value (zero in this case). Repeat this five times, so that you obtain five values of the voltage (nearly the same) for the zero-mass case.

Now add the 50-gram mass holder. Repeat five times the procedure of pressing “Keep” and entering a mass of 50. (We are using a minus sign for all the masses, since the glider will push the force sensor in the opposite direction during the actual experimental runs.) Perform another 50-gram step, and then perform 100-gram steps up to a total of 500 grams, taking five voltage readings for each mass value. Remember to type in minus signs in front of the mass values. Click the red “Stop” button when you are finished.

Your table now contains a column of five voltage readings for each mass. Call up an Excel worksheet, and copy the voltage column of this table into column A of Excel so you can work on it. (That is, select the data in the Data Studio column, and paste in the data.) In column B, copy your mass values form the Data Studio column. In column C, type in the mass series, just once for each mass value. In column D, convert these mass values to forces in newtons. In column E, take the average of the appropriate voltage readings for each mass. Be sure you average only the voltages that correspond to a given mass. For example, you can use “AVERAGE(A4..A8)” to average the values in cells A4 through A8. It is easy to make a mistake here, so have your lab partner check you entries in the AVERAGE function.

Copy the force values from column D and “Paste Special” (the values) into column F, select the data in columns E and F, and graph the force as a function of voltage. Remember that you need to choose “XY scatter”. (If you had not copied column D over to F, you would have obtained a graph of voltage as a function of force.) Insert a trendline with linear regression, and include its equation on the chart. This equation gives the required conversion from voltage to force. Write the equation in the “Data” section.

Level the air track carefully. Mount the force sensor horizontally on the vertical rod of the ring stand at the end of the track so the glider bumper will strike the sensor as the glider moves down the track. Replace the screw hook of the force sensor with the rubber bumper. Set up the photogate so the glider flag clears the gate by a few centimeters before the bumper strikes the sensor. Refer to the (animated) picture below, which shows the glider traveling from left to right (over unequal time intervals, and then it repeats).

The glider then bounces off the sensor and passes through the photogate again. In an experimental run, you should record the two photogate velocity readings and the force-versus-time curve from the sensor.

Weigh the glider, and record its mass (in kilograms) in the “Data” section.

Measure the length of the glider flag, and record its length (in meters) in the “Data” section.

We want to set up the photogate to measure the velocity of the glider. We will measure the time the photogate is blocked by the glider flag, and then do a calculation to find the velocity.

• Disengage “Keep data only when commanded” in the “Sampling Options” button.
• Set up the photogate in Data Studio, and plug it in.
• Click “Timers” in the setup window.
• Under “Timing Sequence Choices”, choose “Blocked”, and then “Unblocked”.
• Pictures of a blocked and an unblocked gate should appear in the window.
• Close the window.
• Click the “Calculate” button in the top tool bar.
• \(d\) is a constant: the measured length of the glider flag (in meters).
• \(t\) is a data measurement variable: timer 1.
• Click “Accept” again, and close this window.

We want the computer to start recording data when the glider first enters the photogate.

• Click the photogate icon, and in “Measurements”, check “State”.
• Click the “Sampling Options” button in the setup tool bar.
• Click the “Delayed Start” tab.
• Check “Data Measurement”.
• Set the first box (which may start with “Timer 1”) to “State, Ch 1 (V)”.
• Set the next box to “Is Below”.
• In the voltage text box, type in “4.9”.

(The photogate outputs a voltage of 5.0 V when unblocked, and 0 V when blocked. The instructions above delay the start of data collection until the photogate voltage drops below 4.9 V, i.e., until it is first blocked.)

Double-click on the force sensor icon. Set the sampling rate to 2000 Hz. Note that the computer will then take a force reading every 1/2000, or 0.0005, second. Drag a table over to the force sensor, and set it to read a column of voltages and a column of velocities (by dragging over the appropriate data). You do not need the time measurement column and can get rid of it by clicking the clock symbol on the table tool bar.

Turn on the air track, and click “Start”. Push the “Tare” button on the side of the force sensor to zero its readings. Send the glider down the track so it passes through the photogate, strikes and bounces off the force sensor, and crosses the gate again. Click “Stop”. Your table should show a long list of force and voltage readings, as well as the two velocity readings at the top. You may need to use the scroll bar on the table to see all the readings, particularly the second velocity reading. The second velocity value is smaller since the collision is inelastic.

Drag a graph to the force sensor on the computer, set it to read the force sensor voltages. You should see a nice graph of the impulsive force. Use the “Scale-to-Fit” button, if necessary, to locate the impulse. Use the “Zoom Select” button so you can see the impulse clearly.

If all is well, you are ready to take data. If not, check over your steps.

Have each lab partner make three measurements with different glider speeds to check the relation \(\Delta\textbf{p} = \int \textbf{F}_{\textrm{net}} \textrm{d}t\). To calculate the impulse, click on the column of the table with the voltage readings, and select an area of the graph that just covers the impulsive force (i.e., when the graph leaves the base line and returns to the base line, before any oscillations are encountered). These voltage readings are automatically highlighted in the table and correspond to the area chosen in the graph. Pull down the “Edit” menu to “Copy”, and paste the selected voltage data on an Excel sheet. (The time readings may copy over also, but we do not need them.) The figure below shows the Data Studio graph and table overlying an Excel sheet. The appropriate area of the graph has been selected, and you can see part of the highlighted table below. Only one velocity value is shown in the Data Studio table; you would need to scroll it to see the other.

In the next Excel column, use the force calibration equation determined earlier to convert the voltages to forces. The impulse is the area under the force-versus-time curve:

\begin{eqnarray} \int \textbf{F}_{\textrm{net}} \textrm{d}t &=& \sum \textbf{F}_i \,\Delta t_i = \Delta t \sum F_i, \end{eqnarray}

since the time intervals \(\Delta t_i = 0.0005\) second are all equal. You can obtain the force sum with an Excel function such as “=Sum(b3..b147)”. When you calculate the change in momentum from \(mv_1\) to \(mv_2\), should you add or subtract these two numbers?

In the “Data” section, record the three values of change in momentum and impulse, as well as the percentage difference between each set of values.

Initial Setup, step 6: Write in your voltage-to-force conversion equation:

Mass of glider (kg) =                                              

Length of glider flag (m) =                                              

In the table below, enter your measured changes in momentum, impulses, and the percentage differences.

ADDITIONAL CREDIT (3 mills)

Data Studio itself has an integral function on the “Special” button of the calculator window. This additional credit is for figuring out how to obtain the integral of just the area under the impulse without significant TA assistance. The procedure is not trivial, and may require some thinking on how to formulate the integral function, use of the smart cursor and its delta function, and reference to the help section of Data Studio. Don't forget conversion of voltage to force. When you get all this worked out, display a table comparing the Data Studio integrals with the Excel integrals.

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