Center of Mass, Momentum, Impulse, and Collision PDF

SH1685 Center of Mass, Momentum, Impulse, and But how about this figure? Collision I. Center of Mass Have you ever encountered the phrase, “center of mass?” Maybe you did, in science shows, or maybe in documentaries on TV. Yo...

SH1685 Center of Mass, Momentum, Impulse, and But how about this figure? Collision I. Center of Mass Have you ever encountered the phrase, “center of mass?” Maybe you did, in science shows, or maybe in documentaries on TV. You may have even noticed, or even saw it in action -- it's just that you’ve never known that it was called as such. So, what is it? Definition and Concept As defined in Physics, the center of mass is a position relative in an object or system, wherein it is the average position of all parts of that object or system, which are weighted according to their collective masses. Simply put, the center of mass simply shows where the object or system focus its entire mass on. For a simple solid object, Source: http://www.clipartbest.com the center of mass is located at its center, called a centroid, as seen below. In order to determine the center of mass in irregularly- shaped objects, we can do it in two ways. One is by using mathematics, while the other methods favors the use of vectors, although both methods can be used at the same time, as explained by the third method. Source: https://www.khanacademy.org The mathematical method is favored over the vector As seen above, the regular solid figures all have their method if numerical values are the only values present centers of mass located at the centroid -- even the ring. in the scenario or problem. Solving for the center of Even though there is nothing at its central hole, the mass, we use the basic equation, center of mass is still there because it only relies at the object’s focus, where all of its mass must be. ∑𝑚𝑚𝑖𝑖 𝑙𝑙𝑖𝑖 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑀𝑀 07 Handout 1 *Property of STI Page 1 of 14  SH1685 The equation for the center of mass requires two draw lines from each individual points from the object, things: mass and dimension. In the equation, first by making sure that the figure is slowly slid across a square or rectangular table’s edge. The object is made ∑𝑚𝑚𝑖𝑖 𝑙𝑙𝑖𝑖 to cross that edge up to a point where it is halfway from 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑀𝑀 falling. Get a ruler and marker, then mark off that line, repeating the same process, making sure that each line it requires the following, bisects the object, meeting at a common point. This is the shape’s center of 𝑚𝑚𝑖𝑖 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 mass 𝑙𝑙𝑖𝑖 = 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑟𝑟 ′ 𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑀𝑀 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 The equation simply states that, in order to get the irregular object’s center of mass, the sum of the products of all points of mass and its respective dimensions must be divided by the total mass of the body itself. The object’s dimension relies on what axis it lies upon: 𝑥𝑥 and 𝑦𝑦. If the object lies on the x-axis, ∑𝑚𝑚𝑖𝑖 𝑥𝑥𝑖𝑖 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 𝑀𝑀 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑚𝑚𝑖𝑖 𝑥𝑥𝑖𝑖 = 𝑚𝑚1 𝑥𝑥1 + 𝑚𝑚2 𝑥𝑥2 + ⋯ + 𝑚𝑚𝑛𝑛 𝑥𝑥𝑛𝑛 𝑀𝑀 = 𝑚𝑚1 + 𝑚𝑚2 + ⋯ + 𝑚𝑚𝑛𝑛 The same equation applies as well to the y-axis. When This is widely used on flat figures, whereas the combined together with 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 , they form the mathematical method is used over solid objects. coordinates where the center of mass may be located. Sometimes, the plumb line also uses