Physics 121 for Majors Schedule HW #5 is due Friday Quiz #2 is due 9/29 Lab #2 is due Monday Midterm 1 is 10/2 in the classroom. Class 6 and Collisions Relative Velocity Last Class Natural motion is straight-line motion at constant speed. Experimentally, motion is momentum, p = mv. Momentum is conserved in all collisions. There are three types of collision problem: Totally inelastic -- objects stick together (or bomb -- an object breaks into pieces) -- kinetic energy is also conserved Some kinetic energy is lost, but the masses don t stick together We learned about totally inelastic collisions Impulse is the change in momentum of an object Physical particles such as electrons have wave characteristics. Momentum Conservation Momentum is conserved in all collisions! Today s Class Kinetic energy is the energy of motion, K = mv. Momentum is conserved in all collisions. There are three types of collision problem: Totally inelastic -- objects stick together some kinetic energy is lost, but objects don t stick together -- kinetic energy is also conserved Impulse is equal and opposite for two colliding particles, no matter what type of collision occurs. 1
Section 1 Kinetic Energy Gottfried Wilhelm von Leibniz 1646-1716 Co-invented calculus with Isaac Newton Newton s bitter rival Vis Viva The force of life Vis viva = mass velocity 2 Vis viva is always positive Kinetic Energy We now recognize that vis viva and kinetic energy are (essentially) the same thing. K K = 1 2 mv But only conserved in certain collisions called elastic collisions! But only conserved in certain collisions called elastic collisions! We ll learn a lot more about energy later. Units of Energy K = 1 2 mv The SI unit of energy is the joule (J) kg m 1 J = 1 s Collision Types After the collision, there are three different types of outcomes: Totally inelastic collision - the objects stick together. Collision Kinetic energy is conserved. Collision Kinetic energy in = Kinetic energy out + Energy lost 2
Section 2 Mathematica Solutions Collision Equations in 1-D m v + m v = p = (m + m )v and m v + m v = p = m v + m v 1 2 m v + 1 2 m v = K = 1 2 m v + 1 2 m v + E E = 0 for elastic collisions Collision Equations in 1-D p = (m + m )v and p = m v + m v K = 1 2 m v + 1 2 m v + E Mathematica Solutions The conservation equations for onedimensional collisions are a little hard to solve because one is linear and the other is quadratic. When using Mathematica, it is best to substitute in numbers early on. E = 0 for elastic collisions Scattering This is really easy with or without Mathematica as only one linear equation is needed. Scattering 3
Scattering Paper Solutions While Mathematica is easiest, it s instructive to look at how these elastic and totally inelastic equations can be solved by hand. These methods have application to 2-D and 3-D collisions and also help us understand some of the foundations of relativity. Section 3 Zero-Momentum Frame Zero Momentum Frame Collisions are one of the most confusing topics in Physics 121. We ll start out with the simplest possible collisions, ones in which the total momentum is zero before the collision. This is called the zero-momentum frame or the center of mass frame. Don t worry about what center of mass means yet. What is the total momentum after the collision? Reference Frames A frame of reference is a set of observers with rulers and synchronized clocks all moving together. An inertial reference frame is one where the observers are not accelerating. We ll only consider inertial reference frames. Collision Types p=0 frame After the collision, there are three different types of outcomes: Collision - the objects bounce back with the same speed they came in. Collision - the objects bounce back slower than they came in. (We ll do this one later - it s a bit hard.) Totally inelastic collision - the objects stick together. This is the maximum possible energy loss. 4
