When two objects interact with each other, they exert forces of equal magnitude and opposite direction on each other. This idea reconsiders N3L much the same way Kinetic Energy and the Work Energy Theorem got us thinking about N2L! An important way to express this concept is in terms of the momentum, p, of an object which is the product of the object’s mass times its velocity!

Since velocity is a vector, so is momentum. We can describe the behavior of two interacting objects or systems, like a baseball and a bat or a squid ejecting water as a source of motility. Each object in a pair undergoes a change in velocity and, hence, a change in momentum and momentum changes are equal in magnitude and opposite and direction. This leads us down the glorious path towards another Great Conservation Law, The Law of Conservation of Momentum! This Great Law is essential in describing what happens when two objects or systems interact. Interestingly, this law happens to harbor the fundamental physical principle from which Newton’s Three Laws of Motion arise! Momentum can also be used in more general situations than those in which we are able to apply Newton’s Laws, including cases when mass changes instead of velocity!

To discuss forces on systems and the Work Energy Theorem, we had to understand the notion of the CoM. The object model is founded on whether every point in the system we are considering could be approximated to move the same displacement as the CoM. In the case of a system of objects, this special point moves as though all of the mass of the system were focused there. Now, we can see that systems behave as if all forces on the system are exerted at this special point and we will finally expand our idea of the CoM by expanding our understanding of momentum!

Let’s take the example of Paula jumping off her skateboard. Paula is standing atop a stationary board and then jumps to the left off the board and the board rolls off to the right. Obviously, the skateboard flies off at a much faster speed than Paula. But why? Our first introduction to the Impulse-Momentum Theorem gets us pointed in the right direction, pun intended!

F(delta)t = (delta)p

This theorem will apply to both the skateboard and Paula and N3L tells us that the forces will be equal in magnitude and opposite in direction. Also, the time of each interaction will clearly be the same. Ergo, we are able to set the two impulses equal to each other and, by definition, their changes in momentum become equal!

m_(sb)v_(sb) = – m_(p)v_(p)

(1) The minus sign tells us that the the skateboard and Paula are moving in opposite directions. To push the board to right, she must move to the left.

(2) If we look at the magnitudes on both sides of the equation, the speed of the skateboard is equal to Paula’s speed multiplied by the ratio of the masses of Paula and the skateboard, m_(p)/m_(s). Since Paula is more massive than the skateboard, the board moves off at a much faster pace.

(3) If Paula pushes off the board with greater force, both Paula and the skateboard will fly off at faster speeds. But this force doesn’t appear in any equation and therefore, indicating, the ratio of the speed’s will be the same without regard to how hard Paula pushes off the board. An object’s mass multiplied by its velocity is called the objects linear momentum.

p = mv

We use a lowercase p for momentum, as per the Latin, petere.

The units for momentum, p, are simply the units of mass times the units for velocity or kg(m/s) or N(s). Unfortunately, there is no special name for this unit, so we just have to roll with it. Now we can easily say that the formula for Paula and the Skateboard is as below.

p_(paula) = -p_(skateboard)

A block of mass M is moving such that its kinetic energy has a value of K₁ and its momentum, p₁. The speed of the block is increased such that its kinetic energy K₂ = 3K₁. In terms of p1, what is the new magnitude of the momentum of the block? Use your proportionate reasoning to find the ratio of the velocities! (p₂ = sqrt(3)p₁)

Pro Tip

The magnitude of momentum for an object is mass times the magnitude of the object’s velocity. At any instant, the magnitude of the object’s velocity is its speed. Since Kinetic Energy depends on speed, we can use this relationship to find the Kinetic Energy in terms of the magnitude of momentum!

K = 1/2 mv² = 1/2 m(p/m)² = p²/2m

Let’s jump back to the equations that say that the momentum object acquires during the push off the skateboard is equal to the F_net multiplied by the time over which the force is exerted. N3L says that Paula and the skateboard exert forces of equal magnitude on each other and that when one is present, the other has to be present as well. Therefore, both forces are exerted for the same time and Paula and the skateboard have equal magnitudes of momentum, p.

The kinetic energy each object acquires is equal to the work done on that object during the push off and work is equal to force times displacement of the object in the direction of the force. So, we treat Paula and the skateboard as individual objects, the forces Paula and the skateboard exert on each other are external forces and can therefore do work. The skateboard moves faster than Paula during the push-off, so it travels a greater distance than Paula and so has more work done on it by the same amount of force! Therefore the skateboard ends up with more kinetic energy! The momenta of the two objects is the same so therefore the lighter of the two must have the greater speed. The kinetic energy of the object is 1/2 mv² = 1/2 pv, so for the same mass, the object with the greater speed has the greater magnitude of momentum.

