#set document(title: "8.1 Linear Momentum and Force", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 8.1#h(0.6em)Linear Momentum and Force === Learning Objectives By the end of this section, you will be able to: - Define linear momentum. - Explain the relationship between momentum and force. - State Newton’s second law of motion in terms of momentum. - Calculate momentum given mass and velocity. === Linear Momentum The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. #strong[Linear momentum] is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as #math.equation(block: true, alt: "p equals m v .")[$p = m "v" .$] Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum #math.equation(block: false, alt: "p")[$p$] is a vector having the same direction as the velocity #math.equation(block: false, alt: "v")[$"v"$]. The SI unit for momentum is #math.equation(block: false, alt: "kg times m/s")[$"kg" · "m/s"$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Linear Momentum] Linear momentum is defined as the product of a system’s mass multiplied by its velocity: #math.equation(block: true, alt: "p equals m v .")[$p = m "v" .$] ] #examplebox("Example 1")[Calculating Momentum: A Football Player and a Football][ (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Strategy No information is given regarding direction, and so we can calculate only the magnitude of the momentum, #math.equation(block: false, alt: "p")[$p$]. (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes #math.equation(block: true, alt: "p equals mv")[$p = "mv"$] when only magnitudes are considered. Solution for (a) To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation. #math.equation(block: true, alt: "p sub player equals 110 kg 8 . 00 m/s equals 880 kg times m/s")[$p_("player") = "110 kg" 8 "." "00 m/s" = "880 kg" · "m/s"$] Solution for (b) To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation. #math.equation(block: true, alt: "p sub ball equals 0.410 kg 25.0 m/s equals 10.3 kg times m/s")[$p_("ball") = "0.410 kg" "25.0 m/s" = "10.3 kg" · "m/s"$] The ratio of the player’s momentum to that of the ball is #math.equation(block: true, alt: "the fraction p sub player over p sub ball equals the fraction 880 over 10 . 3 equals 85.4 .")[$frac(p_("player"), p_("ball")) = frac("880", "10" "." 3) = 85.4 .$] Discussion Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections. A more precise calculation of the ratio without first rounding the intermediate results gives #math.equation(block: false, alt: "the fraction open parenthesis 110 times 8.00 close parenthesis over open parenthesis 0.410 times 25.0 close parenthesis equals 85.9")[$frac(( 110 × 8.00 ), ( 0.410 × 25.0 )) = 85.9$] . ] === Momentum and Newton’s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his #strong[second law of motion] in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is #math.equation(block: true, alt: "F sub net equals the fraction Δ p over Δ t ,")[$F_("net") = frac(Δ p, Δ t) ,$] where #math.equation(block: false, alt: "F sub net")[$F_("net")$] is the net external force, #math.equation(block: false, alt: "Δ p")[$Δ p$] is the change in momentum, and #math.equation(block: false, alt: "Δ t")[$Δ t$] is the change in time. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Newton’s Second Law of Motion in Terms of Momentum] The net external force equals the change in momentum of a system divided by the time over which it changes. #math.equation(block: true, alt: "F sub net equals the fraction Δ p over Δ t")[$F_("net") = frac(Δ p, Δ t)$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Making Connections: Force and Momentum] Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics. ] This statement of Newton’s second law of motion includes the more familiar #math.equation(block: false, alt: "F sub net = m a")[$F_("net") " = " m a$] as a special case. We can derive this form as follows. First, note that the change in momentum #math.equation(block: false, alt: "Δ p")[$Δ p$] is given by #math.equation(block: true, alt: "Δ p equals Δ open parenthesis m v close parenthesis .")