General Physics Module Phys 1011 Linear Momentum PDF

General Physics Module Phys 1011 AAU 7 Linear Momentum Learning Outcome After completing this Chapter, students are expected to: Define linear momentum, Explain the relationship between...

General Physics Module Phys 1011 AAU 7 Linear Momentum Learning Outcome After completing this Chapter, students are expected to: Define linear momentum, Explain the relationship between momentum and force, State kinetic energy of motion in terms of momentum Relate collision and momentum Solve problems related to various types of collision Introduction What happens when two automobiles collide? To begin answering such questions, we introduce momentum. Intuitively, anyone or anything that has a lot of momentum is going to be hard to stop. Physically, the more momentum an object has, the more force has to be applied to stop it in a given time. This concept leads to one of the most powerful principles in physics: conservation of momentum. Using this law, complex collision problems can be solved without knowing much about the forces involved during contact. We’ll also be able to derive information about the average force delivered in an impact. With conservation of momentum, we’ll have a better understanding of what choices to make when designing an automobile or a rocket, or when addressing a football or basketball training. 7.1 The Concept of Momentum and Impulse Learning outcome After completing this section, students are expected to: Define linear impulse. Explain the relationship between momentum and impulse. Calculate momentum given mass and velocity. Describe effects of impulses in everyday life. Solve simple problems involving momentum and impulse ⃗⃗of a body of mass m moving with velocity 𝒗 The linear momentum (or simply momentum)𝒑 ⃗⃗ is defined as ⃗⃗ = 𝒎𝒗 𝒑 ⃗⃗ (7-1) Linear Momentum 174 General Physics Module Phys 1011 AAU Doubling either the mass or the velocity of an object doubles its momentum; doubling both quantities quadruples its momentum. Momentum is a vector quantity. Its components in Cartesian coordinates are 𝒑𝒙 = 𝒎𝒗𝒙 , 𝒑𝒚 = 𝒎𝒗𝒚 and 𝒑𝒛 = 𝒎𝒗𝒛 (7-2) where 𝒑𝒙 is the momentum of the object in the x – direction, 𝒑𝒚 its momentum in the y – direction and 𝒑𝒛 its momentum in the z – direction. The magnitude of the momentum 𝒑 = √𝒑𝒙 𝟐 + 𝒑𝒚 𝟐 + 𝒑𝒛 𝟐 (7-3) of an object of mass m can be related to its kinetic energy 𝑬𝑲 : 𝒑𝟐 𝑬𝑲 = 𝟐𝒎 (7-4) This relationship is easy to prove using the definitions of kinetic energy and momentum and is valid for objects travelling at speeds much less than the speed of light. Curiously, there is no named SI unit for measuring momentum. Momentum in SI units is measured in units of kg m/s. Changing the momentum of an object requires the application of a force. This is, in fact, how Newton originally stated his second law of motion. Starting from the more common version of the second law, we have ⃗⃗ 𝚫(𝒎𝒗 𝚫𝒗 ⃗⃗) ⃗𝑭⃗𝒏𝒆𝒕 = 𝒎𝒂 ⃗⃗ = 𝒎 = 𝚫𝒕 𝚫𝒕 where the mass m and the force are assumed constants. The quantity in parentheses is just the momentum, so we have the following result: ⃗⃗ divided by the elapsed time 𝚫𝒕 equals the constant net The change in an object’s momentum 𝚫𝒑 ⃗⃗ force 𝑭𝒏𝒆𝒕 acting on the object: ⃗⃗ 𝚫𝒑 𝐜𝐡𝐚𝐧𝐠𝐞 𝐨𝐟 𝐦𝐮𝐦𝐞𝐧𝐭𝐮𝐦 𝚫𝒕 = 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 = ⃗𝑭⃗𝒏𝒆𝒕 (7-5) This equation is also valid when the forces are not constant, provided the limit is taken as Δ𝑡 becomes infinitesimally small. Equation (7.5) says that if the net force on an object is zero, the object’s momentum doesn’t change. In other words, the linear momentum of an object is conserved when 𝐹⃗𝑛𝑒𝑡 = 0. Equation (7.5) also shows us that changing an object’s momentum requires the continuous application of a force over a period of time Δ𝑡 leading to the definition of impulse. If a constant force 𝐹⃗ acts on an object, the impulse 𝐼⃗ delivered to the object over a time interval Δ𝑡 is given by 𝑰⃗ = ⃗𝑭⃗𝚫𝒕 (7-6) Linear Momentum 175 General Physics Module Phys 1011 AAU Impulse is a vector quantity with the same direction as the constant force acting on the object. When a single constant force 𝑭⃗⃗ acts on an object, (7.6) can be written as 𝑰⃗ ≡ 𝑭 ⃗⃗𝚫𝒕 = 𝚫𝒑 ⃗⃗ = 𝒎𝒗 ⃗⃗𝐟 − 𝒎𝒗 ⃗⃗𝐢 (7-7) which is a special case of the impulse–momentum theorem. Equation (7.7) shows that the impulse of the force acting on an object equals the change in momentum of that object. That equality is true even if the force is not constant, as long as the time interval 𝚫𝒕 is taken to be arbitrarily small. In real-life situations, the force on an object is only rarely constant. To analyze the case of variable force with rather complex interaction of bodies, it’s useful to define an average force ⃗𝑭⃗𝒂𝒗. The average force is the constant force delivering the same impulse to the object in the time interval 𝚫𝒕 as the actual time-varying force. We can then write the impulse–momentum theorem as ⃗𝑭 ⃗⃗⃗𝒂𝒗 𝚫𝒕 = 𝚫𝒑 ⃗⃗ (7-8) Eample7.1 Teeing off: A golf ball with mass 𝟓. 𝟎 × 𝟏𝟎−𝟐 kg is struck with a club as in Figure 7.1. The force on the ball varies from zero when contact is made up to some maximum value (when the ball is maximally deformed) and then back to zero when the ball leaves the club. Assume that the ball leaves the club face with a velocity of 44 m/s. (a) Find the magnitude of the impulse due to the collision. (b) Estimate the duration of the collision and the average force acting on the ball. Figure 7-1 (Example 7.1) During impact, the club head momentarily flattens the side of the golf ball Solution (a) Find the impulse delivered to the ball. The problem is essentially one dimensional. Note that vi= 0, and calculate the change in momentum, which equals the impulse: m 𝐼 = ∆𝑝 = 𝑝𝑓 − 𝑝𝑖 = (5.0 × 10−2 kg) (44 ) − 0 = 2.2 kg m/s s (b) Estimate the duration of the collision and the average force acting on the ball. Estimate the time interval of the collision, Δ𝑡, using the approximate displacement (radius of the ball) and its average speed (half the maximum speed): ∆𝑥 2 × 10−2 m ∆𝑡 = = = 9.1 × 10−4 s 𝑣𝑎𝑣 22 m/s Linear Momentum 176 General Physics Module Phys 1011 AAU Estimate the average force from Equation (7.8): Δ𝑝 2.2 kg m/s 𝐹𝑎𝑣 = = = 2.4 × 103 N Δ𝑡 9.1 × 10−4 s REMARKS: This estimate shows just how large such contact forces can be. A good golfer achieves maximum momentum transfer by shifting weight from the back foot to the front foot, transmitting the body’s momentum through the shaft and head of the club. This timing, involving a short movement of the hips, is more effective than a shot powered exclusively by the arms and shoulders. Following through with the swing ensures that the motion isn’t slowed at the critical instant of impact. Exercises 1. What average club speed would double the average force? (Assume the final velocity is unchanged.) 