The total momentum of a system is always conserved and it is constant for closed and isolated systems. In the case of the skateboard, the total momentum is equal to the vector sum of the momentum of Paula and the skateboard. During the interaction of Paula pushing off the skateboard, nothing else was interacting with Paula or the skateboard with anywhere near a significant force. F_g and F_N balance and F_f can be considered negligible in comparison, therefore the system of Paula and the skateboard can be considered closed and isolated. The total momentum of the system just before and just after the interaction is constant. Just like the total energy of the system is always conserved but will be constant under certain circumstances.

Forces

(1) Forces exerted by other members of the system are considered internal forces.

(2) Forces exerted by objects outside the system are considered external forces.

Can we rewrite N2L in terms of momentum? Of course we can!

F = ma —> a = (delta)v/(delta)t —> F = m(delta)v/(delta)t —> F = (delta)p/t —> F (delta)t = (delta)p

For two billiard balls, the internal forces are the ones that the ball exerts on another when they collide. External forces are the F_N and F_f exerted by the table and F_g exerted by Earth and the force you exert on the cue. Internal forces can either pull the components of a system together or push them apart and can allow energy to convert from one form to another, while keeping the total energy constant. In order for two true objects to interact, one of them would already have to be moving, since objects only have kinetic energy! The internal forces that one object in our system exerts on another and the details of these forces is irrelevant. We can calculate the effects of a collision with knowing the details.

Impulse Momentum Theorem

sum F_(ext) (delta)t = (delta)p

This is truly remarkable! Only the external forces exerted on a system can affect the system’s total momentum. The internal forces of one object on another allow momentum to be transferred between objects (the cue ball strikes the eight-ball at rest and sends the eight-ball flying) but they don’t affect the total momentum. This parallels our results that internal forces can allow energy to be converted or transferred between objects in a system but cannot change the total energy of the system. This is the most general expression of the conservation of momentum. Just like the Work Energy Theorem told us that any change in energy of the system must equal the transfer of energy into or out of the system, any change in momentum of a system must equal the transfer of momentum, F_(ext)(delta)t. Just as F_(ext)(delta)x is called Work, F_(ext)(delta)t is called Impulse.

The Law of Conservation of Momentum for the special case of no external forces can also be used in all collisions and proves to be so remarkably useful!

p_i = p_f

Tennis anyone?

One morning, I was running on the treadmill whilst watching the Australian Open on the telly! I looked up to see Sasha Zverev clock a serve at 217 km/h. Now, we are all taught to follow through when swinging anything, including a tennis serve. Let’s analyze this phenomena in terms of the Impulse Momentum Theorem! (See Google Classroom)

If we treat two objects of systems A and B as a single system, the interaction between A and B is most easily described as equal and opposite internal forces. Recall external forces can do work, changing the total energy but the internal forces can only rearrange energy! The Impulse Momentum Theorem reckons that to change the momentum of an object, a certain amount of impulse is required. The left side of the equations, F (delta)t, is actually referred to as the “impulse” that is exerted on the system. This can be a result of either a large external force over a short time or a smaller external force over a longer time. The Impulse Momentum Theorem is very much a statement of conservation as any change in momentum of a system has to be due to a transfer of momentum into or out of the system by an external force. If momentum changes due to a collision, the net external force on an object is predominantly due to the other object with which it is colliding. Any other forces (friction) are weak in comparison.

In F (delta)t = (delta)p, the (delta)t is representative of the contact time or the amount of time that the colliding objects interact. If a car comes to stop suddenly, the momentum of its occupants changes from a large initial value to a final value of zero. So the momentum change is very large and, in turn, the Impulse (F (delta)t) is also large. If there is no air bag, the passengers have a hard collision with the dash or steering wheel and come to a sudden stop. In this case, (delta)t is very short and the force of the collision is again very large. Therefore, the structure of the car exerts a tremendous force on the passengers. On the other hand, if the car is equipped with air bags, the air bags compress much more than the dash or the steering wheel as the driver strikes it so the pair are in contact for a greater amount of time. The force of the collision, represented as the air bag on the driver, is greatly reduced and injury can be avoided!! By the same accord, when you jump down from an elevation, or ski a set of moguls, you bend your knees to “absorb” the force. Flexing your legs during the collision between your feet and the ground maximizes the contact time that it takes your CoM to come to rest!!

