# Introduction to Momentum

When Isaac Newton originally formulated his Laws, he defined the “quantity of the motion” as the mass times the velocity; we now know this quantity as momentum, the subject of this chapter.

[notice]

new physical quantity: momentum

measured in: $latex \frac{[kilogram] [meter]}{[second]} \large $

new equation: $latex \vec{p} = m \vec{v}$

[/notice]

Momentum fits into Newton’s 2nd Law as follows:

old equation: $latex \displaystyle\sum \vec{F} = \frac{d\vec{p}}{dt} $

With some basic understanding of vectors along with the above definition of momentum and with Newton’s 2nd Law, we can infer some important things about momentum.

The momentum $latex \vec{p}$ is in the direction of the velocity $latex \vec{v}$, and the magnitude of the momentum is proportional to the magnitude of the velocity through the factor of the object’s mass. That helps us calculate the momentum of an object; it is a combination of two quantities, mass and velocity.

Now that we have defined what the momentum is, how can it be changed? The change in momentum is in the direction of the net force (we know this from the second law shown above). We haven’t talked about how to figure out the direction and magnitude of the net force. In this chapter we will relate a known net force to a change in momentum.

Starting with the second law, we can derive that the magnitude of the change in momentum ($latex \Delta \vec{p}$) is the integral with respect to time of the force. This latter statement is a rearrangement of Newton’s second law,

[notice]

$$ \begin{equation}

\int \sum \vec{F}_{ext} dt = \Delta \vec{p}

\label{eq:int_2nd_law}

\tag{Integral 2nd Law}

\end{equation} $$

[/notice]

when does it apply? always

### Table Cloth Trick

Let’s summarize some of the things in the video and the above equation. Let’s start with a question,

QUESTION

[formidable id=21 ]

Your previous responses:

[frm-entry-links id=21 type=list field_key=8e8gin]

It’s important to clarify the difference between *a force* and *the net force.* What’s the difference? The net force is a combination of all of the forces on an object and that’s an important distinction! If someone tells you there is a huge force on an object so that momentum must be changing, you should say, “Hey wait!, what’s the net force?” because that’s all that actually matters.

The next important ingredient above is the time the force is applied. In the video with the table cloth trick, time is the key factor that makes it work. There is a force between the dishes (and cubs and whatever else is on the table) and the tablecloth. The relevant force in the horizontal direction is friction, and it’s the only force in the problem so that tells us there is a net force. If we look at the equation worked out above, we can see the net force is one ingredient in calculating the change in momentum. The other is the $latex \Delta t$ which is the change in time (from the start of the application of the force to the end) or the **interval of time** the force is applied. In the tablecloth demonstration, $latex \Delta t$ is tiny, a fraction of a second when done right, so the the left hand side of the equation is small. It follows that the change in momentum is also tiny and the plates remain safely on the table.

QUESTION

[formidable id=26 ]

Your previous responses:

[frm-entry-links id=26 type=list field_key=qqjt7s]

Let’s distinguish the two versions of Newton’s second law by calling the one with the derivative the Differential 2nd Law and the one with the integral the $latex \eqref{eq:int_2nd_law}$. Sometimes the lefthand side of the Integral second law is called the Impulse. We are not going to emphasize this name “impulse” but it’s good to mention the word so you are familiar as it may appear when other people are talking about physics.

This is going to be the first chapter applying physical concepts to real situations with substantial mathematical language, and we have to remind ourselves about some common trends in physics calculations. Taking a look at Newton’s second law, we have to notice it is a vector equation. What do the component equations for the Integral second law equation above for the x, y, and z components?

$latex \int \sum F_x dt = \Delta p_x $

$latex \int \sum F_y dt = \Delta p_y $

$latex \int \sum F_z dt = \Delta p_z $

The components of the forces cause changes in the corresponding components of the momentum. This will become crucial in the discussion that follows, so make sure you know why we can separate vector components.

I promised at the onset we would consider simple cases when tackling new ideas. With regard to momentum in the above integral form of Newton’s 2nd law, what would be the simplest case to consider? Many times when a physicist asks you to consider the simplest situation, they want you to consider when something is zero or unity (^a fancy way to say one). Let’s get to the punchline first and then explain why we did it. In this instance, we want to consider when the left hand side (i.e. the net force) is zero. We just agreed we have a vector equation right? That means separately we can consider if the net force along x is zero, if the net force along y is zero, or if the net force along z is zero. Of course, if the net force is just plain zero, then all the components are zero as well. Usually we will consider momentum components separately so let’s consider a single component, e.g. x, and answer a question.