a pin, and a weighted string. The pin is first pierced somewhere on The second method, the plumb line method, requires the object. By tying the weighted string on the free end sketching vector lines on the figure itself. Because of its of the pin, and by letting it swing until it stops, the capability of determining the point of an object’s balance rather than the collected mass, this method is easier. Just 07 Handout 1 *Property of STI Page 2 of 14  SH1685 position where the weighted string stops is then traced, with a forming common point becoming the object’s center of mass. The third method, the subdivision, divides the figure into smaller, more manageable units, provided that each 4m unit is a perfect geometric figure. This method combines 4m the first two, with computing the area of the figure, and plumb line tracing, giving this a very precise form of 3m getting the center of mass. However, this method is only applicable to figures that can be divided into symmetrical geometric figures. 2m 5m All of these do not limit itself to fully-formed (meaning, no holes or gaps) figures, however. Gaps and spaces within the figures are also counted, but they account for the missing mass. When that center of mass is now influenced by gravity, it is now called center of 9m gravity, a point where gravity acts the strongest on an object or system. 4m In Physics, determining the center of mass is vital due 14 m to its many applications, such as determining an object’s toppling stability. Suppose we have a figure projected on the Cartesian plane, as illustrated at right. Solving for the figure’s center of gravity, we can divide it into six (6) unit figures: two (2) rectangles, two right triangles, a In this scenario, we can treat the area of each figure as square, and a circle. Then, applying the formula for x, its own mass. Each figure has its own formula for determining its centroid. The following formulas are ∑𝑚𝑚𝑖𝑖 𝑥𝑥𝑖𝑖 shown at the next page. 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 𝑀𝑀 07 Handout 1 *Property of STI Page 3 of 14  SH1685 FIGURE FORMULA Semicircle Depends on how the circle is Rectangle / Square 1 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝐴𝐴O bisected. If the circle is (𝐴𝐴∎ = 𝑥𝑥𝑥𝑥) 2 bisected along the x-axis, use 𝑦𝑦𝑖𝑖 formula in quarter circle, 1 due to x being at the origin. 𝑥𝑥𝑖𝑖 = 𝑥𝑥 2 If bisected along y, use 𝑥𝑥𝑖𝑖 formula in quarter circle due 1 to y being at the origin. 𝑦𝑦𝑖𝑖 = 𝑦𝑦 2 If bisected along y, Semicircular arc 2𝑟𝑟 𝑥𝑥𝑖𝑖 = 𝜋𝜋 Triangle 𝑏𝑏ℎ 𝑦𝑦𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝐴𝐴∆ = 𝑥𝑥 2 𝑥𝑥𝑖𝑖 = 3 If bisected along x, 𝑥𝑥𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑦𝑦 𝑦𝑦𝑖𝑖 = 3 2𝑟𝑟 𝑦𝑦𝑖𝑖 = 𝜋𝜋 Circle 𝐿𝐿 = 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝐴𝐴O = 𝜋𝜋𝑟𝑟 2 ) Quarter Circle 1 𝐴𝐴𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄 = 𝐴𝐴O 4 4𝑟𝑟 𝑥𝑥𝑖𝑖 = 𝑟𝑟𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 3𝜋𝜋 4𝑟𝑟 𝑦𝑦𝑖𝑖 = 3𝜋𝜋 07 Handout 1 *Property of STI Page 4 of 14  SH1685 Sector of a Circle Quarter Ellipse (𝐴𝐴𝑆𝑆𝑆𝑆𝑆𝑆 = 𝑟𝑟 2 𝜃𝜃𝑟𝑟𝑟𝑟𝑟𝑟 ) 1 𝐴𝐴𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄𝑄 = 𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 4 4𝑥𝑥 2𝑟𝑟 sin 𝜃𝜃 𝑥𝑥𝑖𝑖 = 𝑥𝑥𝑖𝑖 = 3𝜋𝜋 3𝜃𝜃𝑟𝑟𝑟𝑟𝑟𝑟 4𝑦𝑦 𝑦𝑦𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑦𝑦𝑖𝑖 = 3𝜋𝜋 Arc of a Circle Half Ellipse 1 Depends on how the ellipse is 𝑟𝑟 sin 𝜃𝜃 𝐴𝐴𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = 𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 bisected. If the ellipse is 𝑥𝑥𝑖𝑖 = 2 bisected along the x-axis, use 𝑦𝑦𝑖𝑖 𝜃𝜃𝑟𝑟𝑟𝑟𝑟𝑟 formula in quarter ellipse, due to x being at the origin. 