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Chapter 6 Linear Momentum and Collisions In everyday language, the term “momentum” refers to a tendency to “keep moving”. The same is true in classical mechanics, where momentum (and the conservation thereof) is often used to analyze situations where two or more objects interact with each other. In the photo, a runaway train (with lots of momentum) interacted with the wall of the station, with a predictable outcome. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 1 6.1 Linear Momentum Linear momentum, given the symbol p, is a property of mass that is in motion. By definition, the linear momentum of an object is the product of its mass and velocity Or, in mathematical terms, p = 𝑚v Since it is the product of a mass and a velocity, linear momentum has SI units of kg·m/s. Unlike the Joule or Watt, there is no common name for this unit. A couple of notes on momentum: (1) although it involves both m and v, linear momentum isn’t the same as kinetic energy, which depends on the square of the speed. (2) Unlike kinetic energy, linear momentum is a vector property of motion. It has the same direction as velocity. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 2 6.1 Linear Momentum Since p = 𝑚v is a vector equation, we can split it into equivalent scalar equations in each direction, just as we did with Newton’s second law: 𝑝𝑥 = 𝑚𝑣𝑥 , 𝑝𝑦 = 𝑚𝑣𝑦 , 𝑝𝑧 = 𝑚𝑣𝑧 It’s worth noting that linear momentum is often referred to simply as “momentum”. We will try to avoid this lazy terminology, since there is another type of momentum (angular momentum), which we will cover in the next chapter. We will often need to find the total linear momentum P of a system of several masses. This is simply the vector sum of each of the individual linear momenta: P = p1 + p2 + p3 + ⋯ PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 3 6.1 Linear Momentum Many physical problems can be analyzed by considering changes in linear momentum. A non-zero ∆p can arise from a change in speed or a change in direction (or a change in mass…this is rare, but in some circumstances it may be relevant). Two examples of changes in momentum (due to an object rebounding from a wall) are shown here. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 4 6.1 Linear Momentum Force and Momentum Let’s return to Newton’s 2nd law. Assuming that mass remains constant, we know that – at least over a short time – acceleration is a = (v − v0 )/∆𝑡. Therefore, F𝑛𝑒𝑡 𝑚 v − v0 𝑚v − 𝑚v0 𝐩 − 𝐩0 ∆𝐩 = 𝑚a = = = = ∆𝑡 ∆𝑡 ∆𝑡 ∆𝑡 In other words, the net external force acting on an object is equal to the time rate of change of the object’s momentum. The preceding equation represents an equivalent form of Newton’s second law – in fact, this is the form in which Newton first presented the law. It’s also applicable in more situations, for reasons that we won’t need in PC141. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 5 6.1 Linear Momentum Force and Momentum The preceding equation tells us that the following statements are equivalent: 1. A non-zero net force results in a non-zero acceleration 2. A non-zero net force results in a change in momentum. Projectile motion provides an example. As we know, the horizontal velocity is constant. Therefore, the horizontal linear momentum is constant, which is required, since there is no horizontal force in this case. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 6 Problem #1: Net Force WBL LP 6.3 A net force on an object can cause… PC141 Intersession 2013 A …an acceleration B …a change in momentum C …a change in velocity D …all of the preceding Day 12 – June 4 – WBL 6.1-6.3 Slide 7 Problem #2: Baseball WBL Ex 6.5 A 0.150-kg baseball traveling with a horizontal speed of 4.50 m/s is hit by a bat and then moves with a speed of 34.7 m/s in the opposite direction. What is the change in the ball’s momentum? Note: It’s always good to get a sense for “reasonable” numbers…if a baseball is traveling toward home plate at 4.50 m/s, it certainly wasn’t pitched…a pitched baseball would be traveling closer to 40 m/s. Perhaps this one was lobbed by a nearby coach. Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 8 Problem #3: Bad Truck Driver WBL Ex 6.13 A loaded tractor-trailer with a total mass of 5000 kg traveling at 3.0 km/h hits a loading dock and comes to a stop in 0.64 s. What is the magnitude of the average force exerted on the truck by the dock? Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 9 6.2 Impulse Much of this chapter is concerned with collisions, in which two objects (at least one of which is initially in motion), are in contact for a period of time†. Examples are the contact between a baseball and bat, or between two balls on a snooker table. In most of these cases, the force is not constant as a function of time – it increases from zero to some maximum value, then decreases back to zero. The total time during which the force is non-zero is usually on the order of milliseconds. The figure shows one example of a force – vs – time graph. †technically speaking, a “collision” doesn’t have to actually involve physical contact between objects. As we know from chapter 4, there exist “noncontact” forces as well. The math is the same, but we’ll avoid these situations in PC141. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 10 6.2 Impulse Even though the force may vary with time, we can still use the “new” version of Newton’s 2nd law from a few slides ago, if we use average values: F𝑎𝑣𝑔 = ∆𝐩 . ∆𝑡 This can be rearranged as F𝑎𝑣𝑔 ∆𝑡 = ∆𝐩 The term F𝑎𝑣𝑔 ∆𝑡 is known as the impulse of the force. It is a vector property, with the symbol 𝐈. As the product of a force and a time, it has units of N·s. The preceding equation indicates that during a collision, the impulse exerted on an object is equal to the change in the object’s linear momentum. This is referred to as the impulsemomentum theorem. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 11 6.2 Impulse Recall from chapter 5 that the area under a graph of force vs. position is equal to the net work done on an object (which, furthermore, is equal to the change in the object’s kinetic energy). Here, the area under a graph of force vs. time is seen to be equal to the impulse on the object, which is furthermore equal to the change in its linear momentum. Although in most cases the actual relation between force and time is complicated, using the average force facilitates finding the area under the curve. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 12 Problem #4: Astronaut WBL Ex 6.25 An astronaut (mass of 100 kg, with equipment) is headed back to her space station at a speed of 0.750 m/s but at the wrong angle. To correct her direction, she fires rockets from her backpack at right angles to her motion for a brief time. These directional rockets exert a constant force of 100.0 N for only 0.200 s. a) What is the magnitude of the impulse delivered to the astronaut? b) What is her new direction (relative to the initial direction)? c) What is her new speed? Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 13 Problem #5: Pop Fly WBL Ex 6.36 A baseball player pops a pitch straight up. The ball (mass 200 g) was traveling horizontally at 35.0 m/s just before contact with the bat, and 20.0 m/s just after contact. Determine the direction and magnitude of the impulse delivered to the ball by the bat. Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 14 6.3 Conservation of Linear Momentum In chapter 5, we learned that energy is a conserved quantity (in that the total energy of the universe is constant). For systems smaller than the entire universe, there are conditions to this conservation – it only holds true for systems that are closed, and that contain no nonconservative forces. We also learned that when nonconservative forces are present, the total mechanical energy changes only by the extent that the nonconservative forces do net work on the system. Likewise, under certain conditions, the total linear momentum of a system is a conserved quantity. This fact allows us to analyze the dynamics of objects that undergo collisions. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 15 6.3 Conservation of Linear Momentum If the linear momentum of an object is to be conserved, we simply require that the net force acting on it be zero. In other words, if ∆𝐩 F𝑛𝑒𝑡 = =0 ∆𝑡 then ∆p = p𝑓 − p𝑖 = 0. This conservation law also holds true for systems of objects, in which we simply write that ∆P = P𝑓 − P𝑖 = 0 if F𝑛𝑒𝑡 = 0. The condition of zero net force is met by closed systems. Note that such systems may have internal forces, but these always act as Newton’s 3rd-law force pairs. Since they are equal and opposite to each other, they always sum to zero when calculating the net force. Examples will be discussed in class. PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 16 Problem #6: Conserved Linear Momentum WBL LP 6.9 The linear momentum of an object is conserved if PC141 Intersession 2013 A the force acting on the object is conservative B a single, unbalanced internal force is acting on the object C the mechanical energy is conserved D none of the preceding Day 12 – June 4 – WBL 6.1-6.3 Slide 17 Problem #7: Really Bad Ice Skaters WBL Ex 6.43 Two ice skaters not paying attention collide in a completely inelastic collision. Prior to the collision, skater 1, with a mass of 60 kg, has a velocity of 50 km/h eastward, and moves at a right angle to skater 2, who has a mass of 75 kg and a velocity of 7.5 km/h southward. What is the velocity of the skaters after the collision? Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 18 Problem #8: Exploding Bomb WBL Ex 6.47 An explosion of a 10.0-kg bomb releases only two separate pieces. The bomb was initially at rest and a 4.00-kg piece travels westward at 100 m/s immediately after the explosion. a) What are the speed and direction of the other piece immediately after the explosion? b) How much kinetic energy was released in this explosion? Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 19 Problem #9: Ballistic Pendulum WBL Ex 6.55 A ballistic pendulum is a device used to measure the velocity of a projectile – for example, the muzzle velocity of a rifle bullet. The projectile is shot horizontally into, and becomes embedded in, the bob of a pendulum. The pendulum swings upward to some height h, which is measured. The masses of the block and bullet are known. Show that the initial velocity of the projectile is given by 𝒎+𝑴 𝒗𝟎 = 𝟐𝒈𝒉 𝒎 Solution: In class PC141 Intersession 2013 Day 12 – June 4 – WBL 6.1-6.3 Slide 20