The Conservation of Momentum!!

If the net external forces on an object or system of objects is zero, the total momentum of the system is constant. That is, the total momentum of the system at the beginning of a time interval is the same as the total momentum of the given system at the end of the time interval. It is so important to remember that the “Impulse”, or left side of the Impulse-Momentum Theorem, is equal to the change in momentum, not the total momentum. So, F(delta)t = (delta)p shows that of the net external force is equal to zero, then the change in momentum is equal to zero. This is what is meant by momentum is “conserved”. Momentum is not constant (it changes) if there is a net external force on the system. Although it is not generally constant when a system is not closed and isolated, momentum is always conserved, i.e., you can always account for the changes.

p_total(i) = p_total(f)

m₁v_1(i) + m₂v_2(i) = m₁v_1(f) + m₂v_2(f)

Collisions

The law of conservation of momentum for closed, isolated systems states that, if the internal forces during an interaction are much greater in magnitude than external forces, the total momentum of the interacting objects has the same value just before and just after the interaction.

When hockey players collide on the ice, the internal forces are much greater than the friction forces the ice exerts on the players. Similarly, the force between a tennis ball and a tennis racket distort the balls configuration so the external forces of gravity and Sasha’s hand on the racket are feeble in comparison. The total momentum of a system of colliding objects just after the collision is the same as before. The definition of a collision, in physics, is any brief interaction where internal forces dominate, like jumping off a skateboard. Similarly, when cars collide, the net vertical forces are zero and we can ignore the frictional forces as they pale in comparison to the enormous internal forces.

In a collision, momentum is constant only during the collision! Once the collision is over, the large internal forces are no longer exerted. If cars collide, the total momentum just before is equal to the total momentum just after the collision. Once this instant is over, the friction of the racetrack and other external forces, cause the car to come to a stop, transferring all the momentum to the Earth. With the mass of the Earth being so great these effects are unobservable!

The law of conservation of momentum is a vector equation. If the momentum vectors are equal before and after the collision, their (x) and (y) components are also equal in magnitude!

The Format of Vintage Collision Problems!!

An object of mass ma = 9.0 kg slides to the right at 10.0 m/s on a surface of negligible friction. It collides with a second object that moves left at 8.0 m/s before the collision but after the collision moves to the right at 12.25 m/s. What the m₂ if m₁ is still moving right at 3.25 m/s after the collision? (m₁ = 3.0 kg, also see hockey collision example)

A stationary object explodes into two pieces. Piece A has a mass m_A and piece B has a mass m_B = 2m_A

(a) Immediately after the explosion, which piece has the larger magnitude of momentum?

(b) Which piece is moving faster?

(c) Which piece has more kinetic energy?

Answer: (a) We know the momenta will be the same because momentum is zero before the explosion. Therefore the momentum after the explosion must be equal and opposite. In (b) the object with the smaller mass must move faster. And, finally, in (c) the smaller object must also have the kinetic energy because speed is squared in the kinetic energy formula.

A 2.0 kg object is moving east at 4.0 m/s when it collides with a 6.0 kg object at rest. After the collision, the larger object moves east at 2.0 m/s. What is the final magnitude and direction of the smaller object after the collision, assuming external forces are negligible? (v = 2 m/s west)

Two ice skaters, Twyla and Ani, face each other while stationary and push against each other’s hands. Twyla’s mass is two times larger that Ani’s. Calculate their relative speeds after they lose contact assuming friction forces are negligible.

mv_total(i) = m_tv_t + m_av_a

0 = m_tv_t + m_av_a

m_tv_t = m_av_a

2v_t = 1v_a

2v_t = v_a

In this case, we are looking for speeds only but we obviously know that the directions will be opposite!

Inelastic Collisions Dissipate Mechanical Energy

Are both mechanical energy and momentum constant throughout collisions? Sometimes!

We learned that mechanical energy of a system is constant only if conservative forces (external) do work and there are only conservative interactions inside the system. An example may be a system made up of two small systems that behave like ideal springs. They collide and compress, converting kinetic energy to spring potential and then relax, converting spring potential back to kinetic. A collision of this kind, in which internal forces between colliding objects are conservative, is an elastic collision. In an elastic collision, both total mechanical energy AND total momentum are constant. During the interaction, all the kinetic energy not required to conserve momentum is converted to potential energy. After, all the potential energy is restored to kinetic. If the initial momentum of a system is not zero, then not all the kinetic energy can be stored because enough kinetic energy must remain to keep momentum constant.