𝑦𝑦𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 If bisected along y, use 𝑥𝑥𝑖𝑖 formula in quarter ellipse due to 𝐿𝐿 = 2𝑟𝑟𝜃𝜃𝑟𝑟𝑟𝑟𝑟𝑟 y being at the origin. Parabolic Segment If parabola lies along y, Ellipse 2𝑏𝑏ℎ 3𝑥𝑥 𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑥𝑥𝑖𝑖 = (𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜋𝜋𝜋𝜋𝜋𝜋) 3 8 3𝑦𝑦 𝑦𝑦𝑖𝑖 = 5 𝑥𝑥𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 If parabola lies along x, 𝑦𝑦𝑖𝑖 = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 3𝑥𝑥 𝑥𝑥𝑖𝑖 = 5 3𝑦𝑦 𝑦𝑦𝑖𝑖 = 8 07 Handout 1 *Property of STI Page 5 of 14  SH1685 Spandrel 𝐶𝐶𝐶𝐶𝑀𝑀𝑦𝑦 = 𝑥𝑥𝑥𝑥 {[(36)(2)]+[(32)(6)]+[(8){(12)+(4/3)}]+[(6.5)(25)]+ 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑛𝑛 + 1 𝑥𝑥 [(10)(4/3)]-[(12.57)(6.5)]} / (98.43) 𝑥𝑥𝑖𝑖 = 𝑛𝑛 + 2 𝐶𝐶𝐶𝐶𝑀𝑀𝑦𝑦 𝑛𝑛 + 1 72 + 192 + 106.67 + 162.5 + 13.3 − 81.71 𝑦𝑦𝑖𝑖 = 𝑦𝑦 = 4𝑛𝑛 + 2 98.43 546.44 − 81.71 𝐶𝐶𝐶𝐶𝑀𝑀𝑦𝑦 = Source: http://www.mathalino.com/reviewer/engineering-mechanics/centroids-and-centers-gravity 98.43 Then, solve for the center of mass, using the area of the 464.73 𝐶𝐶𝐶𝐶𝑀𝑀𝑦𝑦 = subdivided figures and their respective centroids, 98.43 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 𝑪𝑪𝑪𝑪𝑴𝑴𝒚𝒚 = 𝟒𝟒. 𝟕𝟕𝟕𝟕 {[(36)(4.5)]+[(32)(2)]+[(8)(4/3)]+[(6.5)(25)]+[(10){(9) +(5/3)]-[(12.57)(6.5)]} / (98.43) Combining together for the coordinates of the center of mass (COM), we get that 𝐶𝐶𝐶𝐶𝐶𝐶 = (4.31, 4.72). 162 + 64 + 10.64 + 162.5 + 106.7 − 81.71 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 98.43 If we are to plot the center of mass, we get something like the figure in the next page. But, upon closer 505.84 − 81.71 inspection at the solution, notice that the 𝑥𝑥𝑖𝑖 and 𝑦𝑦𝑖𝑖 values 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 98.43 are a bit off compared to their computed centroid locations. That is because the farther the centroid is to 424.13 the plotted origin, the higher the value of the centroid’s 𝐶𝐶𝐶𝐶𝑀𝑀𝑥𝑥 = 98.43 location because the distance between the centroid and the origin are given consideration in the subdivision 𝑪𝑪𝑪𝑪𝑴𝑴𝒙𝒙 = 𝟒𝟒. 𝟑𝟑𝟑𝟑 method. If the given system is linear, with at least two given masses, the formula still applies with respect to the Solving for y, point you have chosen. ∑𝑚𝑚𝑖𝑖 𝑦𝑦𝑖𝑖 𝐶𝐶𝐶𝐶𝑀𝑀𝑦𝑦 = 𝑀𝑀 Let us say that you have a wooden bar that is 3 meters long. On its leftmost end is a counterweight with a mass 07 Handout 1 *Property of STI Page 6 of 14  SH1685 sample scenario, as well as providing a visual aid, we get that, 𝑚𝑚1 𝑥𝑥1 + 𝑚𝑚2 𝑥𝑥2 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚1 + 𝑚𝑚2 Solving for COM, 𝑚𝑚1 𝑥𝑥1 + 𝑚𝑚2 𝑥𝑥2 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑚𝑚1 + 𝑚𝑚2 (2)(0) + (4)(3) 𝐶𝐶𝐶𝐶𝐶𝐶 = 2+4 of 2 kg, while the rightmost end has a 4-kg 12 counterweight. To solve for the system’s center of mass, 𝐶𝐶𝐶𝐶𝐶𝐶 = 6 you need to determine first where you must place your point of reference. It could be at the leftmost end, the 𝑪𝑪𝑪𝑪𝑪𝑪 = 𝟐𝟐 𝒎𝒎 middle, or to the right. So, for convenience, let us choose the left side as our reference. In the given problem, the center of mass is located at 2 meters, meaning it is closer to the heavier side than it is Rewriting our equation to match the given in our to the lighter one. 07 Handout 1 *Property of STI Page 7 of 14  SH1685 As mentioned earlier, one of the applications of center is used in sports -- a sports team that is on the move of mass is identifying a system’s toppling stability, (meaning, gaining the advantage) has momentum! which is the system’s capability of maintaining balance at a certain angle. If the toppling stability is higher than Momentum is also referred as mass in motion. If an the center of mass, then the object will lose its stability, object has mass, then it has momentum. Mathematically and will eventually topple. This is due to gravity pulling speaking, it down. 