College Physics for AP® Courses

# 8.3Conservation of Momentum

College Physics for AP® Courses8.3 Conservation of Momentum

### Learning Objectives

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

• Describe the law of conservation of linear momentum.
• Derive an expression for the conservation of momentum.
• Explain conservation of momentum with examples.
• Explain the law of conservation of momentum as it relates to atomic and subatomic particles.

The information presented in this section supports the following AP® learning objectives and science practices:

• 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2)
• 5.D.1.4 The student is able to design an experimental test of an application of the principle of the conservation of linear momentum, predict an outcome of the experiment using the principle, analyze data generated by that experiment whose uncertainties are expressed numerically, and evaluate the match between the prediction and the outcome. (S.P. 4.2, 5.1, 5.3, 6.4)
• 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2)
• 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two-object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1)
• 5.D.3.1 The student is able to predict the velocity of the center of mass of a system when there is no interaction outside of the system but there is an interaction within the system (i.e., the student simply recognizes that interactions within a system do not affect the center of mass motion of the system and is able to determine that there is no external force). (S.P. 6.4)

Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and Force, where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved?

The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on him that causes him to bounce backward. However, the Earth also recoils —conserving momentum—because of the force applied to it through the goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any practical sense, but it is real nevertheless.

Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—for example, one car bumping into another, as shown in Figure 8.6. Both cars are coasting in the same direction when the lead car (labeled $m2)m2) size 12{m rSub { size 8{2} } \) } {}$ is bumped by the trailing car (labeled $m1).m1). size 12{m rSub { size 8{1} } \) "." } {}$ The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant.

Figure 8.6 A car of mass $m1m1 size 12{m rSub { size 8{1} } } {}$ moving with a velocity of $v1v1 size 12{v rSub { size 8{1} } } {}$ bumps into another car of mass $m2m2 size 12{m rSub { size 8{2} } } {}$ and velocity $v2v2 size 12{v rSub { size 8{2} } } {}$ that it is following. As a result, the first car slows down to a velocity of $v′1v′1 size 12{ { {v}} sup { ' } rSub { size 8{1} } } {}$ and the second speeds up to a velocity of $v′2v′2 size 12{ { {v}} sup { ' } rSub { size 8{2} } } {}$. The momentum of each car is changed, but the total momentum $ptotptot size 12{p rSub { size 8{"tot"} } } {}$ of the two cars is the same before and after the collision (if you assume friction is negligible).

Using the definition of impulse, the change in momentum of car 1 is given by

$Δ p 1 = F 1 Δ t , Δ p 1 = F 1 Δ t , size 12{Δp rSub { size 8{1} } =F rSub { size 8{1} } Δt} {}$
8.32

where $F1F1 size 12{F"" lSub { size 8{1} } } {}$ is the force on car 1 due to car 2, and $ΔtΔt size 12{Δt} {}$ is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects travelling near the speed of light, without affecting the result that momentum is conserved.

Similarly, the change in momentum of car 2 is

$Δ p 2 = F 2 Δ t, Δ p 2 = F 2 Δ t, size 12{Δp rSub { size 8{1} } =F rSub { size 8{1} } Δt} {}$
8.33

where $F2F2$ is the force on car 2 due to car 1, and we assume the duration of the collision $ΔtΔt size 12{?t} {}$ is the same for both cars. We know from Newton’s third law that $F2=–F1F2=–F1 size 12{F rSub { size 8{2} } = - F rSub { size 8{1} } } {}$, and so

$Δp2=−F1Δt=−Δp1.Δp2=−F1Δt=−Δp1. size 12{Δp rSub { size 8{2} } = - F rSub { size 8{1} } Δt= - Δp rSub { size 8{1} } } {}$
8.34

Thus, the changes in momentum are equal and opposite, and

$Δp1+Δp2=0.Δp1+Δp2=0. size 12{Δp rSub { size 8{1} } +Δp rSub { size 8{2} } =0} {}$
8.35

Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is,

$p1+p2= constant,p1+p2= constant, size 12{p rSub { size 8{1} } +p rSub { size 8{2} } =" constant"} {}$
8.36
$p1+p2=p′1+p′2,p1+p2=p′1+p′2, size 12{p rSub { size 8{1} } +p rSub { size 8{2} } = { {p}} sup { ' } rSub { size 8{1} } + { {p}} sup { ' } rSub { size 8{2} } } {}$
8.37

where $p′1p′1 size 12{ { {p}} sup { ' } rSub { size 8{1} } } {}$ and $p′2p′2 size 12{ { {p}} sup { ' } rSub { size 8{2} } } {}$ are the momenta of cars 1 and 2 after the collision. (We often use primes to denote the final state.)