Elastic collisions are happening all around us. When oxygen and nitrogen in the air collide, the collisions are always elastic. The molecules compress slightly and undergo a reversible change in configuration, storing energy as potential. As we know, in the observable universe, the collision between billiard balls on a pool table is very nearly elastic!

Something very different happens when autos collide. The bodies of the autos undergo irreversible changes in configuration. These nonconservative forces, like heat and sound, dissipate mechanical energy. In a collision in racing, this is by design! By converting mechanical energy to internal energy as it deforms, the structure of the race car prevents all that energy from being used to do potentially harmful work on the driver. A collision in which mechanical energy is not constant is called an inelastic collision. (Even simply pushing off the skateboard, you convert internal energy to kinetic.)

In inelastic collisions, during the interaction, all the initial kinetic energy not required to conserve momentum is converted to internal forms but not all is converted to potential energy. Some is directly converted into internal energy. After the interaction, any any energy converted to potential energy is restored to kinetic but this internal energy is considered dissipated.

If we know the masses of colliding objects and their velocities before and after the collision, its straight forward to determine whether the collision is elastic inelastic. If the kinetic energy has the same value before and after the collision, the collision is elastic. If there is less total kinetic energy after the collision, the collision is inelastic and is the more common result.

A 20.0 g bee bee is fired into a 1.50 kg wood block initially at rest. The speed of the bee bee is 200 m/s just before it strikes the wooden block.

(a) Find the speed fo the block – bee bee system just after the collsion.

(b) Calculate the amount of mechanical energy dissipated in the collision.

(a) m_bv_b = m_wv_w = (m_b + m_w)v_f —> .020(200) + 0 = 1.52v_f —> 4 = 1.52v_f —> v_f = 2.63 m/s

(b) K_(before) = 1/2mv² = 1/2(.020)(200)² = 400 J; K_(after) = 1/2mv² = 1/2(1.52)(2.63)² = 5.23 J

400 J -5.23 J = 395 J

Completely Inelastic Collisions

The type of collision where the most mechanical energy is dissipated is a completely inelastic collision in which two objects stick together after they collide. A collision between two cars is completely inelastic if the two cars lock together and do not separate. An odd example is the fertilized egg!

In a completely inelastic collision, none of the initial kinetic energy is stored as potential energy. The amount of kinetic energy needed to keep the momentum constant remains, but all the rest of the initial kinetic energy is converted to internal energy. When an object of mass m_a and velocity v_a(i) undergoes a completely inelastic collision with a second object of mass m_b and velocity v_b(i), we can regard what remain after the collision as a single object of mass (m_a + m_b). Momentum conservation can then give us an equation for the velocity v_f of the combined object!

m_av_a(i) + m_bv_b(i) = (m_a + m_b)v_f

Because all of the mass of the systems travels off together after a completely inelastic collision, the final velocity in such a collision is velocity of the CoM of the system!

A bit about Newton’s Cradle

The cascade of events, collisions, back and forth continue for quite some time!

(a) Explain why the system behaves in such a manner? For each collision, momentum is conserved and mechanical energy is constant. No energy is transferred out of the system and the rigid metal balls do not acquire internal energy.

(b) Consider Newton’s Cradle consisting of only two balls. One ball is raised and released. Justify the claim that the height after the collision of the ball initially at rest can be no greater than the height from which the first ball was released. No energy is being transferred into the system, so total mechanical energy cannot increase. At the beginning of the swing all the energy is Ug, so at the end, the swing cannot be greater!

(c) Given this conclusion, justify the claim that the velocity of the ball colliding with the ball initially at rest cannot be negative after the collision. If momentum after the collision of the raised ball were negative, then the momentum of the ball initially at rest would have to increase by this amount. This would violate the conservation of mechanical energy.

(d) Given the conclusion in (b) and (c), justify the claim that all of the momentum from the raised ball must be transferred to the ball initially at rest. The ball that is initially raised cannot have a velocity after the collision. Because momentum is conserved, the momentum of the system after the collision must therefore be the momentum of the ball that was initially at rest!