QUESTION

[formidable id=27 ]

Your previous responses:

[frm-entry-links id=27 type=list field_key=t38sd2]

In this case we arrive at a special relationship which we will call the momentum conservation law:

new equation: $latex \Delta p_x = 0 $

when does it apply? when the net force along x is zero

There is nothing special about the label “x” and we could certainly replace it with the y or z directions, given that the respective y and z net forces were zero, and get another conservation law. In some sense, this equation is implied by the integral 2nd law, but usually you will see it written along the different **spatial** components.

What we have done is really at the heart of what a scientist does all the time! We looked at a situation under a special circumstance (the net force being zero) and we made a specific conclusion about the system we are considering. In the process we have created a new equation. Be cautious! The momentum conservation equation is not Newton’s second law! We can only apply momentum conservation in a restricted case, namely when the net force along a specific direction is zero.

These equations that have a limited range of truth don’t really have a special name, but you have to know whether an equation is always true or only “sometimes true” and in the latter case, the circumstances leading to its validity.

QUESTION

[formidable id=28 ]

Your previous responses:

[frm-entry-links id=28 type=list field_key=hjn1ma]

Let’s firm up our understanding by considering a real scenario when there is no external force. Keep in mind your answer to the above equation as you think about an ice skater moving in a straight line on ice. Suppose a skater named Alice has gotten herself up to speed and just wants to coast along the ice for a bit (see figure). She moves at constant speed in the x direction. The y direction will be perpendicular that, and positive up toward the sky. We have not yet talked about what forces are, but use your experience to try to come up with some that may exist in this problem.

QUESTION

[formidable id=29 ]

Your previous responses:

[frm-entry-links id=29 type=list field_key=rerlep]

Have you thought back to Newton’s Laws? We have discussed the 2nd quite a bit, but let’s recall the 1st Law again. If Alice is moving at constant speed, what must the net force be, according to Newton’s 1st law? The net force must be zero! If that is true, then $latex \Delta p_x = 0$. You may hesitate and protest, “But what about the y direction!” If she doesn’t move up or down (if she is not doing any sort of fancy jump) then her velocity in the y direction is zero and the net force is zero as well. But the truth is, it does not matter at all. We can consider the x direction as though what’s going on in the y direction does not exist at all. The reason for this is the vector form of Newton’s 2nd Law.

This example illustrates how a zero net force can be inferred from constant velocity motion and how we can then conclude that the change in momentum is zero. That might not seem like much but it gets us started on momentum and after the discussion below and our next example, we will really see why momentum is such a powerful tool. Later, when we discuss forces in detail, we will return to the idea of momentum and see how things really tie together. For now, a good working definition of force is, something that changes momentum.

# Momentum Changes Proportional to Net Force

The next simplest case to consider is like above when there is a single object, and a net force. The unit for momentum is simply the combined units of its constituents, mass [kg] and velocity [ meter / second] as introduced above. Forces and the net force (which is a combination of all forces on an object or system of objects) have a special name for their unit, the Newton.

new physical quantity: **force**

new unit: **Newton, N**

Just as momentum has constituent units from mass and velocity, one could work out what the constituent units are for force.

QUESTION

[formidable id=30 ]

Your previous responses:

[frm-entry-links id=30 type=list field_key=1iaojo]

Suppose Ben and Alice are kicking a soccer ball around. Alice kicks the ball over to Ben at a speed of 15 m/s and Ben kicks it back to her. Of course, the momentum of the soccer ball changes. First the ball is going toward Ben and after the kick it is going toward Alice. The source of this change is the force exerted by Ben’s foot and the time of Ben’s contact with the ball we can estimate as 20 milliseconds If the ball returns to Alice at a speed of 17 m/s moving flat along the ground, how much force does Ben exert with his foot? Let’s idealize the kick a bit and assume the force is constant (which is probably not the case in real life). The mass of a soccer ball is 0.43kg.