𝒑𝒑 = 𝑚𝑚𝑣𝑣̅ II. Momentum Suppose we have an object moving at a certain amount 𝒑𝒑 is the standard symbol used for momentum. As seen of velocity, with a certain amount of mass. This object from the equation earlier, momentum has two moves really fast, thus we can safely assume that it has components: velocity and mass. Yes, it uses velocity energy, and will do a very high amount of work to because momentum is a vector value. It needs to identify something it might come across with. But, what we do its direction. Also, as one might notice, the formula has not notice is that the object is also gaining another value. quite the similarity with the formula for kinetic energy, That is what we call momentum. But, what is 1 𝑇𝑇 = 𝑚𝑚𝑣𝑣 2 momentum? 2 Definition and Principle The unit for momentum is 𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚/𝑠𝑠. There is no other The term momentum is widely used in sports, as it unit to fill in for momentum. As one might assume as means that things are on the move, and is going to take well, both mass and velocity contribute to the increase of some amounts of effort to stop it. If a team has an object’s momentum, much the same way Newton’s momentum, then it will be really hard to stop them from First Law operates. A heavy object may have a certain winning. But, do you know that momentum is a Physics amount of momentum, but is moving very slowly. A term? lighter car can achieve the same amount of momentum the heavy car has, but it needs to move faster in order to Indeed, it is! compensate for its lighter mass. Momentum is the amount of motion that an object has. Meaning, if an object is moving, it has momentum as well, kind of like moving inertia. Now we know why it 07 Handout 1 *Property of STI Page 8 of 14  SH1685 III. Impulse are readily available. But, if we only have time, we have to Now, we know that momentum is a product of mass and yet determine the strength of the object upon contact. This is velocity. Now, consider the formula of force, as expressed why we have impulse. by Newton’s second law, Definition 𝐹𝐹 = 𝑚𝑚𝑚𝑚 Impulse, written symbolically as 𝑱𝑱, is what we call force in time. It is usually associated with momentum Where F is force, m is mass, and a is acceleration. We also because they share the same relationship. So, rewriting know that acceleration is the change in velocity in a given the equation mathematically, time period. So, if we are to break down acceleration, we get, 𝑱𝑱 = 𝐹𝐹∆𝑡𝑡 ∆𝑣𝑣̅ 𝐹𝐹 = 𝑚𝑚 ∆𝑡𝑡 As stated earlier, it is usually associated with Rewriting this equation yields, momentum. But, there is a catch to it. 𝑚𝑚∆𝑣𝑣̅ If we go back to our example, a heavy truck and a light 𝐹𝐹 = ∆𝑡𝑡 car were moving along an unobstructed highway. If both Hcars were to collide with each other, either one or both And we know that 𝑝𝑝 = 𝑚𝑚𝑣𝑣̅ , so, of them will need to adjust. So, a heavy car has lots of momentum, so, as stated earlier, it will need a greater 𝒑𝒑 amount of force, or a longer time before impact, or both, 𝐹𝐹 = ∆𝑡𝑡 to dampen, or soften, the blow. The same goes with the lighter car. Transposing the equation to get momentum, The Impulse – Momentum Theorem 𝒑𝒑 = 𝐹𝐹∆𝑡𝑡 The Impulse – Momentum Theorem states that an applied impulse is equal to a change in momentum in a That equation right there shows that a change in given system. Mathematically speaking, momentum is equal to a change in force exerted within a certain time. It is certainly related to momentum, but in some 𝑱𝑱 = ∆𝒑𝒑 ways, it isn’t. This equation is used in the study of an average force being applied onto a system. We can always compute This theorem has a logical equivalence to Newton’s for momentum