[$Δ p = Δ ( m v ) .$] If the mass of the system is constant, then #math.equation(block: true, alt: "Δ open parenthesis m v close parenthesis equals m Δ v .")[$Δ ( m v ) = m Δ v .$] So that for constant mass, Newton’s second law of motion becomes #math.equation(block: true, alt: "F sub net equals the fraction Δ p over Δ t equals the fraction m Δ v over Δ t .")[$F_("net") = frac(Δ p, Δ t) = frac(m Δ v, Δ t) "."$] Because #math.equation(block: false, alt: "the fraction Δ v over Δ t equals a")[$frac(Δ v, Δ t) = a$], we get the familiar equation #math.equation(block: true, alt: "F sub net = m a")[$F_("net") " = " m a$] #emph[when the mass of the system is constant]. Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail#strong[;] however, the relationship between momentum and force remains useful when mass is constant, such as in the following example. #examplebox("Example 2")[Calculating Force: Venus Williams’ Racquet][ During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? Strategy This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as #math.equation(block: true, alt: "F sub net equals the fraction Δ p over Δ t .")[$F_("net") = frac(Δ p, Δ t) "."$] As noted above, when mass is constant, the change in momentum is given by #math.equation(block: true, alt: "Δ p equals m Δ v equals m v sub f minus v sub i .")[$Δ p = m Δ v = m v_(f) − v_(i) .$] In this example, the velocity just after impact and the change in time are given; thus, once #math.equation(block: false, alt: "Δ p")[$Δ p$] is calculated, #math.equation(block: false, alt: "F sub net equals the fraction Δ p over Δ t")[$F_("net") = frac(Δ p, Δ t)$] can be used to find the force. Solution To determine the change in momentum, substitute the values for the initial and final velocities into the equation above. #math.equation(block: true, alt: "Δ p, equals, m v sub f minus v sub i; equals, 0.057 kg 58 m/s minus 0 m/s; equals, 3 .306 kg times m/s approximately equals 3.3 kg times m/s")[$Δ p & = & m v_(f) − v_(i) \ & = & "0.057 kg" "58 m/s" − 0 m/s \ & = & 3 ".306 kg" · "m/s" ≈ "3.3 kg" · "m/s"$] Now the magnitude of the net external force can determined by using #math.equation(block: false, alt: "F sub net equals the fraction Δ p over Δ t")[$F_("net") = frac(Δ p, Δ t)$]: #math.equation(block: true, alt: "F sub net, equals, the fraction Δ p over Δ t equals the fraction 3.306 kg times m/s over 5 . 0 times 10 to the power minus 3 s; equals, 661 N approximately equals 660 N,")[$F_("net") & = & frac(Δ p, Δ t) = frac("3.306 kg" ⋅ "m/s", 5 "." 0 × "10"^(− 3) #h(0.25em) s) \ & = & "661 N" ≈ "660 N,"$] where we have retained only two significant figures in the final step. Discussion This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using #math.equation(block: false, alt: "F sub net = ma")[$F_("net") #h(0.15em) "=" #h(0.15em) "ma"$], but one additional step would be required compared with the strategy used in this example. ] === Test Prep for AP Courses A boy standing on a frictionless ice rink is initially at rest. He throws a snowball in the +#emph[x]-direction, and it travels on a ballistic trajectory, hitting the ground some distance away. Which of the following is true about the boy while he is in the act of throwing the snowball? + He feels an upward force to compensate for the downward trajectory of the snowball. + He feels a backward force exerted by the snowball he is throwing. + He feels no net force. + He feels a forward force, the same force that propels the snowball. #solutionbox[ (b) ] A 150-g baseball is initially moving 80 mi/h in the –#emph[x]-direction. After colliding with a baseball bat for 20 ms, the baseball moves 80 mi/h in the +#emph[x]-direction. What is the magnitude and direction of the average force exerted by the bat on the baseball? === Section Summary - Linear momentum (#emph[momentum] for brevity) is defined as the product of a system’s mass multiplied by