If we idealize, we can write out the Integral 2nd Law as

$latex \sum \vec{F}_{const} \Delta t = \Delta \vec{p}$

QUESTION

[formidable id=31]

Your Previous responses:

[frm-entry-links id=31 type=list field_key=x0h378]

As we do more examples, we want you to have an idea of how big quantities are by having some intuition about the size of units. While we just used the mass of a soccer ball, let’s get familiar with some other everyday sized objects’ masses. How many kilograms are you?

# Momentum, Treating a Collection of Objects

So far, we deduced from both the 1st and 2nd Laws that we can understand why a skater (or some other object) moving at constant velocity will keep moving at constant velocity in the absence of a force. And that’s what Newton’s 1st law already told us. The important part was to get a feel for momentum and why delta P was zero. Next we looked at a scenario when there was a net force on a single object. We calculated the net force during a kick based on the initial and final momentum (the change in momentum) and the time of the kick.

We can spice up the discussion a bit more if we consider an object that can break apart or separate into multiple objects. This could be something that is solid and later explodes in fragments (like fireworks), an object that is designed to come apart at specific times (like a space shuttle), and even two objects that come together for a short time and then separate again (like a pair of ice skaters).

Our friends Alice and Ben have taken up skating. Suppose they move together at constant velocity and in the x direction. Can we find the momentum of the system of objects consisting of Alice and Ben? We can!

momentum of Alice = mass of Alice times velocity of Alice

momentum of Ben = mass of Ben times velocity of Ben

momentum of the system = momentum of Alice plus momentum of Ben

We used the definition of momentum, $latex \vec{p} = m\vec{v}$, and the fact that the moment of a system of objects is the sum of their individual momentum,

$latex \vec{p}_{system} = \displaystyle\sum_i \vec{p}_i $

In our example, their velocities are the same because they are moving together. What would happen if either Alice or Ben pushed away from the other while they moved at constant velocity? “What would happen” means how does the momentum of Ben and Alice change? To know how the momentum changes we need to know if there is a net force and what that force is. Whenever there is more than one object, it is useful to think about forces they may be exerting on each other, and that may require us to use Newton’s 3rd Law.

QUESTION

[formidable id=32 ]

Your previous responses:

[frm-entry-links id=32 type=list field_key=slaew1]

If Alice pushes on Ben, then by Newton’s 3rd Law Ben pushes back on Alice, even if he doesn’t want to. These are hte blue vectors in the figure above. If they both push, then maybe the size of the forces are bigger, but that doesn’t matter. The point is there is a force of Alice on Ben ($latex \vec{F}_{A on B}$) and another of Ben on Alice ($latex \vec{F}_{B on A}$), and these forces are equal in magnitude and opposite in direction. And one force is on Ben and the other force is on Alice, don’t forget that part.

Now, let’s look again at the integral 2nd law in the x direction

$latex \int (\sum \vec{F})_x dt = \Delta p_x $

where in this case P_x is the momentum of the system of Alice and Ben. Let’s take a look at the interval of time where Alice pushes on Ben. The two forces ($latex \vec{F}_{AonB}$ and $latex \vec{F}_{BonA})$ only have components along the x direction and can be inserted into the sum F_x as the only forces on the two skaters as they push on each other.

$latex \int (\vec{F}_{AonB} + \vec{F}_{BonA})_x dt = \Delta \vec{p}_x $

We already concluded the two forces are equal in magnitude and opposite in direction, which means we can replace one of the forces in the equation above with the negative of the other. Everything happens along the x direction so we can look at just a single component of the Momentum Conservation Law.

$latex \int (\vec{F}_{AonB})_x + -(\vec{F}_{AonB})_x dt = (\Delta \vec p)_x $

$latex 0 = \Delta \vec{p}_x $

This tells us that again, the change in momentum is zero, in this case for the system. If we just considered Ben as the object, his momentum would be different after the push. That is to say, velocity would be different and his mass the same, so his momentum different. Same goes for Alice. For Alice and Ben individually, momentum is not conserved. Let’s summarize what we just did by bringing to light how you should think about the Integral 2nd Law when trying to use momentum conservation on a system of objects.