2. A 0.150-kg baseball, thrown with a speed of 40.0 m/s, is hit straight back at the pitcher with a speed of 50.0 m/s. (a) What is the magnitude of the impulse delivered by the bat to the baseball? (b) Find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 2.00 × 10−3 s. ANSWERS (a) 13.5 kg m/s (b) 6.75 kN BOXING AND BRAIN INJURY (application of impulse) Boxers in the nineteenth century used their bare fists. In modern boxing, fighters wear padded gloves. How do gloves protect the brain of the boxer from injury? Also, why do boxers often “roll with the punch”? EXPLANATION: The brain is immersed in a cushioning fluid inside the skull. If the head is struck suddenly by a bare fist, the skull accelerates rapidly. The brain matches this acceleration only because of the large impulsive force exerted by the skull on the brain. This large and sudden force (large 𝐹𝑎𝑣 and small Δ𝑡) can cause severe brain injury. Padded gloves extend the time Δ𝑡 over which the force is applied to the head. For a given impulse 𝐹𝑎𝑣 Δ𝑡, a glove results in a longer time interval than a bare fist, decreasing the average force. Because the average force is decreased, the acceleration of the skull is decreased, reducing (but not eliminating) the chance of brain injury. The same argument can be made for “rolling with the punch”: If the head is held steady while being struck, the time interval over which the force is applied is relatively short and the average force is large. If the head is allowed to move in the same direction as the punch, the time interval is lengthened and the average force reduced. Linear Momentum 177 General Physics Module Phys 1011 AAU Eample7.2 How good are the bumpers? In a crash test, a car of mass 1.5 × 103 kg collides with a wall and rebounds as in Figure 7.2a. The initial and final velocities of the car are vi =-15.0 m/s and vf = 2.60 m/s, respectively. If the collision lasts for 0.150 s, find (a) the impulse delivered to the car due to the collision and (b) the magnitude and direction of the average force exerted on the car. Figure 7-2 Example 7.2) (a) This car’s momentum changes as a result of its collision with the wall. (b) In a crash test (an inelastic collision), much of the car’s initial kinetic energy is transformed into the energy it took to damage the vehicle Solution: (a) Find the impulse delivered to the car. Calculate the initial and final momenta of the car 𝒑𝒊 = 𝒎𝒗𝒊 = (𝟏. 𝟓 × 𝟏𝟎𝟑 𝐤𝐠)(−𝟏𝟓. 𝟎 𝐦/𝐬) = −𝟐. 𝟐𝟓 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 𝒑𝒇 = 𝒎𝒗𝒇 = (𝟏. 𝟓 × 𝟏𝟎𝟑 𝐤𝐠)(𝟐. 𝟔. 𝟎 𝐦/𝐬) = 𝟎. 𝟑𝟗 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 The impulse is just the difference between the final and initial momenta: 𝑰 = 𝒑𝒇 − 𝒑𝒊 = [(𝟎. 𝟑𝟗) − (−𝟐. 𝟐𝟓)] × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 = 𝟐. 𝟔𝟒 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 (b) Find the average force exerted on the car. Apply Equation (7.8), the impulse–momentum theorem 𝚫𝒑 𝟐. 𝟔𝟒 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 𝑭𝒂𝒗 = = = 𝟏. 𝟕𝟔 × 𝟏𝟎𝟓 𝐍 𝚫𝒕 𝟎. 𝟏𝟓𝟎 𝐬 REMARKS: If the car doesn’t rebound off the wall, the average force exerted on the car is smaller than the value just calculated. With a final momentum of zero, the car undergoes a smaller change in momentum. During the collision the state of the wall is hardly changed due to the fact it has much larger mass compared to that of the car. Similarly, although a ball has greater velocity, a player has much greater mass, and 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. Exercises Linear Momentum 178 General Physics Module Phys 1011 AAU 1. When a person is involved in a car accident, why is the likelihood of injury greater in a head- on collision as opposed to being hit from behind? Answer using the concepts of relative velocity, momentum, and average force. 2. Suppose the car doesn’t rebound off the wall, but the time interval of the collision remains at 0.150 s. In this case, the final velocity of the car is zero. Find the average force exerted on the car. ANSWER +𝟏. 𝟓 × 𝟏𝟎𝟓 𝐍 7.2 Conservation of Momentum Learning outcome After completing this section, students are expected to: Describe the principle of conservation of momentum. Derive an expression for the conservation of momentum. Explain conservation of momentum with examples. Solve simple problems involving conservation of momentum When a collision occurs in an isolated system, the total momentum of the system doesn’t change with the passage of time. Instead, it remains constant both in magnitude and in direction. The momenta of the individual objects in the system may change, but the vector sum of all the momenta will not change. The total momentum is therefore said to be conserved. In this section, we will see how the laws of motion lead us to this important conservation law. A collision may be the result of physical contact between two objects, as illustrated in Figure 7.3a. This is a common macroscopic event, as when a pair of billiard balls or a baseball and a bat strike each other. By contrast, because contact on a submicroscopic scale is hard to define accurately, the notion of collision must be generalized to that scale. Forces between two objects arise from the electrostatic interaction of the electrons in the surface atoms of the objects. Electric charges are either positive or negative. Charges with the same sign repel each other, while charges with opposite sign attract each other. To understand the distinction between macroscopic and microscopic collisions, consider the collision between two positive charges, as shown in Figure 7.3b. Because the two particles in the figure are both positively charged, they repel each other. During such a microscopic collision, particles need not touch in the normal sense in order to interact and transfer momentum. Figure 7-3(a) A collision between