Collision Collision The objects stick together after the collision. Collision There are two objects with masses m 1 =1 kg and m 2 = 3 kg. m 1 moves to the right and has a velocity of 6 m/s. m 2 moves to the left and has a velocity of 2 m/s. The system initially has zero total momentum. Collision In the zero-momentum frame, the final object must have zero momentum.... That s not too hard! m v + m v = 0 If the objects stick together making one object of mass M=m 1 +m 2 = 4 kg, how fast does it move? Note: velocity can be negative and momentum can be negative, but the equation has a plus sign! Collision Collision There are two objects with masses m 1 =1 kg and m 2 = 3 kg. m 1 moves to the right and has a velocity of 6 m/s. m 2 moves to the left and has a velocity of 2 m/s. The system initially has zero total momentum. After an elastic collision, the speeds are the same as before the collision. How can that be? 5
Collision If the total momentum is zero for two objects that collide elastically, each has the same speed after the collision as before the collision. Convince yourself that momentum and kinetic energy are conserved in this collision! Collision If the total momentum is zero for two objects that collide elastically, each has the same speed after the collision as before the collision, but the direction reverses. But that s only true when the total momentum is zero! m v + m v = 0 = m v + m v Collision Collision, but not totally inelastic! If the total momentum is zero for two objects that collide inelastically, the final speed of each object is the same fraction of the initial speed. (Each may have 60% of its initial speed, for example. One cannot have 70% of its initial speed and the other 50% of its initial speed. ) But that s only true when the total momentum is zero! Collision If the total momentum is zero for two objects that collide inelastically, after the collision, each has the same fraction of the speed it had before the collision. Collision There are two objects with masses m 1 =1 kg and m 2 = 3 kg. m 1 moves to the right and has a velocity of 6 m/s. m 2 moves to the left and has a velocity of 2 m/s. The system initially has zero total momentum. After an elastic collision, the speeds are 80% of the initial speeds. Is momentum conserved? Is kinetic energy conserved? m v + m v = 0 = m 0.8v + m 0.8v 6
Collision There are two objects with masses m 1 =1 kg and m 2 = 3 kg. m 1 moves to the right and has a velocity of 6 m/s. m 2 moves to the left and has a velocity of 2 m/s. The system initially has zero total momentum. 8.64 J of energy are lost in the collision. Collision The initial kinetic energy is The final kinetic energy is K = 1 2 m v + 1 2 m v K = 1 2 m f v + 1 2 m f v = f K K = f K If the velocity is smaller by a factor f, the kinetic energy is smaller by a factor f 2. Now what do you do? V = fv V = fv So in this problem 8.64 J = 1 f 24 J f = 0.8 Recap Zero Momentum Frame (Upper case for after the collision) Totally inelastic V = 0 (reverse direction) V = v V = v (reverse direction and slow down): Section 4 Velocity Addition V = fv V = fv K = f K But what if the total momentum isn t zero? Find a reference frame where it is zero and do your physics there! Relative Velocity Key Ideas There s no such thing as being at rest. Everything moves with respect to other things. As long as two people don t accelerate, the normal laws of physics must be valid for both. This is called The Principle of Relativity or Classical Relativity. If someone is playing baseball on a trailer bed moving at +50 mph, an observer on the ground behind the truck sees the same physics, but everything is moving with a velocity of +50 mph greater. What About Special Relativity? Let s say a truck is moving at 90% of the speed of light (v 1 =2.70 10 8 m/s) and a baseball is thrown at 90% of the speed of light relative to truck (v 2 =v 1 ). How fast does an observer on the ground see the baseball moving? The correct formula (c is the speed of light) is: v /c + v /c 1.8 v = c = c = 0.9945 c 1 + (v /c)(v /c) 1 + 0.9 We ll learn about this later 7