(e) Predict what happens when two balls are raised and released in a Newton’s Cradle with at least three balls. You can raise and release two balls simultaneously and both the momentum and mechanical energy of the system are constant. When two balls are initially raised, two balls must be raised finally, otherwise mechanical energy would not be conserved.

There are three special cases of Completely Inelastic Collisions. Let’s look at the general scenario where Object A moves toward Object B, initially at rest, and they stick together.

m_av_a(i) + m_bv_b(i) = (m_a + m_b)v_f

v_f = m_av_a(i)/(m_a + m_b)

(1) When object A moving has a much great mass than object B resting! Whenever one quantity in an expression is significantly larger than another, it’s a good approximation to ignore the smaller quantity. So we can replace the denominator in our equation with just m_a.

v_f = m_av_a(i)/m_a ~ v_a(i)

(2) The two objects have the same mass, m_a = m_b.

v_f = v_a(i)/2

This states that when a moving object collides and sticks to a stationary object of the same mass, the combined system moves at half the initial speed.

(3) The moving object has much less mass that the object at rest. In this case, the mass of object A is much less than the mass of object B and the numerator is much less than the denominator. Object B hardly moves when struck by object A.

m_a/(m_a + m_b) ~ 0

The Center of Mass

If a system is rotating, like a gymnast flipping through the air or a barrel rolling down a hill or something changing shape, the object model no longer holds. So the quantity “a” in a = F/m refers to the acceleration of a certain point, the Center of Mass (CoM) of the system. The CoM moves as though all of the object’s mass is squeezed into a tiny blob at that point and all the external forces are exerted on the blob.

The position of a system’s CoM is kind of an average that takes into account the masses and positions of all the objects that make up that system. In a weighted average, each value affects the results to a greater or lesser extent depending on where it appears in that set. In this case, weight doesn’t refer to F_g but refers to the amount that each value contributes to the result. If you hold a lunch tray with only one hand, you will place your hand under the CoM and that position will be closer to the heaviest food or drink!

x_cm = (m₁(x₁))/m_T + (m₂(x₂))/m_T + … (m_n(x_n))/m_T

The more massive an object, the greater the ratio m_n/M_T and the greater the importance of that single object’s position. On the exam, you will not be expected to calculate the CoM of a collection of masses! But you will be expected to understand the concept of CoM, and you should be able to locate the approximate position qualitatively!

Two students are sitting on a seesaw, one at each end, that is pivoted about a point at its center. Claire, on the left end, has a mass of 60 kg and Ben, on the right end, has a mass of 80 kg. If the seesaw has a length of 2.40 m, how far from Ben is the CoM of the system of the students. That is, where do you put the pivot so they balance? We will designate x = 0 where Claire sits.

x_cm = (60(0))/140 + (80(2.4))/140 = 1.37 m from Claire or 1.03 m from Ben!

The CoM is the geometrical center of a symmetrical system.

Let’s justify the use of the idea that the CoM moves as though all of the mass of the system were concentrated there. The total momentum of a system equals the vector sum of the momentum of all the objects in the system and also equals the total mass of the system multiplied by the velocity of the CoM. In other words, the total momentum of a system of objects of total mass, M_T, is the same as if all of the objects were squeezed into a single blob moving at the velocity of the CoM. This means that the total linear momentum of a football doesn’t depend on how fast the football is rotating but only on the velocity of the CoM.

(sum)F_(ext) = m_Ta_cm

The CoM of a system of objects moves exactly as if all the mass were concentrated there and all of the external forces on the system were exerted there. Only external forces affect the motion of the CoM. Forces on various points of the system can affect how those parts move relative to each other but have no effect on the motion of the CoM. This is the case of the complicated movements of a ballet dancer where her CoM follows a simple arc!

Frank launches a canon shell over level ground at a target 200 m away. Frank is perfectly aimed to hit the target, and air resistance can be neglected. At the highest point of the shell’s trajectory, the shell explodes into two identical halves, both of which hit the ground at the same time. One half falls vertically downward from the point of explosion. Where does the other half land?

(a) on the target (b) 50 m short of the target (c) 50 m beyond the target (d) 100 beyond the target (e) 150 m beyond the target

(d) If the shell did not explode, the CoM would hit the target. The force that blows the shell apart is internal to the shell, so it does not affect the motion of the CoM. This means that the CoM must still hit the target. This explosion happens when the shell is exactly halfway along its trajectory, so the half that fall’s vertically lands 100 m short of the target (half the horizontal). Since the CoM of a system of two equal masses is halfway between the masses, the other mass must land 100 m on the other side of the target!!