two objects resulting from direct contact. (b) A collision between two charged objects (in this case, a proton and a helium nucleus). Linear Momentum 179 General Physics Module Phys 1011 AAU Figure 7.4 shows an isolated system of two particles before and after they collide. By “isolated,” we mean that no external forces, such as the gravitational force or friction, act on the system. Before the collision, the velocities of the two particles are𝑣⃗1𝑖 and 𝑣⃗2𝑖 ; after the collision, the velocities are 𝑣⃗1𝑓 and 𝑣⃗2𝑓. The impulse–momentum theorem applied to m1becomes 𝐹⃗21 ∆𝑡 = 𝑚1 𝑣⃗1𝑓 − 𝑚1 𝑣⃗1𝑖 Likewise, for m2, we have 𝐹⃗12 ∆𝑡 = 𝑚2 𝑣⃗2𝑓 − 𝑚2 𝑣⃗2𝑖 where 𝐹⃗21 is the average force exerted by m2 on m1 during the collision and 𝐹⃗12 is the average force exerted by m1 on m2 during the collision, as in Figure 7.3a. Figure7-4: Before and after a head-on collision between two particles. The momentum of each object changes during the collision, but the total momentum of the system is constant. Notice that the magnitude of the change of velocity of the lighter particle is greater than that of the heavier particle, which is true in general. We use average values for 𝐹⃗21 and 𝐹⃗12 even though the actual forces may vary in time in a complicated way. Newton’s third law states that at all times these two forces are equal in magnitude and opposite in direction: 𝐹⃗21 = −𝐹⃗12. In addition, the two forces act over the same time interval. As a result, we have ⃗⃗𝟐𝟏 ∆𝒕 = −𝑭 𝑭 ⃗⃗𝟏𝟐 ∆𝒕 or ⃗⃗𝟏𝒇 − 𝒎𝟏 𝒗 𝒎𝟏 𝒗 ⃗⃗𝟏𝒊 = −(𝒎𝟐 𝒗 ⃗⃗𝟐𝒇 − 𝒎𝟐 𝒗 ⃗⃗𝟐𝒊 ) after substituting the expressions obtained for ⃗𝑭⃗𝟐𝟏 and ⃗𝑭⃗𝟏𝟐. This equation can be rearranged to give the following important result: ⃗⃗𝟏𝒇 +𝒎𝟐 𝒗 𝒎𝟏 𝒗 ⃗⃗𝟐𝒇 = 𝒎𝟏 𝒗 ⃗⃗𝟏𝒊 + 𝒎𝟐 𝒗 ⃗⃗𝟐𝒊 (7-9) This result is a special case of the law of conservation of momentum and is true of isolated systems containing any number of interacting objects. Linear Momentum 180 General Physics Module Phys 1011 AAU When no net external force acts on a system, the total momentum of the system remains constant in time. Defining the isolated system is an important feature of applying this conservation law. A cheerleader jumping upwards from rest might appear to violate conservation of momentum, because initially her momentum is zero and suddenly she’s leaving the ground with velocity v. The flaw in this reasoning lies in the fact that the cheerleader isn’t an isolated system. In jumping, she exerts a downward force on Earth, changing its momentum. This change in Earth’s momentum isn’t noticeable, however, because of Earth’s gargantuan mass compared to the cheerleader’s. When we define the system to be the cheerleader and Earth, momentum is conserved. Action and reaction, together with the accompanying exchange of momentum between two objects, is responsible for the phenomenon known as recoil. Everyone knows that throwing a baseball while standing straight up, without bracing one’s feet against Earth, is a good way to fall over backwards. This reaction, an example of recoil, also happens when you fire a gun or shoot an arrow. Conservation of momentum provides a straightforward way to calculate such effects, as the next example shows. Example 7.3 The Archer: An archer stands at rest Figure 7-5(Example on frictionless ice; his total mass including his bow 7.3) An archer fires and quiver of arrows is 60.00 kg. (See Fig. 7.5.) (a) an arrow If the archer fires a 0.030 0-kg arrow horizontally horizontally to the at 50.0 m/s in the positive x - direction, what is his right. Because he is subsequent velocity across the ice? (b) He then standing on fires a second identical arrow at the same speed frictionless ice, he relative to the ground but at an angle of 30.0° will begin to slide above the horizontal. Find his new speed. (c) to the left across Estimate the average normal force acting on the the ice. archer as the second arrow is accelerated by the bowstring. Assume a draw length of 0.800 m Solution (a) Find the archer’s subsequent velocity across the ice. Write the conservation of momentum equation for the x - direction.𝒑𝒊 = 𝒑𝒇 Let m1 and v1f be the archer’s mass and velocity after firing the arrow, respectively, and m2 and v2fthe arrow’s mass and velocity. Both velocities are in the x - direction. Substitute pi= 0 and expressions for the final momenta: 𝟎 = 𝒎𝟏 𝒗𝟏𝐟 + 𝒎𝟐 𝒗𝟐𝒇 Solve for v1fand substitute m1 = 59.97 kg, m2 = 0.030 0 kg, and v2f = 50.0 m/s: 𝒎𝟐 𝟎. 𝟎𝟑𝟎𝟎 𝐤𝐠 𝐦 𝐦 𝒗𝟏𝐟 = − 𝒗𝟐𝒇 = − ( ) 𝟓𝟎. 𝟎 = −𝟎. 𝟎𝟐𝟓𝟎 𝒎𝟏 𝟓𝟗. 𝟗𝟕 𝐤𝐠 𝐬 𝐬 (b) Calculate the archer’s velocity after he fires a second arrow at an angle of 30.0° above the horizontal. Write the x - component of the momentum equation with m1 again the archer’s mass after firing the first arrow as in part (a) and m2 the mass of the next arrow: 𝒎𝟏 𝒗𝟏𝐢 = (𝒎𝟏 − 𝒎𝟐 )𝒗𝟏𝐟 + 𝒎𝟐 𝒗𝟐𝒇 𝐜𝐨𝐬 𝜽 Linear Momentum 181 General Physics Module Phys 1011 AAU Solve for v1f, the archer’s final velocity, and substitute: m1 = 59.97 kg, m2 = 0.030 0 kg, and v2f = 50.0 m/s, 𝒗𝟏𝐢 = 𝟎. 𝟎𝟐𝟓𝟎 𝐦/𝐬 𝒎𝟏 𝒎𝟐 𝒗𝟏𝐟 = 𝒗𝟏𝐢 − 𝒗 𝐜𝐨𝐬 𝜽 (𝒎𝟏 − 𝒎𝟐 ) (𝒎𝟏 − 𝒎𝟐 ) 𝟐𝒇 𝟓𝟗. 𝟗𝟕 𝐤𝐠 𝐦 𝟎. 𝟎𝟑𝟎𝟎 𝐤𝐠 𝐦 𝐦 𝒗𝟏𝒇 = (−𝟎. 𝟎𝟐𝟓𝟎 ) − (𝟓𝟎 ) 𝐜𝐨𝐬 𝟑𝟎 = −𝟎. 𝟎𝟒𝟔𝟕 𝟓𝟗. 𝟗𝟒 𝐤𝐠 𝐬 𝟓𝟗. 𝟗𝟒 𝐤𝐠 𝐬 𝐬 (c) Estimate the average normal force acting on the archer as the arrow is accelerated by the bowstring. Use kinematics in one dimension to estimate the acceleration of the arrow: 𝒗𝟐 − 𝒗𝟐𝟎 = 𝟐𝒂∆𝒙 Solve for the acceleration and substitute values setting v=v2f, the final velocity of the arrow: 𝒂 = 𝒗𝟐 −𝒗𝟐𝟎 (𝟓𝟎 𝐦/𝐬)𝟐 +𝟎 + = 𝟏. 𝟓𝟔 × 𝟏𝟎𝟑 𝐦/𝐬 𝟐 𝟐∆𝒙 𝟐(𝟎.𝟖𝟎𝟎 𝐦) Find the time the arrow is accelerated using 𝒗 − 𝒗𝟎 𝟓𝟎 𝐦/𝐬 𝒗 = 𝒂𝒕 + 𝒗𝟎 : 𝒕 = = = 𝟎. 𝟎𝟑𝟐𝟎 𝐬: 𝒂 𝟏. 𝟓𝟔 × 𝟏𝟎𝟑 𝐦/𝐬𝟐 Write the y - component of the impulse–momentum theorem: ∆𝒑𝒚 𝒎𝟐 𝒗𝟐𝒇 𝐬𝐢𝐧 𝜽 𝑭𝒚,𝒂𝒗 ∆𝒕 = ∆𝒑𝒚 , 𝑭𝒚,𝒂𝒗 = ∆𝒕 = ∆𝒕 (𝟎. 𝟎𝟑𝟎𝟎 𝐤𝐠)(𝟓𝟎. 𝟎 𝐦/𝐬 ) 𝐬𝐢𝐧 𝟑𝟎𝟎 𝑭𝒚,𝒂𝒗 = = 𝟐𝟑. 𝟒 𝐍 𝟎. 𝟎𝟑𝟐𝟎 𝐬 The average normal force is given by the archer’s weight plus the reaction force R of the arrow on the archer: ∑ 𝑭𝒚 = 𝑵 − 𝒎𝒈 − 𝑹 = 𝟎, 𝐦 𝑵 = 𝒎𝒈 + 𝑹 = (𝟓𝟗. 𝟗𝟒 𝐤𝐠) (𝟗. 𝟖𝟎 ) + (𝟐𝟑. 𝟒 𝐍) = 𝟔. 