its velocity. - In symbols, linear momentum #math.equation(block: false, alt: "p")[$p$] is defined to be #math.equation(block: true, alt: "p equals m v ,")[$p = m v ,$] where #emph[#math.equation(block: false, alt: "m")[$m$]] is the mass of the system and #math.equation(block: false, alt: "v")[$v$] is its velocity. - The SI unit for momentum is #math.equation(block: false, alt: "kg times m/s")[$"kg" · "m/s"$]. - Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. - In symbols, Newton’s second law of motion is defined to be #math.equation(block: true, alt: "F sub net equals the fraction Δ p over Δ t ,")[$F_("net") = frac(Δ p, Δ t) ","$]#math.equation(block: false, alt: "F sub net")[$F_("net")$] is the net external force, #math.equation(block: false, alt: "Δ p")[$Δ p$] is the change in momentum, and #math.equation(block: false, alt: "Δ t")[$Δ t$] is the change time. === Conceptual Questions An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? Professional Application Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. How can a small force impart the same momentum to an object as a large force? === Problems & Exercises (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of #math.equation(block: false, alt: "7 . 50 m/s")[$7 "." "50 m/s"$]. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of #math.equation(block: false, alt: "600 m/s")[$"600 m/s"$]. (c) What is the momentum of the 90.0-kg hunter running at #math.equation(block: false, alt: "7 . 40 m/s")[$7 "." "40 m/s"$] after missing the elephant? #solutionbox[ (a) #math.equation(block: false, alt: "1.50 times 10 to the power 4 kg times m/s")[$"1.50" × "10"^(4) #h(0.25em) "kg" ⋅ "m/s"$] (b) 625 to 1 (c) #math.equation(block: false, alt: "6 . 66 times 10 squared kg times m/s")[$6 "." "66" × "10"^(2) #h(0.25em) "kg" ⋅ "m/s"$] ] (a) What is the mass of a large ship that has a momentum of #math.equation(block: false, alt: "1 . 60 times 10 to the power 9 kg times m/s")[$1 "." "60" × "10"^(9) #h(0.25em) "kg" · "m/s"$], when the ship is moving at a speed of #math.equation(block: false, alt: "48.0 km/h?")[$"48.0 km/h?"$] (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of #math.equation(block: false, alt: "1200 m/s")[$"1200 m/s"$]. (a) At what speed would a #math.equation(block: false, alt: "2 . 00 times 10 to the power 4 -kg")[$2 "." "00" × "10"^(4) "-kg"$] airplane have to fly to have a momentum of #math.equation(block: false, alt: "1 . 60 times 10 to the power 9 kg times m/s")[$1 "." "60" × "10"^(9) #h(0.25em) "kg" · "m/s"$] (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of #math.equation(block: false, alt: "60.0 m/s")[$"60.0 m/s"$]? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. #solutionbox[ (a) #math.equation(block: false, alt: "8 . 00 times 10 to the power 4 m/s")[$8 "." "00" × "10"^(4) #h(0.25em) " m/s"$] (b) #math.equation(block: false, alt: "1 . 20 times 10 to the power 6 kg times m/s")[$1 "." "20" × "10"^(6) #h(0.25em) " kg" · "m/s"$] (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be #math.equation(block: false, alt: "minus 0 . 0100 m/s")[$− 0 "." "0100 m/s"$], which is probably not noticeable. ] (a) What is the momentum of a garbage truck that is #math.equation(block: false, alt: "1 . 20 times 10 to the power 4 kg")[$1 "." "20" × "10"^(4) #h(0.25em) " kg"$] and is moving at #math.equation(block: false, alt: "30 . 0 m/s")[$30 "." "0 m/s"$]? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck? A runaway train car that has a mass of 15,000 kg travels at a speed of #math.equation(block: false, alt: "5 .4 m/s")[$5 ".4 m/s"$] down a track. Compute the time required for a force of 1500 N to bring the car to rest. #solutionbox[ 54 s ] The mass of Earth is #math.equation(block: false, alt: "5 . 972 times 10 to the power 24 kg")[$5 "." "972" × 10^("24") #h(0.25em) " kg"$] and its orbital radius is an average of #math.equation(block: false, alt: "1 . 496 times 10 to the power 11 m")[$1 "." "496" × 10^("11") #h(0.25em) " m"$]. Calculate its linear momentum.