Grinnell stands on one end of a long wooden plank with a length L that rests on an icy surface. The ice exerts a negligible friction force on the plank. The plank and Grinnell have the same mass, m. Grinnell’s mom stands to the side of the plank, marking the CoM.

(a) Describe the CoM of the Grinnell/Plank system relative to the end of the plank where Grinnell is standing. The CoM of the system is L/4 from Grinnell. It is so important to remember that the CoM of a uniform plank or rod is its geometrical center. Therefore the point halfway between the end, where Grinnell stands, and the geometrical center of the plank is 1/4 the distance of the plank from where Grinnell is standing. We could also use the formula we discovered but like we said, we only need to know the intuition regarding the CoM!

x_cm = m_g(x_g)/m_T + m_p(x_p)/m_T —> the plank and Grinnell have the same mass, m —> x_cm = 0 + m(1/2)L/2m = L/4

(b) Grinnell runs to the other end of the plank with a speed v_o while Mom maintains her position on the ice. Predict how the CoM of the system moves while the child runs. The CoM doesn’t change because there are no external forces!

(c) Predict the location of the middle of the plank and the location of Grinnell relative to Mom when Grinnell has reached the other end of the plank. Relative to Mom, Grinnell has moved to a position that is L/2 on the other side from where Grinnell started, whereas the center of the plank has also moved L/2 in the opposite direction.

(d) From Grinnell’s POV, she has been displaced by a distance L. Explain why her displacement from Mom’s POV is NOT equal to L. From Grinnell’s POV, she has moved a distance L (in the coordinate frame of the plank, the displacement is L.) But in the coordinate frame fixed on Mom, she has only moved L/2 because the plank has moved a distance L/2 opposite to the direction of Grinnell’s motion!

A great example of the manipulation of the CoM is the Fosbury Flop that redefined the high jump and thereby the Olympics forever! Have a look at this great video!

Bowery, Kimchi and the CoM!!

Bowery and her dog, Kimchi, sit at opposite ends of a 3 m long boat on which there are negligible horizontal forces exerted on the flat bottom boat by the surface of the still pond. The boat is pointed toward the shore with Kimchi in the bow at the front of the boat and Bowery in the stern at the rear of the boat. The bow is 6 m from the shore. Bowery has a mass of 52 kg and Kimchi has a mass of 20 kg. The mass of the boat is 135 kg.

(a) Discuss the location of the CoM relative to the geometrical center of the boat. In the reference frame with the origin at the center of the boat, Kimchi is 1.5 m from the center (m = 20 kg) and Bowery is 1.5 m from the center of the boat in the opposite direction (m = 52 kg). So the CoM must be closer to Bowery than Kimchi. Therefore, the CoM is between the center of the boat and Bowery, but closer to the center of the boat!

(Quantitatively x_cm = (m_bx_b + m_kx_k + m_vx_v)/m_T = ((52(0)) + 20(3) + 135(1.5))/207 = 1.26 m from Bowery and therefore, the CoM is 0.24 m from the center of the boat.)

Kimchi suddenly walks rapidly rapidly towards Bowery!

(b) Justify the claim that, in the reference frame that moves with the boat, momentum is not constant. Initially, the total momentum is zero. In the reference frame fixed on the boat, the boat does not move and Bowery remains seated. However, Kimchi is moving in the reference frame of the boat and the total momentum becomes the momentum of Kimchi and is no longer zero!

(c) Justify the claim that in the reference frame fixed on the shore, momentum is constant. Momentum of a the Bowery-Kimchi-boat system is constant in the absence of a net external force. Since there are negligible forces in the plane of the surface of the water exerted on the boat by the water, the momentum of the Bowery-Kimchi-boat system must be constant. From this reference frame, as Kimchi moves backwards, Bowery moves forwards and the CoM does not move!

(d) Predict the direction that the boat moves as Kimchi is moving. Kimchi moves in a direction away from the shore. The CoM of the system must remain constant in the absence of an external force, so the boat and Bowery must move toward the shore. In the reference frame fixed on the shore, Kimchi moves in a direction away from the shore. The CoM of the system must remain constant in the absence of an external force, so the boat must move toward the shore, although the CoM of the system does not move.

Por lo General

In a system of masses, the ratio of the distances from the CoM is inversely proportional to the ratio of the masses.