readily if we have velocity and mass, which Second Law, 𝐹𝐹 = 𝑚𝑚𝑚𝑚. Meaning, any applied force to an 07 Handout 1 *Property of STI Page 9 of 14  SH1685 object will have significant changes in its momentum 𝐽𝐽 𝐽𝐽𝑠𝑠𝑠𝑠 = and impulse values. Given an object with a standard 𝑚𝑚𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑔𝑔 mass, 𝐹𝐹∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣̅ This equation yields time, wherein it shows how efficient the propellant is being used within a specific This equation implies that if mass is constant, then time. everything else is constant. This can be best described by an ice hockey puck sliding along the ice, only to collide IV. Collision with the goalkeeper’s legs. If the formula we found uses We all have seen it in the news, in TV, and even in a changing, or variable mass, the formula should be, movies, and written and illustrated works. We usually associate it with vehicular accidents, because these are 𝐹𝐹∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 + 𝑣𝑣∆𝑚𝑚 examples of this concept at work. If an object has momentum, and demonstrates impulse before impact, Specific Impulse the moment that object impacts with another object, the Specific impulse is the measurement used in object is demonstrating collision. determining the efficacy of rocket propellants. It is widely used in rocket science. There are two ways of So, what is it? defining specific impulse. Definition 1. If the system requires that the specific impulse be Collision is best defined as the brief, simultaneous defined as impulse per mass, we get, interaction between two or more bodies, which is due to the internal force acting between the bodies. It involves 𝐽𝐽 energy, force, velocity, momentum, and impulse. As 𝐽𝐽𝑠𝑠𝑠𝑠 = 𝑚𝑚 such, this is the direct application of momentum and impulse that involves forces, mass, and velocity. It When simplified, what we will get is the exhaust applies another Conservation Law: The Law of velocity (𝑣𝑣𝑒𝑒 ). Exhaust velocity is the velocity required to Conservation of Momentum, wherein momentum is a create an effective thrust and efficient use of propellant. constant in all things that have mass. It also defines that the vector sum of all given momenta (sing. momentum) 2. If the system requires that the specific impulse be of all objects within a system, that sum cannot be defined as impulse per weight, we simply add the changed by any other interactions within that system. constant 𝑔𝑔, Simply put, if one object has a momentum within a certain direction, the other objects must have the same 07 Handout 1 *Property of STI Page 10 of 14  SH1685 momentum, but traveling in directions opposite that of The scenarios are: the other. 1. One moving (projectile), one at rest (target): Collision does not only imply the hard-hitting impact between two cars. Even the simplest ones, such as (𝑚𝑚1 𝑣𝑣1 )0 = (𝑚𝑚1 𝑣𝑣1 )𝑓𝑓 + (𝑚𝑚2 𝑣𝑣2 )𝑓𝑓 playing Billiards (a classic example), playing volleyball, -- even walking! All these examples demonstrate This equation demonstrates that since the other collision. There are two kinds of collisions, both of object is not moving (𝑣𝑣20 = 0), the conservation of which conserve momentum. They will only differ momentum will be evident after the collision takes whether energy (implying to kinetic energy) will be place. Consequently, the energy conserved is, conserved or not. 1 1 1 Elastic Collision 𝑚𝑚𝑣𝑣 2 = 𝑚𝑚𝑣𝑣 2 + 𝑚𝑚𝑣𝑣 2 2 10 2 1𝑓𝑓 2 2𝑓𝑓 Elastic collision is defined as a type of collision where the concerned bodies conserve both energy and momentum upon contact with each other. The best way And, should the target be at rest initially, in order to describe this type of collision is a ball bouncing off to find the final velocities, using both conservation the floor. Sure, the ball will bounce upon hitting the equations, ground, and it will conserve momentum and energy. But, 𝑚𝑚1 − 𝑚𝑚2 each bounce will conserve less and less energy due to the 𝑣𝑣1𝑓𝑓 = 𝑣𝑣 force being weaker with each succeeding bounce, until it 𝑚𝑚1 + 𝑚𝑚2 10 will stop altogether. This kind of elastic collision is 2𝑚𝑚1 called a contact collision, and is not a perfect collision. 