$latex \int \sum \vec{F}_{external} dt = \Delta \vec{p} $

First thing, this is the vector version, which means this applies to each and every component of net force and change in momentum. The change we made is the subscript on the forces on the left, $latex \vec{F}_{external}$. We just showed that we don’t need to consider *internal* forces, like the ones between Alice and Ben, because of Newton’s 3rd Law. Of course, if you consider internal forces, things will still work out just fine, because they will cancel out when considering the system and all forces. This notation just simplifies how we can think about setting up the problem. It’s not much hassle with two skaters but think about considering the atoms in a balloon! We don’t want to write down all those pairs of forces!

What good is all this? We have just shown that we can consider a system when applying the momentum conservation law. Momentum conservation may not always apply to single objects, but choosing a system of those single objects may have momentum conservation. Let’s take a look at the pair of skaters again, this time with some numbers to see how this works.

Alice and Ben are skating together and get up to a constant speed of 5 m/s. They are facing each other as in figure X and Alice has her back in the direction of motion(positive x direcction). Alice decides she wants to push Ben away and gives him a shove so that Ben moves from her in the opposite direction(the negative x direction). Her shove causes Ben to slow down to 1 m/s moving in the same direction he started and she picks up speed after the push. If Ben’s mass is 85kg and Alice’s mass is 55kg, what is Alice’s velocity after the push? There are no external forces in the x direction when defining the system as Alice and Ben. Is it possible for Ben to have a negative velocity after the push (keeping our convention that the +x direction is the direction Ben is facing)?

What physical concept are we using? Momentum conservation for a system of objects.

If we are doing momentum conservation, When we hear “conservation” we need to think “before” and “after”. Well there is before the push and after the push, and that will get us started. Let’s now try to write the momentum before the the push

$latex p_{Alice} = m_{Alice} u_{Alice} = (55Kg)(5m/s) = 275 kgm/s$

$latex p_{Ben} = m_{Ben} u_{Ben} = (85Kg)(5m/s) = 425kgm/s$

$latex p_{System} = p_{Alice} + p_{Ben} = 700 kg*m/s$

Now comes the application of Conservation of Momentum. After the push, P_System = 700 kg*m/s because we already decided the system’s momentum did not change (no external forces). If we can just write down an expression for the system after the push we can relate it back to the the initial momentum. The system momentum is still just Alice and Ben’s individual momentum so let’s right as much as we know.

$latex p_{System} = m_{Alice} v_{Alice} + m_{Ben} v_{Ben} = (55Kg) v_{Alice} + (85kg)(1m/s)$

There is only one unknown quantity, the velocity of Alice and that’s precisely what we were after, so we can now equate this P_System after the push to P_system before the push

$latex m_{Alice} v_{Alice} = 700 kgm/s \rightarrow v_{Alice} = 11.18 m/s$

Alice may be confused as to why her momentum changed (it got quite a bit larger). She says “I understand Momentum Conservation just fine and there was no force on me, because I was the one who pushed on Ben.”

QUESTION

[formidable id=33 ]

Your previous responses:

[frm-entry-links id=33 type=list field_key=4tfpgh]

Although we are studying momentum this gives a chance to build our insight on Newton’s Laws, in this case his third law. Alice failed to realize that when she pushed on Ben, Ben had to push back on her with equal magnitude.

You may be thinking this is cool (I hope you are!) but ask yourself, how many situations are there with people and other important objects on ice with no friction. Probably not so many or are there?

QUESTION

[formidable id=34 ]

Your previous responses:

[frm-entry-links id=34 type=list field_key=qjzmd7]

The “collision” in the example above is the interval of time in which Alice pushes on Ben (and Ben’s simultaneous force on Alice). As long as we know the momentum right before a “collision”, then we can calculate the momentum after the “collision” if momentum if during the collision, the net force is zero.

QUESTION

[formidable id=35 ]

Your previous responses:

[frm-entry-links id=35 type=list field_key=3c54j5]

After this whole skating incident, Alice and Ben talk to their friend Chloe who watched the whole thing happen from a bench near the ice skating rink. Chloe and Ben compare notes on how fast Alice was going after she pushed him but there is a problem. Chloe says Alice was moving 11.18 m/s, exactly what we calculate above, but Ben says Alice was moving 10.18 m/s. Who is right?