𝟏𝟏 × 𝟏𝟎𝟐 𝐍 𝐬𝟐 REMARKS: The negative sign on 𝒗𝟏𝐟 indicates that the archer is moving opposite the arrow’s direction, in accordance with Newton’s third law. Because the archer is much more massive than the arrow, his acceleration and velocity are much smaller than the acceleration and velocity of the arrow. A technical point: the second arrow was fired at the same velocity relative to the ground, but because the archer was moving backwards at the time, it was travelling slightly faster than the first arrow relative to the archer. Velocities must always be given relative to a frame of reference. Notice that conservation of momentum was effective in leading to a solution in parts (a) and (b). The final answer for the normal force is only an average because the force exerted on the arrow is unlikely to be constant. If the ice really were frictionless, the archer would have trouble standing. In general, the coefficient of static friction of ice is more than sufficient to prevent sliding in response to such small recoils. Exercise 1. Would firing a heavier arrow necessarily increase the recoil velocity? Explain. 2. A 70.0-kg man and a 55.0-kg woman holding a 2.50-kg purse on ice skates stand facing each other. (a) If the woman pushes the man backwards so that his final speed is 1.50 m/s, with what average force did she push him, assuming they were in contact for 0.500 s? (b) What is the woman’s recoil speed? (c) If she now throws her 2.50-kg purse at him at a 20.0° angle Linear Momentum 182 General Physics Module Phys 1011 AAU above the horizontal and at 4.20 m/s relative to the ground, what is her subsequent speed? ANSWERS (a) 2.10 × 102 N(b) 1.83 m/s (c) 2.09 m/s 7.3 Collisions in One Dimension and in Two Dimensions Learning outcome After completing this section, students are expected to: Identify the types of collision Define inelastic collision. Explain perfectly inelastic collision Apply an understanding of collisions to everyday life Describe an elastic collision of two objects in one dimension. Solve simple problems involving collisions We have seen that for any type of collision, the total momentum of the system just before the collision equals the total momentum just after the collision as long as the system may be considered isolated. The total kinetic energy, on the other hand, is generally not conserved in a collision because some of the kinetic energy is converted to internal energy, sound energy, and the work needed to permanently deform the objects involved, such as cars in a car crash. We define an inelastic collision as a collision in which momentum is conserved, but kinetic energy is not. The collision of a rubber ball with a hard surface is inelastic, because some of the kinetic energy is lost when the ball is deformed during contact with the surface. When two objects collide and stick together, the collision is called perfectly inelastic. For example, if two pieces of putty collide, they stick together and move with some common velocity after the collision. If a meteorite collides head on with Earth, it becomes buried in Earth and the collision is considered perfectly inelastic. Only in very special circumstances is all the initial kinetic energy lost in a perfectly inelastic collision. An elastic collision is defined as one in which both momentum and kinetic energy are conserved. Billiard ball collisions and the collisions of air molecules with the walls of a container at ordinary temperatures are highly elastic. Macroscopic collisions such as those between billiard balls are only approximately elastic, because some loss of kinetic energy takes place—for example, in the clicking sound when two balls strike each other. Perfectly elastic collisions do occur, however, between atomic and subatomic particles. Elastic and perfectly inelastic collisions are limiting cases; most actual collisions fall into a range in between them. Linear Momentum 183 General Physics Module Phys 1011 AAU As a practical application, an inelastic collision is used to detect glaucoma, a disease in which the pressure inside the eye builds up and leads to blindness by damaging the cells of the retina. In this application, medical professionals use a device called a tonometer to measure the pressure inside the eye. This device releases a puff of air against the outer surface of the eye and measures the speed of the air after reflection from the eye. At normal pressure, the eye is slightly spongy, and the pulse is reflected at low speed. As the pressure inside the eye increases, the outer surface becomes more rigid, and the speed of the reflected pulse increases. In this way, the speed of the reflected puff of air can measure the internal pressure of the eye. We can summarize the types of collisions as follows: ▪ In an elastic collision, both momentum and kinetic energy are conserved. ▪ In an inelastic collision, momentum is conserved but kinetic energy is not. ▪ In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same. In the remainder of this section, we will treat perfectly inelastic collisions and elastic collisions in one dimension Figure 7-6(a) Before and (b) after a perfectly inelastic head-on collision between two objects. 7.3.1 Perfectly Inelastic Collisions Consider two objects having masses m1 and m2 moving with known initial velocity components v1iand v2ialong a straight line, as in Figure 7.6a. If the two objects collide head-on, stick together, and move with a common velocity component vf after the collision, then the collision is perfectly inelastic (Fig. 7.6b). Because the total momentum of the two-object isolated system before the collision equals the total momentum of the combined-object system after the collision, we can solve for the final velocity using conservation of momentum alone: 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = (𝑚1 + 𝑚2 )𝑣𝑓 (7-10) 𝑚1 𝑣1𝑖 +𝑚2 𝑣2𝑖 𝑣𝑓 = (𝑚1 +𝑚2 ) (7-11) It’s important to notice that 𝒗𝟏𝒊 , 𝒗𝟐𝒊 , and 𝒗𝒇 represent the x - components of the velocity vectors, so care is needed in entering their known values, particularly with regard to signs. For example, in Figure 7.6a, 𝒗𝟏𝒊 would have a positive value (m1 moving to the right), whereas 𝒗𝟐𝒊 would have a Linear Momentum 184 General Physics Module Phys 1011 AAU negative value (m2 moving to the left). Once these values are entered, Equation (7.11) can be used to find the correct final velocity, as shown in Examples 7.4 and 7.5. Example 7.4 A Truck Versus a Compact: A pickup truck with mass 𝟏. 𝟖𝟎 × 𝟏𝟎𝟑 𝐤𝐠 traveling eastbound at +15.0 m/s, while a compact car with mass 𝟗. 