𝑣𝑣2𝑓𝑓 = 𝑣𝑣 The other type, which is noncontact collision, is almost 𝑚𝑚1 + 𝑚𝑚2 10 perfect. The noncontact collision is a kind of elastic collision where the objects interact with each other, even 2. Both objects are in motion: without direct contact. A perfect example would be a planet slinging a space shuttle in another direction. (𝑚𝑚1 𝑣𝑣1 )0 + (𝑚𝑚2 𝑣𝑣2 )0 = (𝑚𝑚1 𝑣𝑣1 )𝑓𝑓 + (𝑚𝑚2 𝑣𝑣2 )𝑓𝑓 There are two scenarios involving this kind of This is also evident in the energy equation, collision, with three major cases, and demonstrate two setups. 1 1 1 1 𝑚𝑚𝑣𝑣 2 + 𝑚𝑚𝑣𝑣 2 = 𝑚𝑚𝑣𝑣 2 + 𝑚𝑚𝑣𝑣 2 2 10 2 20 2 1𝑓𝑓 2 2𝑓𝑓 07 Handout 1 *Property of STI Page 11 of 14  SH1685 These formulas can be used to determine the final This case represents the one where the projectile velocities of objects moving in one dimension, is more massive than the target. The result is pretty straightforward: the larger projectile will simply 𝑚𝑚1 − 𝑚𝑚2 2𝑚𝑚2 move at approximately the same speed it is in, while 𝑣𝑣1𝑓𝑓 = 𝑣𝑣10 + 𝑣𝑣 𝑚𝑚1 + 𝑚𝑚2 𝑚𝑚1 + 𝑚𝑚2 20 the smaller target gets double, in estimate, the projectile’s speed. 2𝑚𝑚1 𝑚𝑚2 − 𝑚𝑚1 𝑣𝑣2𝑓𝑓 = 𝑣𝑣10 + 𝑣𝑣 3. 𝑚𝑚1 < 𝑚𝑚2 𝑚𝑚1 + 𝑚𝑚2 𝑚𝑚1 + 𝑚𝑚2 20 After covering the two scenarios, there are two 𝑖𝑖𝑖𝑖 𝑣𝑣10 = 𝑣𝑣1 , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣20 = 0, outcomes and three cases to cover. The cases are the 𝑣𝑣1𝑓𝑓 ≈ −𝑣𝑣1 , 𝑣𝑣2𝑓𝑓 ≈ 0 running conditions in which it covers both the two scenarios, which demonstrates one of the two types of The final case is also straightforward, this time the elastic collision. situation is reversed. The smaller projectile will rebound at the same speed, but will move in the opposite The three cases are as follows: direction. 1. 𝑚𝑚1 = 𝑚𝑚2 And finally, the two setups of elastic collision. The setups, in truth, uses both the scenarios and cases to 𝑖𝑖𝑖𝑖 𝑣𝑣10 = 𝑣𝑣1 , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣20 = 0, create one of the two outcomes based on its build. 𝑣𝑣1𝑓𝑓 ≈ 0, 𝑣𝑣2𝑓𝑓 ≈ 𝑣𝑣1 1. Head-on collision is the type of elastic collision This case demonstrates that objects of equal where the projectile is moving directly along a masses will have a clean transfer of velocity straight line. The target, upon colliding with the between objects, resulting for the projectile to stop, projectile, will then move in the direction the and the target to move at the same speed. projectile is moving at. 2. 𝑚𝑚1 > 𝑚𝑚2 2. The non-head-on collision is the type of collision where the projectile hits the target at a certain angle, 𝑖𝑖𝑖𝑖 𝑣𝑣10 = 𝑣𝑣1 , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣20 = 0, in which the resulting angle after impact is dependent on the masses of the given objects in 𝑣𝑣1𝑓𝑓 ≈ 𝑣𝑣1 , 𝑣𝑣2𝑓𝑓 ≈ 2𝑣𝑣1 collision. Take note of the differences of each 07 Handout 1 *Property of STI Page 12 of 14  SH1685 resulting angle: Inelastic collision occurs more often than elastic collision because, in reality, energy is lost more than be a. 𝜃𝜃 = 90° 𝑖𝑖𝑖𝑖 𝑚𝑚1 = 𝑚𝑚2 conserved. Inelastic collision has the equation, b. 𝜃𝜃 < 90° 𝑖𝑖𝑖𝑖 𝑚𝑚1 > 𝑚𝑚2 c. 