# Relative Velocity

To answer the question above about who is measuring the correct velocity of Alice, we need to discuss a new concept called a frame of reference. In the above discussion of Alice and Ben ice skating, what point of view are you taking to visualize and measure what is happening? Does your point of view match up with Chloe’s point of view from the stationary bench near the skating rink? We usually phrase problems from this perspective; that is, we do problems from a stationary vantage point in a laboratory. For this reason, the usual scientist viewpoint is called the laboratory frame of reference. Think back to the situation above and imagine how the whole thing looks from either Alice’s or Ben’s perspective. How is Chloe (someone sitting still on Earth) measuring velocities? She measures all her velocities relative the Earth. In this case, the Earth frame, the lab frame, Chloe’s frame, and any stationary onlooker’s frame would be akin to a lab frame of reference.

You are free to do physics from any frame you choose, however; physical laws only take the same form in inertial frames. This means that in order to use the equations the way we have been writing them in this text, you have to choose an inertial frame and stick with it while solving a problem. Once you get an answer, it’s easy to convert that answer to one in another frame and we will get to that in a moment. So what is an inertial frame? The simplest definition is that an inertial frame is any frame with constant velocity.

Velocity measurements depend on who is measuring and which frame of reference you observe from. Anyone moving in Ben’s frame of reference ( 1 m/s in the positive x direction), will see things just as Ben does just as anyone standing still near the skating rink would agree with Chloe. If you know the velocity in one frame, you can get the velocity in another frame from the following,

$latex \vec{v}_{A \text{ in frame } C} = \vec{v}_{A \text{ in frame } B} + \vec{v}_{\text{frame }B\text{ relative frame }C} $

Let’s try to answer the question about who is right posed above. Ben measures velocity relative his own frame, which is moving according to Chloe (and probably you imagining the problem) to the left at 1 m/s after the collision. Let’s plug in to the equation,

$latex \vec{v}_{\text{Alice relative Chloe}} = \vec{v}_{\text{Alice relative Ben}} + \vec{v}_{\text{Ben relative Chloe}}$

$latex 11.18 m/s (\text{Alice relative Chloe}) = 10.18 m/s (\text{Alice relative Ben}) + 1.00 m/s (\text{Ben relative Chloe})$

True statement! But maybe we just got lucky somehow, so let’s double check with by asking what Alice measures for Ben’s velocity. Alice has some measurements of her own, she says Ben is moving -10.18m/s and Chloe is moving – 11.18 m/s (she measures her own velocity in her frame and that’s zero).

$latex \vec{v}_{\text{Ben relative Alice}} = \vec{v}_{\text{Ben relative Chloe}} + \vec{v}_{\text{Chloe relative Alice}}$

$latex -10.18 m/s = 1.00 m/s + -11.18 m/s$

.

Still works! You can keep checking as many cases as you find necessary to convince yourself that this is the proper velocity transformation equation between inertial frames. Right now we should motivate a new question, which frames are good ones to choose? To do this, we need to consider a new situation, two objects initially apart, that collide and separate. The animation below shows four panels. The first panel is the laboratory frame. The second and third panels are frames moving with the initial velocity of the red and blue masses respectively. The final frame is a custom frame and you can choose the velocity from which to observe the collision.

It’s important to note that Newton’s Laws apply in any frame that moves at constant velocity so any of the frames used above are good ones and we could always work out how in fact, everyone is in agreement. These constant velocity frames are so important, they are given a special name, intertial frames, and unless we say otherwise, always assume we have to do physics in an inertial frame for the remainder of this text.

### Interactive momentum demo

Open interactive momentum demo in a new window

QUESTION

[formidable id=36 ]

Your previous responses:

[frm-entry-links id=36 type=list field_key=3vf3os]

In the second panel, we say we are using a frame with the initial velocity of the red ball (in the lab frame).

QUESTION

[formidable id=37 ]

Your previous responses:

[frm-entry-links id=37 type=list field_key=llqg06]

1D Momentum Summary: Write a brief summary describing what you learned from the interactive demo or list questions if you have them. Did this help you understand the Ice Skating example?

QUESTION

[formidable id=38 ]

Your previous responses:

[frm-entry-links id=38 type=list field_key=bbs3br]

I hope seeing a few different situations gives you a feel for what’s happening from the point of view of either Alice or Ben. In the 1D momentum demo, the lab frame is the top most panel; that’s like you sitting on the bench in the park watching Alice and Ben. Alice and Ben each have their own frame of reference, and so do the red and blue masses, but they are not inertial frames. In the demo, panels 2 and 3 are inertial because they are the initial velocity f the red and blue masses (and these are constant).