𝟎𝟎 × 𝟏𝟎𝟐 𝐤𝐠 is traveling westbound at -15.0 m/s. (See Fig. 7.7.) The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision. (b) Find the change in the velocity of each vehicle. (c) Find the change in the kinetic energy of the system consisting of both vehicles. Figure 7-7(Example 7.4) Solution (a) Find the final speed after collision. Let m1 and v1i represent the mass and initial velocity of the pickup truck, while m2 and v2ipertain to the compact. Apply conservation of momentum: 𝒑𝒊 = 𝒑𝒇 Substitute the values and solve for the final velocity, 𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = (𝒎𝟏 + 𝒎𝟐 )𝒗𝒇 𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 (𝟏. 𝟖𝟎 × 𝟏𝟎𝟑 𝐤𝐠)(𝟏𝟓. 𝟎 𝐦/𝐬) + (𝟗. 𝟎𝟎 × 𝟏𝟎𝟐 𝐤𝐠)(𝟐𝟏𝟓. 𝟎 𝐦/𝐬) 𝒗𝒇 = = (𝒎𝟏 + 𝒎𝟐 ) (𝟏. 𝟖𝟎 × 𝟏𝟎𝟑 𝐤𝐠 + 𝟗. 𝟎𝟎 × 𝟏𝟎𝟐 𝐤𝐠) = 𝟓. 𝟎 𝐦/𝐬 (b) Find the change in velocity for each vehicle. Change in velocity of the pickup truck is 𝐦 𝐦 ∆𝒗𝟏 = 𝒗𝒇 − 𝒗𝟏𝒊 = 𝟓. 𝟎𝟎 − 𝟏𝟓. 𝟎 = −𝟏𝟎. 𝟎 𝐦/𝐬 𝐬 𝐬 Change in velocity of the compact car is 𝐦 𝐦 ∆𝒗𝟐 = 𝒗𝒇 − 𝒗𝟐𝒊 = 𝟓. 𝟎𝟎 − (−𝟏𝟓. 𝟎) = +𝟐𝟎. 𝟎 𝐦/ 𝐬 𝐬 (c) Find the change in kinetic energy of the system. Calculate the initial kinetic energy of the system: 𝟏 𝟏 𝑬𝑲𝒊 = 𝟐(𝒎𝟏 𝒗𝟐𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝟐𝒊 ) = 𝟐 ((𝟏. 𝟖𝟎 × 𝟏𝟎𝟑 𝐤𝐠)(𝟏𝟓. 𝟎 𝐦/𝐬)𝟐 + (𝟗. 𝟎 × 𝟏𝟎𝟐 𝐤𝐠)(𝟐𝟏𝟓. 𝟎 𝐦/𝐬)𝟐 ) Linear Momentum 185 General Physics Module Phys 1011 AAU 𝑬𝑲𝒊 = 𝟑𝟑. 𝟎𝟒 × 𝟏𝟎𝟓 𝐉 Calculate the final kinetic energy of the system and the change in kinetic energy, 𝟏 𝟏 𝑬𝑲𝒇 = 𝟐(𝒎𝟏 + 𝒎𝟐 )𝒗𝟐𝒇 = 𝟐 ((𝟏. 𝟖𝟎 × 𝟏𝟎𝟑 𝐤𝐠)+(𝟗. 𝟎 × 𝟏𝟎𝟐 𝐤𝐠)(𝟓. 𝟎 𝐦/𝐬)𝟐 ) 𝑬𝑲𝒇 = 𝟑. 𝟑𝟖 × 𝟏𝟎𝟒 𝐉 ∆𝑬𝑲𝒊 = 𝑬𝑲𝒇 − 𝑬𝑲𝒊 = −𝟐. 𝟕𝟎 × 𝟏𝟎𝟓 𝐉 REMARKS: During the collision, the system lost almost 90% of its kinetic energy. The change in velocity of the pickup truck was only 10.0 m/s, compared to twice that for the compact car. This example underscores perhaps the most important safety feature of any car: its mass. Injury is caused by a change in velocity, and the more massive vehicle undergoes a smaller velocity change in a typical accident. Exercises 1. If the mass of both vehicles were doubled, how would the final velocity be affected? The change in kinetic energy? 2. Suppose the same two vehicles are both travelling eastward, the compact car leading the pickup truck. The driver of the compact car slams on the brakes suddenly, slowing the vehicle to 6.00 m/s. If the pickup truck travelling at 18.0 m/s crashes into the compact car, find (a) the speed of the system right after the collision, assuming the two vehicles become entangled, (b) the change in velocity for both vehicles, and (c) the change in kinetic energy of the system, from the instant before impact (when the compact car is travelling at 6.00 m/s) to the instant right after the collision. ANSWERS (a) 14.0 m/s (b) pickup truck: ∆𝒗𝟏 = −𝟒. 𝟎 𝐦/𝐬 compact car: ∆𝒗𝟐 = 𝟖. 𝟎 𝐦/𝐬(c) −𝟒. 𝟑𝟐 × 𝟏𝟎𝟒 J Example 7.5 The Ballistic Pendulum The ballistic pendulum (Fig. 7.8) is a device used to measure the speed of a fast - moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet embeds in the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the Figure 7-8 Example 7.5(Example 7.5) (a) technique, assume that the mass of the bullet, m1, is 5.00 Diagram of a ballistic pendulum. Note g, the mass of the pendulum, m2, is 1.000 kg, and h is 5.00 that 𝒗⃗⃗𝒔𝒚𝒔 is the velocity of the system cm. (a) Find the velocity of the system after the bullet just after the perfectly inelastic collision embeds in the block. (b) Calculate the initial speed of the bullet Linear Momentum 186 General Physics Module Phys 1011 AAU Solution (a) Find the velocity of the system after the bullet embeds in the block. Apply conservation of energy to the block–bullet system after the collision: (𝑬𝑲 + 𝑬𝑷 )𝐚𝐟𝐭𝐞𝐫 𝐜𝐨𝐥𝐥𝐢𝐬𝐢𝐨𝐧 = (𝑬𝑲 + 𝑬𝑷 )𝐭𝐨𝐩 Substitute expressions for the kinetic and potential energies. Note that both the potential energy at the bottom and the kinetic energy at the top are zero: 𝟏 𝟐 (𝒎𝟏 + 𝒎𝟐 )𝒗𝟐𝒔𝒚𝒔 + 𝟎 = 𝟎 + (𝒎𝟏 + 𝒎𝟐 )𝒈𝒉 Solve for the final velocity of the block–bullet system, 𝒗𝒔𝒚𝒔 from 𝒗𝟐𝒔𝒚𝒔 = 𝟐𝒈𝒉, 𝒗𝒔𝒚𝒔 = √𝟐(𝟗. 𝟖 𝐦/𝐬 𝟐 )𝟓 × 𝟏𝟎−𝟐 𝐦 = 𝟎. 𝟗𝟗 𝐦/𝐬 (b) Calculate the initial speed of the bullet. Write the conservation of momentum equation and substitute expressions. 𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = (𝒎𝟏 + 𝒎𝟐 )𝒗𝒔𝒚𝒔 Solve for the initial velocity of the bullet, and substitute values: 𝒎𝟏 + 𝒎𝟐 (𝟏. 𝟎𝟎𝟓 𝐤𝐠)(𝟎. 𝟗𝟗𝟎 𝐦/𝐬) 𝒗𝟏𝒊 = 𝒗𝒔𝒚𝒔 = = 𝟏𝟗𝟗 𝐦/𝐬 𝒎𝟏 𝟓. 𝟎𝟎 × 𝟏𝟎−𝟑 𝐤𝐠 REMARKS Because the impact is inelastic, it would be incorrect to equate the initial kinetic energy of the incoming bullet to the final gravitational potential energy associated with the bullet–block combination. The energy isn’t conserved! Exercises 1. List three ways by which mechanical energy can be lost from the system in this experiment. 2. A bullet with mass 5.00 g is fired horizontally into a 2.000-kg block attached to a horizontal spring. The spring has a constant 6.00 × 102 N/m and reaches a maximum compression of 6.00 cm. (a) Find the initial speed of the bullet–block system. (b) Find the speed of the bullet. ANSWERS (a) 1.04 m/s (b) 417 m/s 7.3.2 Elastic Collisions Now consider two objects that undergo an elastic head-on collision (Fig. 7.9). In this situation, both the momentum and the kinetic energy of the system of two objects are conserved. We can write these conditions as 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓 (7-12) and 1 2 2 2 2 𝑚1 𝑣1𝑖 2 + 12𝑚2 𝑣2𝑖 = 12𝑚1 𝑣1𝑓 + 12𝑚2 𝑣2𝑓 (7-13) Linear Momentum 187 General Physics Module Phys 1011 AAU where v is positive if an object moves to the right and negative if it moves to the left. In a typical problem involving elastic collisions, there are two unknown quantities, and Equations (7.12) and (7.13) can be solved simultaneously to find them. These two equations are linear and quadratic, respectively. An alternate approach simplifies the quadratic equation to another linear equation, facilitating solution. Cancelling the factor ½ in Equation (7.11), we rewrite the equation as 2 2 2 2 𝑚1 (𝑣1𝑖 −𝑣1𝑓 ) + 𝑚2 = 𝑚2 (𝑣2𝑓 − 𝑣2𝑖 ) Here we have moved the terms containing m1 to one side of the equation and those containing m2 to the other. Next, we factor both sides of the equation: 𝑚1 (𝑣1𝑖 + 𝑣1𝑓 )(𝑣1𝑖 − 𝑣1𝑓 ) = 𝑚2 (𝑣2𝑓 − 𝑣2𝑖 )(𝑣2𝑓 + 𝑣2𝑖 ) (7-14) Now we separate the terms containing m1 and m2 in the equation for the conservation of momentum (7.12) to get 𝑚1 (𝑣1𝑖 − 𝑣1𝑓 ) = 𝑚2 (𝑣2𝑓 − 𝑣2𝑖 ) (7-15) Next, we divide Equation (7.14) by Equation (7.15), producing 𝑣1𝑖 + 𝑣1𝑓 = 𝑣2𝑓 + 𝑣2𝑖 (7-16) Gathering initial and final values on opposite sides of the equation gives 𝑣1𝑖 − 𝑣2𝑖 = −(𝑣1𝑓 − 𝑣2𝑓 ). (7-17) This equation, in combination with Equation (7.12), will be used to solve problems dealing with perfectly elastic head-on collisions. Equation (7.16) shows that the sum of the initial and final velocities for object 1 equals the sum of the initial and final velocities for object 2. According to Equation (7.17), the relative velocity of the two objects before the collision, 𝑣1𝑖 − 𝑣2𝑖 , equals the negative of the relative velocity of the two objects after the collision, −(𝑣1𝑓 − 𝑣2𝑓 ). To better understand the equation, imagine that you are riding along on one of the objects. As you measure the velocity of the other object from your vantage point, you will be measuring the relative velocity of the two objects. In your view of the collision, the other object comes toward you and bounces off, leaving the collision with the same speed, but in the opposite direction. This is just what Equation (7.17) states. Linear Momentum 188 General Physics Module Phys 1011 AAU Before an elastic collision the two objects move After the collision the object velocities change, independently. but both the energy and momentum of the system are conserved. Figure 7-9(a) Before and (b) after an elastic head - on collision between two hard spheres. Unlike an inelastic collision, both the total momentum and the total energy are conserved. ROBLEM-SOLVING STRATEGY - One-Dimensional Collisions The following procedure is recommended for solving one - dimensional problems involving collisions between two objects: 1. Coordinates. Choose a coordinate axis that lies along the direction of motion. 2. Diagram. Sketch the problem, representing the two objects as blocks and labelling velocity vectors and masses. 3. Conservation of Momentum. Write a general expression for the total momentum of the system of two objects before and after the collision, and equate the two, as in Equation (7.12). On the next line, fill in the known values. 4. Conservation of Energy. If the collision is elastic, write a general expression for the total energy before and after the collision, and equate the two quantities, as in Equation (7.13) or (preferably) Equation (7.17). Fill in the known values. (Skip this step if the collision is not perfectly elastic.) 5. Solve the equations simultaneously. Equations (7.12) and (7.17) form a system of two linear equations and two unknowns. If you have forgotten Equation (7.17), use Equation (7.13) instead. Steps 1 and 2 of the problem-solving strategy are generally carried out in the process of sketching and labelling a diagram of the problem. This is clearly the case in our next example, EXAMPLE 7.6 LET’S PLAY POOL Two billiard balls of identical mass move toward each other as in Figure 7.9, with the positive x - axis to the right (steps 1 and 2). Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are 130.0 cm/s and 220.0 cm/s, what are the velocities of the balls after the collision? Assume friction and rotation are unimportant. SOLUTION Write the conservation of momentum equation. Because 𝑚1 = 𝑚2 , we can cancel the masses, then substitute 𝑣1𝑖 = +30.0 cm/s and 𝑣2𝑖 = −20.0 cm/s (Step 3): 𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = 𝒎𝟏 𝒗𝟏𝒇 + 𝒎𝟐 𝒗𝟐𝒇 𝟑𝟎. 𝟎𝐜𝐦/𝐬 − 𝟐𝟎. 𝟎𝐜𝐦/𝐬 = 𝒗𝟏𝒇 + 𝒗𝟐𝒇 Linear Momentum 189 General Physics Module Phys 1011 AAU (1) 𝟏𝟎. 𝟎𝐜𝐦/𝐬 = 𝒗𝟏𝒇 + 𝒗𝟐𝒇 Next, apply conservation of energy in the form of Equation 7.17 (Step 4): (2) 𝒗𝟏𝒊 − 𝒗𝟐𝒊 = −(𝒗𝟏𝒇 − 𝒗𝟐𝒇 ) 𝟑𝟎. 𝟎𝐜𝐦/𝐬 − (−𝟐𝟎. 𝟎𝐜𝐦/𝐬) = −(𝒗𝟏𝒇 − 𝒗𝟐𝒇 ) (3) 𝟓𝟎. 𝟎𝐜𝐦/𝐬 = −(𝒗𝟏𝒇 − 𝒗𝟐𝒇 ) Now solve Equations (1) and (3) simultaneously by adding them together (Step 5): 𝟐𝒗𝟐𝒇 = 𝟔𝟎. 𝟎𝐜𝐦/𝐬𝒗𝟐𝒇 = 𝟑𝟎. 𝟎𝐜𝐦/𝐬 Substitute the answer for 𝒗𝟐𝒇 into Equation (1) to get 𝒗𝟏𝒇 : 𝟏𝟎. 𝟎𝐜𝐦/𝐬 = 𝒗𝟏𝒇 + 𝟑𝟎. 𝟎𝐜𝐦/𝐬𝒗𝟏𝒇 = −𝟐𝟎. 𝟎𝐜𝐦/𝐬 REMARKS Notice the balls exchanged velocities—almost as if they’d passed through each other. This is always the case when two objects of equal mass undergo an elastic head-on collision. QUESTION 7.6 In this example, is it possible to adjust the initial velocities of the balls so that both are at rest after the collision? Explain. EXERCISE 7.6 Find the final velocities of the two balls if the ball with initial velocity 𝑣2𝑖 = −20cm/s has a mass equal to onehalf that of the ball with initial velocity 𝑣1𝑖 = +30.0 cm/s. ANSWER 𝑣1𝑓 = −3.33cm/s; 𝑣2𝑓 = +46.7cm/s 7.3.3 Two dimensional (Glancing) Collisions In Section 7.2 we showed that the total linear momentum of a system is conserved when the system is isolated (i.e., when no external forces act on the system). For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved. However, an important subset of collisions takes place in a plane. The game of billiards is a familiar example involving multiple collisions of objects moving on a two-dimensional surface. We restrict our attention to a single two-dimensional collision between two objects that takes place in a plane, and ignore any possible rotation. For such collisions, we obtain two component equations for the conservation of momentum: 𝒎𝟏 𝒗𝟏𝒊𝒙 + 𝒎𝟐 𝒗𝟐𝒊𝒙 = 𝒎𝟏 𝒗𝟏𝒇𝒙 + 𝒎𝟐 𝒗𝟐𝒇𝒙 𝒎𝟏 𝒗𝟏𝒊𝒚 + 𝒎𝟐 𝒗𝟐𝒊𝒚 = 𝒎𝟏 𝒗𝟏𝒇𝒚 + 𝒎𝟐 𝒗𝟐𝒇𝒚 We must use three subscripts in this general equation, to represent, respectively, (1) the object in question, and (2) the initial and final values of the components of velocity. Now, consider a two-dimensional problem in which an object of mass 𝑚1 collides with an object of mass 𝑚2 that is initially at rest, as in Figure 7.10. After the collision, object 1 moves at an angle 𝜽 with respect to the horizontal, and object 2 moves at an angle φ with respect to the horizontal. This is called a glancing collision. Applying the law of conservation of momentum in component form, and noting that the initial y-component of momentum is zero, we have x - component: x − component 𝑚1 𝑣1𝑖 + 0 = 𝑚1 𝑣1𝑓 cos 𝜃 + 𝑚2 𝑣2𝑓 cos 𝜑 (7-18) Linear Momentum 190 General Physics Module Phys 1011 AAU y − component 0 + 0 = 𝑚1 𝑣1𝑓 sin 𝜃 + 𝑚2 𝑣2𝑓 sin 𝜑 (7-19) If the collision is elastic, we can write a third equation, for conservation of energy, in the form 1 2 2 2 𝑚1 𝑣1𝑖 2 = 12𝑚1 𝑣1𝑓 + 12𝑚2 𝑣2𝑓 (7-20) If we know the initial velocity 𝑣1𝑖 and the masses, we are left with four unknowns (𝑣1𝑓 , 𝑣2𝑓 , 𝜃, and 𝜑). Because we have only three equations, one of the four remaining quantities must be given in order to determine the motion after the collision from conservation principles alone. If the collision is inelastic, the kinetic energy of the system is not conserved, and Equation (7.20) does not apply. Before the collision After the collision Figure 7-10A glancing collision between two objects. Example 7.7: Collision at an Intersection A car with mass 1.50 × 103 kg travelling east at a speed of 25.0 m/s collides at an intersection with a 2.50 × 103 - kg pickup truck travelling north at a speed of 20.0 m/s, as shown in Figure 7.11. Find the magnitude and direction of the velocity of the wreckage immediately after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and Figure 7-11(Example 7.7) A top view of the road can be neglected. a perfectly inelastic collision between a car and a pickup truck Solution Find the x - components of the initial and final total momenta: 𝐦 ∑ 𝒑𝒙𝒊 = 𝒎𝐜𝐚𝐫 𝒗𝐜𝐚𝐫 = (𝟏. 