This result—that momentum is conserved—has validity far beyond the preceding one-dimensional case. It can be similarly shown that total momentum is conserved for any isolated system, with any number of objects in it. In equation form, the conservation of momentum principle for an isolated system is written

$p tot = constant , p tot = constant , size 12{p rSub { size 8{"tot"} } ="constant"} {}$
8.38

or

$ptot=p′tot,ptot=p′tot, size 12{p rSub { size 8{"tot"} } =p' rSub { size 8{"tot"} } } {}$
8.39

where $ptotptot size 12{p rSub { size 8{"tot"} } } {}$ is the total momentum (the sum of the momenta of the individual objects in the system) and $p′totp′tot size 12{ ital "p'" rSub { size 8{"tot"} } } {}$ is the total momentum some time later. (The total momentum can be shown to be the momentum of the center of mass of the system.) An isolated system is defined to be one for which the net external force is zero $Fnet=0.Fnet=0. size 12{ left (F rSub { size 8{ ital "net"} } =0 right ) "." } {}$

### Conservation of Momentum Principle

$p tot = constant p tot = p′ tot ( isolated system ) p tot = constant p tot = p′ tot ( isolated system )$
8.40

### Isolated System

An isolated system is defined to be one for which the net external force is zero $Fnet=0.Fnet=0. size 12{ left (F rSub { size 8{ ital "net"} } =0 right ) "." } {}$

### Making Connections: Cart Collisions

Consider two air carts with equal mass (m) on a linear track. The first cart moves with a speed v towards the second cart, which is initially at rest. We will take the initial direction of motion of the first cart as the positive direction.

The momentum of the system will be conserved in the collision. If the collision is elastic, then the first cart will stop after the collision. Conservation of momentum therefore tells us that the second cart will have a final velocity v after the collision in the same direction as the initial velocity of the first cart.

The kinetic energy of the system will be conserved since the masses are equal and the final velocity of cart 2 is equal to the initial velocity of cart 1. What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case?

Consider the center of mass of this system as the frame of reference. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed $v2v2$. After the collision, the center of mass continues moving in the same direction, away from (now stationary) cart 1 at a speed $v2v2$. How would a graph of center-of-mass velocity vs. time compare to a graph of momentum vs. time?

Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. Again, they have an elastic collision, so after the collision, they exchange velocities (each cart moving in the opposite direction of its initial motion with the same speed). As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time?

Let us return to the example where the first cart is moving with a speed v toward the second cart, initially at rest. Suppose the second cart has some putty on one end so that, when the collision occurs, the two carts stick together in an inelastic collision. In this case, conservation of momentum tells us that the final velocity of the two-cart system will be half the initial velocity of the first cart, in the same direction as the first cart’s initial motion. Kinetic energy will not be conserved in this case, however. Compared to the moving cart before the collision, the overall moving mass after the collision is doubled, but the velocity is halved.

The initial kinetic energy of the system is:

8.41

The final kinetic energy of the two carts (2m) moving together (at speed v/2) is:

$k f = 1 2 (2m) ( v 2 ) 2 = 1 4 m v 2 k f = 1 2 (2m) ( v 2 ) 2 = 1 4 m v 2$
8.42

What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case?

Consider the center of mass of this system. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed $v2v2$. After the collision, the two carts move together at a speed $v2v2$. How would a graph of center-of-mass velocity vs. time compare to a graph of momentum vs. time?

Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. They have putty on the end of each cart so that they stick together after the collision. As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time?

Perhaps an easier way to see that momentum is conserved for an isolated system is to consider Newton’s second law in terms of momentum, $Fnet=ΔptotΔt Fnet=ΔptotΔt$. For an isolated system, $Fnet=0Fnet=0$; thus, $Δptot=0Δptot=0 size 12{?p rSub { size 8{"tot"} } =0} {}$, and $ptotptot$ is constant.

We have noted that the three length dimensions in nature—$xx size 12{x} {}$, $yy size 12{y} {}$, and $zz size 12{z} {}$—are independent, and it is interesting to note that momentum can be conserved in different ways along each dimension. For example, during projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero and momentum is unchanged. But along the vertical direction, the net vertical force is not zero and the momentum of the projectile is not conserved. (See Figure 8.7.) However, if the momentum of the projectile-Earth system is considered in the vertical direction, we find that the total momentum is conserved.