QUESTION

[formidable id=39 ]

Your previous responses:

[frm-entry-links id=39 type=list field_key=wie4x7]

One could see things just like Alice if you move with the same velocity as Alice. That’s what Alice’s friend Barbara sees as she rides her bike past the whole thing at 11.18 m/s (the final velocity of Alice in the lab frame). She sees Alice stationary (velocity equal zero) and Chloe moving away from her at -11.18 m/s along with Ben moving away at 10.18. Those are the same observations Alice makes in her own frame after the collision.

We have looked at momentum collisions in one dimension from multiple frames. Using our definition of momentum for individual objects, we learned to define the momentum of a larger collection of objects. When working with systems of objects we have observed the power of momentum conservation for collision type problems. What we discuss next is nothing new conceptually but is an extension of all the ideas we have covered so far.

# 2D Momentum Conservation

As we saw through a few examples, there are many real situations involving momentum conservation in only one dimension. That’s great, but there are just as many important scenarios involving momentum conservation in two dimensions! The integral form of Newton’s 2nd Law is a vector equation and if more than one component of the external force is zero, we can conserve more than one component of the momentum.

Nothing has changed from the previous sections. We can examine the vector components of the Integral 2nd Law. Taking two such components (generally x and y) let’s us solve problems in 2 dimensions. These kinds of problems are very common. Cars, boats and people all move around on surfaces. Flat surfaces define a nice two dimensional landscape for momentum conservation problems. Remember that for momentum conservation to be true, the net external force on the system needs to be zero, but only for the duration of the collision under consideration. We don’t always need to consider situations on ice (with no friction). We will again take a real example (this one a bit more involved than what we have considered so far) to illustrate how momentum conservation plays out in two dimensions.

Alice and Ben are relaxing on a Sunday afternoon watching some American football. They have a longstanding debate about whether offensive players or defensive players run faster. A receiver catches a pass and is running from midfield toward the sideline at a 45 degree angle. In the figure above, the sideline is parallel the left edge of the image. The safety breaks on the play and chases him at a 30 degree angle as shown in the figure. The safety makes the tackle near the sideline and they go out of bounds together exactly perpendicular the sideline. To settle their debate, Alice and Ben decide to calculate how fast each player is moving before the tackle and are able to get all angles and the receivers velocity (8 m/s) by replaying some of the game on their TV. They also look up the mass of the players online. The receiver is 80kg and the safety is 95kg. They will need to calculate the speed of the safety before the tackle. How fast was the safety moving right before the tackle?

To answer the above question, we need to think about both the x and y components of the momentum. Which one do we need and what do we already know?

We don’t know the velocity of the “pile” as it goes out of bounds, but we do know a component of that velocity! If we put the x axis perpendicular the sideline (going horizontal in the picture) and the y axis parallel the sideline,

QUESTION

[formidable id=40 ]

Your previous responses:

[frm-entry-links id=40 type=list field_key=bcrhws]

Let’s write the components of the momentum for the receiver, safety, and pile. Let’s call to the left positive x, and up positive y.

Before

$latex \vec{p}_{receiver} = (80kg)(8\frac{m}{s})\text{ at } 45^\circ = (640 cos45^\circ ,-640sin45^\circ ) kg\frac{m}{s}$

$latex \vec{p}_{safety} = (95kg)(|\vec{v}_{safety}|)\text{ at } 30^\circ = (95kg|\vec{v}_{safety}| cos30^\circ ,95Kg|\vec{v}_{safety}|sin30^\circ)$

After

$latex \vec{p}_{pile} = (175kg)|\vec{v}_{pile}| out of bounds = (175kg|\vec{v}_{pile}|, 0)$

Momentum is a vector, and now we see why that matters. During the collision, there is no external force applied for a very long time, so momentum is conserved in both x and y directions. The relationships are,

x-component

$latex \Delta p_x = 0 \rightarrow p_{x,initial} = p_{x,final}$

$latex 640 cos45^\circ kg\frac{m}{s} + 95kg|\vec{v}_{safety}| cos30^\circ = 175kg|\vec{v}_{pile}| $

y-component

$latex \Delta p_y = 0 \rightarrow p_{y,initial} = p_{y,final}$

$latex ( -640sin45^\circ ) kg\frac{m}{s} + 95kg(|\vec{v}_{safety}|sin30^\circ) = 0 $

Looking at these two equations, there are two variables, $latex \vec{v}_{safety}$ and $latex \vec{v}_{pile}$. In the first equation for the x-component, they both appear so we can not solve immediately. In the y-component equation, there is only one variable, which is incidentally what we set out to find!