𝟓𝟎 × 𝟏𝟎𝟑 𝐤𝐠) (𝟐𝟓. 𝟎 ) = 𝟑. 𝟕𝟓 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 𝐬 Linear Momentum 191 General Physics Module Phys 1011 AAU ∑ 𝒑𝒙𝒇 = (𝒎𝐜𝐚𝐫 + 𝒎𝐭𝐫𝐮𝐤 )𝒗𝒇 𝐜𝐨𝐬 𝜽 = (𝟒 × 𝟏𝟎𝟑 𝐤𝐠)𝒗𝒇 𝐜𝐨𝐬 𝜽 Set the initial x - momentum equal to the final x - momentum: (1) 𝟑. 𝟕𝟓 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 = (𝟒 × 𝟏𝟎𝟑 𝐤𝐠)𝒗𝒇 𝐜𝐨𝐬 𝜽 Set the initial y - momentum equal to the final y - momentum: ∑ 𝒑𝒚𝒊 = 𝒎𝐭𝐫𝐮𝐤 𝒗𝐭𝐫𝐮𝐤 = (𝟐. 𝟓 × 𝟏𝟎𝟑 𝐤𝐠)(𝟐𝟎. 𝟎 𝐦/𝐬) = 𝟓. 𝟎𝟎 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 ∑ 𝒑𝒚𝒇 = (𝒎𝐜𝐚𝐫 + 𝒎𝐭𝐫𝐮𝐤 )𝒗𝒇 𝐬𝐢𝐧 𝜽 = (𝟒 × 𝟏𝟎𝟑 𝐤𝐠)𝒗𝒇 𝐬𝐢𝐧 𝜽 Set the initial y - momentum equal to the final y - momentum: (2) 𝟓. 𝟎𝟎 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 = (𝟒 × 𝟏𝟎𝟑 𝐤𝐠)𝒗𝒇 𝐬𝐢𝐧 𝜽 Divide Equation (2) by Equation (1) and solve for 𝜽: 𝐦 𝟓.𝟎𝟎 ×𝟏𝟎𝟒 𝐤𝐠 𝐭𝐚𝐧 𝛉 = 𝐦 𝐬 = 𝟏. 𝟓𝟓, 𝛉 = 𝟓𝟑. 𝟏° 𝟑.𝟕𝟓 ×𝟏𝟎𝟒 𝐤𝐠 𝐬 Substitute this angle back into Equation (2) to find vf: 𝟓. 𝟎𝟎 × 𝟏𝟎𝟒 𝐤𝐠 𝐦/𝐬 𝒗𝒇 = = 𝟏𝟓. 𝟔 𝐦/𝐬 (𝟒 × 𝟏𝟎𝟑 𝐤𝐠) 𝐬𝐢𝐧 𝟓𝟑. 𝟏° REMARKS: It’s also possible to first find the x - and y - components vfx and vfy of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, 𝒗𝒇 = √𝒗𝟐𝒇𝒙 + 𝒗𝟐𝒇𝒚 , and the inverse tangent function 𝜽 = 𝐭𝐚𝐧−𝟏 (𝒗𝒇𝒚 /𝒗𝒇𝒙 ). Setting up this alternate approach is a simple matter of substituting 𝒗𝒇𝒙 = 𝒗𝒇 𝐜𝐨𝐬 𝜽and 𝒗𝒇𝒚 = 𝒗𝒇 𝐬𝐢𝐧 𝜽 into Equations (1) and (2). Exercise 1. If the car and truck had identical mass and speed, what would the resultant angle have been? 2. A 3.00 - kg object initially moving in the positive x - direction with a velocity of +5.00 m/s collides with and sticks to a 2.00-kg object initially moving in the negative y - direction with a velocity of –3.00 m/s. Find the final components of velocity of the composite object. Answer 𝑣𝑓𝑥 = 3.00 m/s; 𝑣𝑓𝑥 = −1.20 m/s 7.4 The Concept of Center of Mass Learning outcome After completing this section, students are expected to: Define point masses. Linear Momentum 192 General Physics Module Phys 1011 AAU Define the center of mass of an object. Apply the concept of center of mass to everyday life Solve simple problems involving momentum and collisions of rigid bodies In this section, we describe the overall motion of a system in terms of a special point called the center of mass of the system. The system can be either a small number of particles or an extended, continuous object, such as a gymnast leaping through the air. We shall see that the translational motion of the center of mass of the system is the same as if all the mass of the system were concentrated at that point. That is, the system moves as if the net external force were applied to a single particle located at the center of mass. This model, the particle model, was introduced in Chapter 2. This behavior is independent of other motion, such as rotation or vibration of the system or deformation of the system (for instance, when a gymnast folds her body). Consider a system consisting of a pair of particles that have different masses and are connected by a light, rigid rod (Fig. 7.12). The position of the center of mass of a system can be described as being the average position of the system’s mass. The center of mass of the system is located somewhere on the line joining the two particles and is closer to the particle having the larger mass. If a single force is applied at a point on the rod above the center of mass, the system rotates clockwise (see Fig. 7.12a). If the force is applied at a point on the rod below the center of mass, the system rotates counterclockwise (see Fig. 7.12b). If the force is applied at the center of mass, the system moves in the direction of the force without rotating (see Fig. 7.12c). The center of mass of an object can be located with this procedure. Figure 7-13The center of mass of two particles of unequal mass on the x axis is located at 𝑥𝐶𝑀 , a Figure 7-12A force is applied to a system of two particles of unequal point between the mass connected by a light, rigid rod.. particles, closer to the one having the larger mass. The center of mass of the pair of particles described in Figure 7.13 is located on the x axis and lies somewhere between the particles. Its x coordinate is given by 𝒎𝟏 𝒙𝟏 +𝒎𝟐 𝒙𝟐 𝒙𝑪𝑴 = 𝒎𝟏 +𝒎𝟐 (7-21) 2 For example, if 𝑥1 = 0, 𝑥2 = 𝑑, and 𝑚2 = 2𝑚1 , we find that 𝑥𝐶𝑀 = 𝑑. That is, the center of mass 3 lies closer to the more massive particle. If the two masses are equal, the center of mass lies midway between the particles. Linear Momentum 193 General Physics Module Phys 1011 AAU We can extend this concept to a system of many particles with masses mi in three dimensions. The x coordinate of the center of mass of n particles is defined to be 𝒎𝟏 𝒙𝟏 +𝒎𝟐 𝒙𝟐 +𝒎𝟑 𝒙𝟑 +⋯+𝒎𝒏 𝒙𝒏 ∑𝒊 𝒎𝒊 𝒙𝒊 𝟏 𝒙𝑪𝑴 = 𝒎𝟏 +𝒎𝟐 +𝒎𝟑 +⋯+𝒎𝒏 𝒙𝒏 = ∑𝒊 𝒎𝒊 = 𝑴 ∑𝒊 𝒎𝒊 𝒙𝒊 (7-22) where 𝑥𝑖 is the x coordinate of the ith particle and the total mass is 𝑀 = ∑𝑖 𝑚𝑖 where the sum runs over all n particles. The y and z coordinates of the center of mass are similarly defined by the equations 1 1 𝑦𝐶𝑀 = ∑ 𝑚𝑦, 𝑧𝐶𝑀 = ∑ 𝑚𝑧 (7-23) 𝑀 𝑖 𝑖 𝑖 𝑀 𝑖 𝑖 𝑖 The center of mass can be located in three dimensions by its position vector 𝑟⃗CM. The components of this vector are 𝑥𝐶𝑀 , 𝑦𝐶𝑀 , and 𝑧𝐶𝑀 , defined in Equations (7.22) and (7.23). Therefore, 𝟏 ̂= ⃗𝒓⃗𝑪𝑴 = 𝒚𝑪𝑴 𝒊̂ + 𝒚𝑪𝑴 𝒋̂ + 𝒛𝑪𝑴 𝒌 ̂) (∑ 𝒎𝒊 𝒙𝒊 𝒊̂ + ∑ 𝒎𝒊 𝒚𝒊 𝒋̂ + ∑ 𝒎𝒊 𝒛𝒊 𝒌 𝑴 𝒊 𝒊 𝒊 or 1 𝑟⃗𝐶𝑀 = ∑ 𝑚 𝑟⃗ (7-24) 𝑀 𝑖 𝑖 𝑖 Where 𝑟⃗𝑖 is the position vector of the ith particle, defined by 𝑟⃗𝑖 = 𝑦𝑖 𝑖̂ + 𝑦𝑖 𝑗̂ + 𝑧𝑖 𝑘̂. It’s often possible