Figure 8.7 The horizontal component of a projectile’s momentum is conserved if air resistance is negligible, even in this case where a space probe separates. The forces causing the separation are internal to the system, so that the net external horizontal force $Fx–netFx–net$ is still zero. The vertical component of the momentum is not conserved, because the net vertical force $Fy–netFy–net$ is not zero. In the vertical direction, the space probe-Earth system needs to be considered and we find that the total momentum is conserved. The center of mass of the space probe takes the same path it would if the separation did not occur.

The conservation of momentum principle can be applied to systems as different as a comet striking Earth and a gas containing huge numbers of atoms and molecules. Conservation of momentum is violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not.

### Making Connections: Take-Home Investigation—Drop of Tennis Ball and a Basketball

Hold a tennis ball side by side and in contact with a basketball. Drop the balls together. (Be careful!) What happens? Explain your observations. Now hold the tennis ball above and in contact with the basketball. What happened? Explain your observations. What do you think will happen if the basketball ball is held above and in contact with the tennis ball?

### Making Connections: Take-Home Investigation—Two Tennis Balls in a Ballistic Trajectory

Tie two tennis balls together with a string about a foot long. Hold one ball and let the other hang down and throw it in a ballistic trajectory. Explain your observations. Now mark the center of the string with bright ink or attach a brightly colored sticker to it and throw again. What happened? Explain your observations.

Some aquatic animals such as jellyfish move around based on the principles of conservation of momentum. A jellyfish fills its umbrella section with water and then pushes the water out resulting in motion in the opposite direction to that of the jet of water. Squids propel themselves in a similar manner but, in contrast with jellyfish, are able to control the direction in which they move by aiming their nozzle forward or backward. Typical squids can move at speeds of 8 to 12 km/h.

The ballistocardiograph (BCG) was a diagnostic tool used in the second half of the 20th century to study the strength of the heart. About once a second, your heart beats, forcing blood into the aorta. A force in the opposite direction is exerted on the rest of your body (recall Newton’s third law). A ballistocardiograph is a device that can measure this reaction force. This measurement is done by using a sensor (resting on the person) or by using a moving table suspended from the ceiling. This technique can gather information on the strength of the heart beat and the volume of blood passing from the heart. However, the electrocardiogram (ECG or EKG) and the echocardiogram (cardiac ECHO or ECHO; a technique that uses ultrasound to see an image of the heart) are more widely used in the practice of cardiology.

### Applying Science Practices: Verifying the Conservation of Linear Momentum

Design an experiment to verify the conservation of linear momentum in a one-dimensional collision, both elastic and inelastic. For simplicity, try to ensure that friction is minimized so that it has a negligible effect on your experiment. As you consider your experiment, consider the following questions:

• Predict how the final momentum of the system will compare to the initial momentum of the system that you will measure. Justify your prediction.
• How will you measure the momentum of each object?
• Should you have two objects in motion or one object bouncing off a rigid surface?
• Should you verify the relationship mathematically or graphically?
• How will you estimate the uncertainty of your measurements? How will you express this uncertainty in your data?

When you have completed each experiment, compare the outcome to your prediction about the initial and final momentum of the system and evaluate your results.

### Making Connections: Conservation of Momentum and Collision

Conservation of momentum is quite useful in describing collisions. Momentum is crucial to our understanding of atomic and subatomic particles because much of what we know about these particles comes from collision experiments.

### Subatomic Collisions and Momentum

The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles. Giant machines hurl subatomic particles at one another, and researchers evaluate the results by assuming conservation of momentum (among other things).

On the small scale, we find that particles and their properties are invisible to the naked eye but can be measured with our instruments, and models of these subatomic particles can be constructed to describe the results. Momentum is found to be a property of all subatomic particles including massless particles such as photons that compose light. Momentum being a property of particles hints that momentum may have an identity beyond the description of an object’s mass multiplied by the object’s velocity. Indeed, momentum relates to wave properties and plays a fundamental role in what measurements are taken and how we take these measurements. Furthermore, we find that the conservation of momentum principle is valid when considering systems of particles. We use this principle to analyze the masses and other properties of previously undetected particles, such as the nucleus of an atom and the existence of quarks that make up particles of nuclei. Figure 8.8 below illustrates how a particle scattering backward from another implies that its target is massive and dense. Experiments seeking evidence that quarks make up protons (one type of particle that makes up nuclei) scattered high-energy electrons off of protons (nuclei of hydrogen atoms). Electrons occasionally scattered straight backward in a manner that implied a very small and very dense particle makes up the proton—this observation is considered nearly direct evidence of quarks. The analysis was based partly on the same conservation of momentum principle that works so well on the large scale.

Figure 8.8 A subatomic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed to occasionally scatter straight backward from a proton.
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