$latex (-640sin45^\circ ) kg\frac{m}{s} + 95kg(|\vec{v}_{safety}|sin30^\circ) = 0 \rightarrow |\vec{v}_{safety}| = 9.53 \frac{m}{s} $

For completeness, let’s solve for the velocity of the pile as it goes out of bound. To do that, we can take our result for the safety’s initial velocity and plug into the x-component equation above which we haven’t used yet.

$latex |\vec{v}_{pile}| = \frac{640cos45^\circ kg\frac{m}{s} + 95kg(9.53\frac{m}{s})cos30^\circ}{175kg} = $

# Center of Mass

In the above problems, we defined a system of objects to proceed with solving conservation of momentum problems. We can formalize the idea of a system of objects by defining a new concept, the center of mass. To define the center of mass we first need to review something we have been using to solve problems.

QUESTION

[formidable id=41 ]

Your previous responses:

[frm-entry-links id=41 type=list field_key=kkgho3]

As it turns out, forces and linear momentum do *not* depend on the shape of an object. You may protest that in the real world, the shape of an object makes a huge different to how things move, and that is a valid concern and what you say is true! The shape of an object will be of absolute importance in the next chapter when we think about angular momentum and torque. Even after we come to understand angular momentum however; the facts remain unchanged for linear quantities: the motion of an object’s center of mass has no dependence on shape and we can treat objects as points, with all the mass concentrated at the center of mass.

The center of mass of a system of objects can be calculated if you know the location of all its constituent parts. Let’s label an object with a numeral and then subscript the position coordinates with that numerical value.

$latex \vec{r}_1 = (x_1,y_1) = (3,8)$

$latex \vec{r}_2 = (x_2,y_2) = (3,2)$

$latex \vec{r}_3 = (x_3,y_3) = (-1,2)$

$latex \vec{r}_4 = (x_4,y_4) = (7,-2)$

But that’s just the locations of the objects. We need the masses of each constituent object as well.

$latex m_1 = 7kg$

$latex m_2 = 1kg$

$latex m_3 = 5kg$

$latex m_4 = 3kg$

Now, the center of mass is defined in a pretty sensible way (we use the abbreviation “CM” for center of mass often), using $latex m_{total} = m_1 + m_2 + m_3 + m_4$,

$latex m_{total} \vec{r}_{CM} = m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 + m_4\vec{r}_4$

$latex \vec{r}_{CM} = \frac{ 7kg(3,8) + 1kg(3,2) + 5kg(-1,2) + 3kg(7,-2) }{ 7kg + 1kg + 5kg + 3kg} = (2.5, 3.875) $

The process we just completed above can be summarized whenever you want to find the center of mass of a system of objects. For the system of interest,

(1) Identify all the objects in the system, along with their position vectors and masses.

(2) The center of mass position is

$latex \vec{r}_{CM} = \frac{ \displaystyle\sum_i m_i \vec{r}_i }{ \displaystyle\sum_i m_i } $.

Once you know the center of mass, we can apply what we know about momentum to the entire system. To restate Newton’s second law in integral form for the center of mass, all we need to do is replace the momentum of a single object with the collective momentum of the system as we did in the momentum conservation problems.

$latex \displaystyle\sum \vec{F}_{\text{external to system}} =\frac{d\vec{p}_{CM}}{dt}$

We can see from the above equation that the motion of a system of objects follows the same law of motion as an individual object. The subscript on force remind us that the we only need to consider forces on our system that come from external sources. The subscript on momentum reminds us that we are finding the change in momentum of the center of mass of the system. Concerning our system of objects, we know how to calculate the position of the center of mass and now we can calculate how the momentum changes in time. This is a complete prescription for determining the motion of objects.

The above discussion of center of mass does lead to one additional result that can be used to understand physical scenarios: **if the net force on a system of objects is zero, then the momentum does not change and the location of the center of mass is constant.** In a sense, we could have included this discussion with our first look at momentum conservation, because we saw that the momentum of a system of objects can not change if there are no external forces. That discussion lead us to consider collision type problems which are a different flavor than the problems we are about to explore.