to guess the location of the center of mass. The center of mass of a homogeneous, symmetric body must lie on the axis of symmetry. For example, the center of mass of a homogeneous rod lies midway between the ends of the rod, and the center of mass of a homogeneous sphere or a homogeneous cube lies at the geometric center of the object. Several examples involve homogeneous, symmetric objects where the centers of mass coincide with their geometric centers. A rigid object in a uniform gravitational field can be balanced by a single force equal in magnitude to the weight of the object, as long as the force is directed upward through the object’s center of mass. Example 7.8: Where is the Center Of Mass? (a) Three objects are located in a coordinate system as shown in Figure 7.14a. Find the center of mass. (b) How does the answer change if the object on the left is displaced upward by 1.00 m and the object on the right is displaced downward by 0.500 m (Fig. 7.14b)? Treat the objects as point particles Linear Momentum 194 General Physics Module Phys 1011 AAU Figure 7-14(Example 7.8) Locating the center of mass of a system of three particles. Solution (a) Find the center of mass of the system in Figure 7.13a. Apply Equation (7.22) to the system of three objects: ∑𝒊 𝒎𝒊 𝒙𝒊 𝒎𝟏 𝒙𝟏 +𝒎𝟐 𝒙𝟐 +𝒎𝟑 𝒙𝟑 (1) 𝒙𝑪𝑴 = ∑𝒊 𝒎𝒊 = 𝒎𝟏 +𝒎𝟐 +𝒎𝟑 , Compute the numerator and the denominator of Equation (1): ∑𝒊 𝒎𝒊 𝒙𝒊 = 𝟏. 𝟓𝟎 𝐤𝐠 𝐦, ∑𝒊 𝒎𝒊 = 𝟏𝟏. 𝟎𝟎 𝐤𝐠 , 𝟏.𝟓𝟎 𝐤𝐠 𝐦 substitute into Equation (1).𝒙𝑪𝑴 = 𝟏𝟏.𝟎𝟎 𝐤𝐠 = 𝟎. 𝟏𝟑𝟔 𝐦 b) How does the answer change if the positions of the objects are changed as in Figure 7.13b? Because the x-coordinates have not been changed, the x-coordinate of the center of mass is also unchanged: ∑𝒊 𝒎𝒊 𝒚𝒊 𝒎𝟏 𝒚𝟏 + 𝒎𝟐 𝒚𝟐 + 𝒎𝟑 𝒚𝟑 𝒚𝑪𝑴 = = ∑𝒊 𝒎𝒊 𝒎𝟏 + 𝒎𝟐 + 𝒎𝟑 ∑ 𝒎𝒊 𝒚𝒊 = (𝟓. 𝟎𝟎 𝐤𝐠)(𝟏. 𝟎𝟎 𝐦) + (𝟐. 𝟎𝟎 𝐤𝐠)(𝟎 𝐦) + (𝟒. 𝟎𝟎 𝐤𝐠)(−𝟎. 𝟓𝟎𝟎𝐦) = 𝟓. 𝟎𝟎 𝐤𝐠 𝐦 𝒊 𝟑. 𝟎𝟎 𝐤𝐠 𝐦 𝒙𝑪𝑴 = = 𝟎. 𝟐𝟕𝟑 𝐦 𝟏𝟏. 𝟎𝟎 𝐤𝐠 REMARKS: Notice that translating objects in the y - direction doesn’t change the x-coordinate of the center of mass. The three components of the center of mass are each independent of the other two coordinates. Exercises 1. If 1.00 kg is added to the masses on the left and right in Figure 7.13a, does the center of mass (a) move to the left, (b) move to the right, or (c) remain in the same position? 2. If a fourth particle of mass 2.00 kg is placed at (0, 0.25 m) in Figure 7.13a, find the x- and y - coordinates of the center of mass for this system of four particles. Answer xcm 5 0.115 m; ycm 5 0.038 5 m Linear Momentum 195 General Physics Module Phys 1011 AAU Example 7.9 Tug of War at The Ice Fishing Hole Bob and Sherry are lying on the ice, a fishing hole of radius 1.00 m cut in the ice halfway between them. A rope of length 10.0 m lies between them, and they both grip it and begin pulling, as in Figure 7.15a. Bob has mass of mB= 85.0 kg and Sherry has mass of mS= 48.0 kg, so Sherry reaches the hole, first. Where is Bob at that time? Assume the hole is centered on the origin and that Bob and Sherry start at xB= 5.00 m and xS= –5.00 m, respectively. Neglect forces of friction. Figure 7-15 (Example 7.9) (a) Bob and Sherry engage in a tug - of war.(b) (Exercise 7.9) Solution Calculate the center of mass using the initial positions: 𝒎𝑺 𝒙𝑺 + 𝒎𝑩 𝒙𝑩 (𝟒𝟖. 𝟎 𝐤𝐠)(−𝟓. 𝟎𝟎𝐦) + (𝟖𝟓. 𝟎 𝐤𝐠)(𝟓. 𝟎𝟎𝐦) 𝒙𝑪𝑴 = = = 𝟏. 𝟑𝟗 𝐦 𝒎𝑺 + 𝒎𝑩 𝟒𝟖. 𝟎 𝐤𝐠 + 𝟖𝟓. 𝟎 𝐤𝐠 Find Bob’s position using the center of mass equation and the fact that Sherry reaches the hole first: 𝒎𝑺 𝒙𝑺 + 𝒎𝑩 𝒙𝑩 (𝟒𝟖. 𝟎 𝐤𝐠)(−𝟏. 𝟎𝟎𝐦) + (𝟖𝟓. 𝟎 𝐤𝐠)𝒙𝑩 𝟏. 𝟑𝟗 𝐦 = = 𝒎𝑺 + 𝒎𝑩 𝟏𝟑𝟑 𝐤𝐠 𝒙𝑩 = 𝟐. 𝟕𝟒 𝐦 REMARKS: So when Sherry reaches the edge of the fishing hole, Bob is still nearly two meters away from the edge on his side of the hole. Exercises 1. How do the speeds vS of Sherry and vB of Bob compare during the motion? 2. A man of mass M 5 75.0 kg is standing in a canoe of mass 40.0 kg that is 5.00 m long, as in Figure 7.15b. The far end of the canoe is next to a dock. From a position 0.500 m from his end of the canoe, he walks to the same position at the other end of the canoe. (a) Find the center of mass of the canoe–man system, taking the end of the dock as the origin. (b) Neglecting drag forces, how far is he from the dock? (Hint: the final location of the canoe’s center of mass will be 2.00 m farther from the dock than the man’s final position, which is unknown.) ANSWERS (a) 3.80 m (b) 3.10 m Linear Momentum 196 General Physics Module Phys 1011 AAU 7.5 Summary ⃗⃗ of an object of mass m moving with velocity 𝒗 The linear momentum 𝒑 ⃗⃗ is defined as 𝑝⃗ ≡ 𝑚𝑣⃗ Momentum carries units of kg # m/s. The impulse 𝑰⃗of a constant force 𝑭⃗⃗ delivered to an object is equal to the product of the force and the time interval during which the force acts: 𝐼⃗ ≡ 𝐹⃗ Δ𝑡 These two concepts are unified in the impulse– momentum theorem, which states that the impulse of a constant force delivered to an object is equal to the change in momentum of the object: 𝐼⃗ = 𝐹⃗ Δ𝑡 = Δ𝑝⃗ = 𝑚𝑣⃗f − 𝑚𝑣⃗i Solving problems with this theorem often involves estimating speeds or contact times (or both), leading to an average force. When no net external force acts on an isolated system, the total momentum of the system is constant. This principle is called conservation of momentum. In particular, if the isolated system consists of two objects undergoing a collision, the total momentum of the system is the same before and after the collision (Fig. 7.4). Conservation of momentum can be written mathematically for this case as 𝑚1 𝑣⃗1𝑓 +𝑚2 𝑣⃗2𝑓 = 𝑚1 𝑣⃗1𝑖 + 𝑚2 𝑣⃗2𝑖 Collision and recoil problems typically require finding unknown velocities in one or two dimensions. Each vector component gives an equation, and the resulting equations are solved simultaneously. In an inelastic collision, the momentum of the system is conserved, but kinetic energy is not. In a perfectly inelastic collision, the colliding objects stick together and the common final velocity is given by 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 𝑣𝑓 =. (𝑚1 + 𝑚2 ) In an elastic collision, both the momentum and the kinetic energy of the system are conserved. A one - dimensional elastic collision between two objects can be solved by using the conservation of momentum and conservation of energy equations: 𝒎𝟏 𝒗𝟏𝒊 + 𝒎𝟐 𝒗𝟐𝒊 = 𝒎𝟏 𝒗𝟏𝒇 + 𝒎𝟐 𝒗𝟐𝒇 and 𝟏 𝟏 𝟏 𝟏 𝒎 𝒗𝟐 𝟐 𝟏 𝟏𝒊 + 𝟐𝒎𝟐 𝒗𝟐𝟐𝒊 = 𝟐𝒎𝟏 𝒗𝟐𝟏𝒇 + 𝟐𝒎𝟐 𝒗𝟐𝟐𝒇 Linear Momentum 197