Section 5 Applying Velocity Addition Before the Collision Two masses approach each other in the lab: v 1 =9 m/s v 2 = 3 m/s What is the total momentum? 4 kg 9 m s 2kg 3 m/s = 30 kg m s If one object of mass m 1 +m 2 =6 kg had this momentum, what would be its velocity? 30 kg m s 6 kg = 5 m/s Before the Collision Two masses approach each other in the lab: v 1 =9 m/s v 2 = 3 m/s V=? Before the Collision Two masses approach each other in the lab: v 1 =9 m/s v 2 = 3 m/s V=? An observer is walking by with velocity V = 2 m/s. What velocity does she measure for each mass? 9 m s 2 m s = 7 m s 3 m s m s 2 = 5 m s The observer measures the total momentum to be zero. How fast is she walking? 4 kg 9 m s V + 2 kg 3 m s V = 0 V = 5 m s Before the Collision Two masses approach each other in the lab: v 1 =9 m/s v 2 = 3 m/s V=? Now what velocities does the observer measure? Verify that the momentum is zero. 4 kg 4 m s + 2 kg 8 m s = 0 Collision in the CM Frame v 1 =4 m/s v 2 = 8 m/s Before: V=0 What happens after each type of collision? 8
CM Collision v 1 =0 m/s v 2 = 0 m/s V=0 Lab Collision v 1 =5 m/s v 2 = 5 m/s V=5 m/s What happens in the lab frame? The two masses stick and both move with the velocity of the center of mass (the 5 m/s we found before). Collision Energy Table Frame K 1 K 2 K tot K loss Lab before 162 J 9 J 171 J Lab after 50 J 25 J 75 J 96 J CM before 32 J 64 J 96 J CM after 0 J 0 J 0 J 96 J CM Collision v 1 = 4 m/s v 2 =8 m/s V=0 The velocities reverse direction. What happens in the lab frame? Lab Collision v 1 =+1 m/s V=5 m/s v 2 =13 m/s Add the velocity of the center of mass (5 m/s) to each velocity found in the cm frame. Collision Energy Table Frame K 1 K 2 K tot K loss Lab before 162 J 9 J 171 J Lab after 2 J 169 J 171 J 0 J CM before 32 J 64 J 96 J CM after 32 J 64 J 96 J 0 J 9
CM Collision v 1 = 3 m/s V=0 v 2 =6 m/s Assume the final speeds are ¾ of the original speeds. Each velocity magnitude is reduced and its direction is reversed. What happens in the lab frame? Lab Collision v 1 =+2 m/s V=5 m/s v 2 =11 m/s Add the velocity of the center of mass (5 m/s) to each velocity found in the cm frame. Collision Energy Table Frame K 1 K 2 K tot K loss Lab before 162 J 9 J 171 J Lab after 8 J 121 J 129 J 42 J CM before 32 J 64 J 96 J CM after 18 J 36 J 54 J 42 J CM Collision v 1 = 4f m/s V=0 v 2 =8f m/s What happens in the lab frame? Assume that ½ the energy is lost. (This is harder!) The velocity magnitudes are reduced by a factor f and the directions are reversed. Lab Collision v 1 =5-4f m/s V=5 m/s v 2 =5+8f m/s Add the velocity of the center of mass (5 m/s) to each velocity found in the cm frame. Collision Energy Table Frame K 1 K 2 K tot K loss Lab before 162 J 9 J 171 J Lab after??? 48 J CM before 32 J 64 J 96 J CM after 32f 2 J 64f 2 J 96f 2 J 48 J 96f = 48 f = 1 2 10
CM Collision v 1 = 2.83 m/s V=0 v 2 =5.66 m/s Lab Collision v 1 =2.17 m/s V=5 m/s v 2 =10.66 m/s We use the value of f we found and redo everything! The velocity magnitudes are reduced by a factor f and the directions are reversed. Add the velocity of the center of mass (5 m/s) to each velocity found in the cm frame. Collision Energy Table Frame K 1 K 2 K tot K loss Lab before 162 J 9 J 171 J Lab after 9.43 J 113.57 J 123 J 48 J CM before 32 J 64 J 96 J CM after 16 J 32 J 48 J 48 J Section 6 Impulse Impulse -- Reminder Impulse is the change of an object s momentum (typically due to a collision or interaction with another object). J = p = p p Later on we ll see that impulse is a way of describing the net result of forces. In fact, we define force as F = p t Impulse and Momentum Conservation For now, we should note that in collisions, the impulses of the two colliding objects are equal and opposite so they sum to zero. This is equivalent to saying that momentum is conserved. (Why?) 11
Impulse Example Collison in the lab: v 1 =9 m/s v 2 = 3 m/s Before: p = 28 kg m s p = +28 kg m s Section 7 Recap v 1 =+2 m/s v 2 =11 m/s Big Ideas Kinetic energy is the energy of motion, K = mv. There are three kinds of collisions: p and K are conserved. p is conserved, but some K is lost Totally inelastic p is conserved, and the objects stick together, causing a maximal loss of K Collision problems are most easily done in the zeromomentum frame. Impulse is the change in momentum, J = p. Force is the rate of change of momentum, F = p t 12