Alice and Ben have gone with their friends to a lake and are on a small boat. The lake is calm and there is essentially no friction between the bottom of the boat. Ben is sitting next to Alice when he realizes he wants to get something out of his pack at the other end of the boat. He stands up and walks to where his pack is, then realizing the boat has moved relative the shoreline. He asks Alice what happened and so Alice explains.

The boat has mass 40kg, Alice is 55kg and Ben 85kg. The boat is 2.5 meters long and Alice and Ben are sitting 0.5 meters from one end. Ben’s trip to his pack covers a distance 1.5 meters bringing him 0.5 meters from the other end of the boat. How far does the boat move because of Ben’s motion.

First let’s find the center of mass of the boat. Now, we are going to have to choose a coordinate system in order to give our objects coordinates. It does not matter which coordinates you choose and there is only motion along one axis, so we will put the center of the boat at the origin. Given that choice and the information above, we can write down the center of mass equation

$latex x_{CM} = \frac{1}{m_A+m_B+m_{boat}} (m_Ax_A+m_Bx_B+m_{boat}) $

$latex x_{CM} = \frac{1}{55kg+85kg+40kg}( (55kg)(0.75m)+(85kg)(0.75)+(40kg)(0) ) $

$latex x_{CM} = 0.583 m$

We just calculated the position of the center of mass.

QUESTION

[formidable id=42 ]

Your previous responses:

[frm-entry-links id=42 type=list field_key=n5f4ll]

On the lake, there is no friction and therefore no external forces so the center of mass will remain this value throughout our calculation. Let us proceed by trying to come up with a picture after Ben has moved to the left. If we simply use our intuition about which way the boat will move, can we agree the boat goes to the right as been walks to the left? The boat can’t possibly move to the left with Ben because that would move only move mass to the left in our coordinate system, changing $latex x_{CM}$.

We are trying to solve for the distance the boat moves, and we have agreed it moves to the right. Let us call the displacement the boat moves $latex d$, and note that when Ben stops near the left end of the boat, the center of the boat will be at $latex d$ to the right (see figure). Now are task is to rewrite the locations of Alice and Ben.

QUESTION

[formidable id=43 ]

Your previous responses:

[frm-entry-links id=43 type=list field_key=k1co0m]

Alice’s position is still 0.75m from the center of the boat (that’s how we first wrote her position above) so if the boat has moved to the right by $latex d$, then Alice’s position is $latex d+0.75$. Ben is now 0.75 on the other side of the center, and his position is therefore $latex d-0.75$. Let’s write this down using primed coordinates for the new locations.

$latex x’_{boat} = d$

$latex x’_A = d+0.75m$

$latex x’_B = d-0.75m$

$latex x’_{CM} = \frac{1}{180kg} ( (55kg)(d+0.75m) + (85kg)(d-0.75m) + (40kg)(d) ) $

We already agreed the center of mass does not move, so the equation above may be set equal to the center of mass we calculated before Ben took his short walk on the boat. That’s $latex x_{CM} = 0.583$. Knowing the left hand side of the above equation means we know everything except for the value of $latex d$, which is the distance the boat travels and the thing we set out to find.

If we solve for d, we find $latex d = 0.708m$ Alice’s new position is $latex x’_A=1.458m$. Ben’s position is $latex x’_B=-0.042m$. It is always a good idea to check our work. To check your work, you can take the new positions of Alice, Ben, and the boat and explicitly check that you get the same center of mass that we started with when Alice and Ben were seated next to each other.

That wraps up our chapter on momentum. We covered how a single object’s momentum changes in response to a net force. We learned that by treating a system of objects with no external forces present, momentum is conserved. The treatment of a system of objects lead us to consider the center of mass of a system of objects and we found that a system of object’s momentum obeys the same laws as a single object.

# Problems and Exercises

- (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant?
- Consider the following question:
*A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg.*Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer. - Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.
- A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left?
- (a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from?

**Citations**

OpenStax College. “Linear Momentum and Force.” Connexions. June 20, 2012. http://cnx.org/content/m42156/1.3/.

OpenStax College. “Elastic Collisions in One Dimension.” Connexions. June 13, 2012. http://cnx.org/content/m42163/1.3/