𝜃𝜃 > 90° 𝑖𝑖𝑖𝑖 𝑚𝑚1 < 𝑚𝑚2 𝑚𝑚1 𝑣𝑣10 + 𝑚𝑚2 𝑣𝑣20 = (𝑚𝑚1 + 𝑚𝑚2 )𝑣𝑣𝑓𝑓 One thing to note, however, is that there is no perfectly Inelastic collision also has a coefficient in it, called the elastic collision. Kinetic energy, as stated before, will coefficient of restitution, or COR. This coefficient shows still be lost by some amount. That lost kinetic energy is the ratio between objects before and after impact. There converted into other forms, such as heat, sound, or are numerous equations for COR, but the one we will use potential energy. Always keep in mind that elastic is, collision uses the equation for the Conservation of 𝑣𝑣𝑓𝑓 𝐶𝐶𝐶𝐶𝐶𝐶 = Momentum, which is, 𝑣𝑣0 𝑚𝑚1 𝑣𝑣10 + 𝑚𝑚2 𝑣𝑣20 = 𝑚𝑚1 𝑣𝑣1𝑓𝑓 + 𝑚𝑚2 𝑣𝑣2𝑓𝑓 Normally, this equation is related to inelastic collision, which has a coefficient value of zero (0). But, if the COR Subsequently, all of the equations presented here are has a value of at least one (1) or higher, then the collision just variations of the momentum and energy is elastic. Take note, however, that in inelastic collisions, conservation equations. the two interacting objects will not fuse into one object. They still exist as two objects sticking together long Inelastic Collision enough while in motion, in which the two will split after There is another kind of collision, in which if the target coming to a complete stop. does not bend to the force generated by the incoming projectile, the said target will move alongside the References: Bauer, W., & Westfall, G. D. (2016). General Physics 1 (2nd ed.). already-moving object, or the two interacting objects Columbus, OH: McGraw-Hill Education. will stick together, if enough force and momentum is Bauer, W., & Westfall, G. D. (2016). General Physics 1 (2nd ed.). Quezon generated. This collision is evident in vehicular City: Abiva Publishing House, Inc. accidents, target ranges, or even in simple pastimes like Bautista, D.C. (2013). Science Impact: Integrated Science (3rd ed.). playing football in all its variants (American, soccer, and Antipolo City: Academe Publishing House, Inc. Belleza, R.V., Gadong, E.S.A., …, Sharma, M. PhD (2016). General rugby), and, as young kids, throwing clay at the wall. Physics 1. Quezon City, Vibal Publishing House, Inc. This kind of collision is called inelastic collision. Catchilar, Gerry C. & Malenab, Ryan G., (2003), Fundamentals of Physics, Mandaluyong City, National Book Store. 07 Handout 1 *Property of STI Page 13 of 14  SH1685 CHED (2017). Center of Mass, Impulse, and Momentum. Retrieved 2017, Nave, C. R. (2016). Center of Mass. Retrieved from The Georgia State March 28 from Teach Together: CHED K-12 Curriculum Sharing University’s HyperPhysics: Site: http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html http://teachtogether.chedk12.com/teaching_guides/view/119 Nave, C. R. (2016). Collisions. Retrieved from The Georgia State CHED (2017). Conservation of Linear Momentum and the Types of University’s HyperPhysics: Collisions. Retrieved 2017, March 28 from Teach Together: http://hyperphysics.phy-astr.gsu.edu/hbase/colcon.html#c1 CHED K-12 Curriculum Sharing Site: Nave, C. R. (2016). Force and Momentum. Retrieved from The Georgia http://teachtogether.chedk12.com/teaching_guides/view/120 State University’s HyperPhysics: CHED (2017). Context-Rich Problems Involving Mass, Momentum, http://hyperphysics.phy-astr.gsu.edu/hbase/rocket.html#c1 Impulse, and Collisions. Retrieved 2017, March 28 from Teach Nave, C. R. (2016). Momentum. 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Retrieved 2017, March 22 from Khan Academy: https://www.khanacademy.org/science/physics/linear- momentum/center-of-mass/a/what-is-center-of-mass Khan Academy (2017). What is center of mass?. Retrieved 2017, March 22 from Khan Academy: https://www.khanacademy.org/science/physics/linear- momentum/center-of-mass/a/what-is-center-of-mass 07 Handout 1 *Property of STI Page 14 of 14