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Physics 151: Lecture 19 Today’s Agenda Topics Example problem (conservation of momentum) Impulse Ch. 9.2 Center of Mass Ch. 9.6 Physics 151: Lecture 19, Pg 1 See text: 9-2 Conservation of Momentum rocket science final v+Dv video See Figure 12- Physics 151: Lecture 19, Pg 2 See text: 9-2 Example Problem : A object of mass m=0.200 kg is dropped from l =30.0 cm height above a basket base (M=0.200 kg) which is attached to the ceiling with a spring. If the spring is stretched by x = 0.05 m before, find the maximum distance the spring will stretch ? m M l x d m d = 0.182 m See Figure 12- Physics 151: Lecture 19, Pg 3 See text: 9-2 Force and Impulse The diagram shows the force vs time for a typical collision. The impulse, I, of the force is a vector defined as the integral of the force during the collision. F t I F dt Impulse I = area under this curve ! t Dt Impulse has units of Ns. See Figure 12- ti tf Physics 151: Lecture 19, Pg 4 See text: 9-2 Force and Impulse F Using dP dt the impulse becomes: t I F dt t dP dt dt F t dP Pf Pi DP I DP impulse = change in momentum ! See Figure 12- Dt t Physics 151: Lecture 19, Pg 5 See text: 9-2 Force and Impulse Two different collisions can have the same impulse since I depends only on the change in momentum, not the nature of the collision. F same area F Dt Dt big, F small t Dt t Dt small, F big Physics 151: Lecture 19, Pg 6 See text: 9-2 Force and Impulse soft spring F F stiff spring Dt t Dt big, F small Animation Dt t Dt small, F big Physics 151: Lecture 19, Pg 7 Lecture 19, ACT 1 Force & Impulse Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ? (a) heavier F light (b) lighter F (c) same heavy Physics 151: Lecture 19, Pg 8 Lecture 19, ACT 2 Force & Impulse What is the average force that wall exerts on ball (0.40kg) if duration of wall-ball contact is 0.01 s ? a) 20 N b) 200 N c) 2,000 N d) 20,000 N vi = 30m/s before collision vf = 20m/s after collision Physics 151: Lecture 19, Pg 9 Lecture 19, ACT 3 A 0.20 kg stone you throw rises 20.3 m in the air. The magnitude of the impulse the stone received from your hand while being thrown is a. 0.27 Ns. b . 2.7 Ns. c . 4.0 Ns. d . 9.6 Ns. e . 34.3 Ns. Physics 151: Lecture 19, Pg 10 System of Particles: Until now, we have considered the behavior of very simple systems (one or two masses). But real life is usually much more interesting ! For example, consider a simple rotating disk. An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object! Physics 151: Lecture 19, Pg 11 System of Particles: Center of Mass How do we describe the “position” of a system made up of many parts ? Define the Center of Mass (average position): For a collection of N individual pointlike particles whose masses and positions we know: N mi ri RCM i 1 M m2 m1 r1 RCM r2 y x (In this case, N = 2) Physics 151: Lecture 19, Pg 12 System of Particles: Center of Mass If the system is made up of only two particles: N mi ri RCM i 1 M So: m1r1 m2 r2 m1 m2 m1 m2 r1 m2 r2 r1 m1 m2 RCM r1 m2 r2 r1 M where M = m1 + m2 (r1 - r2) m1 r1 RCM m2 r2 y x Physics 151: Lecture 19, Pg 13 System of Particles: Center of Mass If the system is made up of only two particles: RCM r1 m2 r2 r1 M where M = m1 + m2 If m1 = m2 RCM 1 r1 r2 r1 2 the CM is halfway between the masses. r2 - r1 + m1 r1 RCM m2 r2 y x Physics 151: Lecture 19, Pg 14 System of Particles: Center of Mass The center of mass is where the system is balanced ! m1 + m2 m1 + m2 Physics 151: Lecture 19, Pg 15 See text: 9-6 Example Calculation: Consider the following mass distribution: i mi xi (m)0 (2 m)12 (m)24 12 M 4m i mi y i (m)0 (2 m)12 (m)0 6 M 4m X CM YCM RCM = (12,6) (12,12) 2m m (0,0) m (24,0) Physics 151: Lecture 19, Pg 16 See text: 9-6 System of Particles: Center of Mass For a continuous solid, we have to do an integral. dm y RCM r x rdm rdm M dm where dm is an infinitesimal mass element. video See example 8-4, A Triangular Plate Physics 151: Lecture 19, Pg 17 Example: Astronauts & Rope :-) :-( A male astronaut and a female astronaut are at rest in outer space and 20 meters apart. The male has 1.5 times the mass of the female. The female is right by the ship and the male is out in space a bit. The male wants to get back to the ship but his jet pack is broken. Conveniently, there is a rope connected between the two. So the guy starts pulling in the rope. Does he get back to the ship? Does he at least get to meet the woman? m M = 1.5m Physics 151: Lecture 19, Pg 18 Lecture 19, ACT 4 Center of Mass Motion A woman weighs exactly as much as her 20 foot long boat. Initially she stands in the center of the motionless boat, a distance of 20 feet from shore. Next she walks toward the shore until she gets to the end of the boat. What is her new distance from the shore. (There is no horizontal force on the boat by the water). 20 ft (a) 10 ft before 20 ft (b) 15 ft (c) 16.7 ft ? ft after Physics 151: Lecture 19, Pg 19 See text: 9.6 Center of Mass Motion: Review We have the following law for CM motion: FEXT MACM This has several interesting implications: It tell us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate F and A like we are used to doing. It tells us that if FEXT = 0, the total momentum of the system does not change. As the woman moved forward in the boat, the boat went backward to keep the center of mass at the same place. Physics 151: Lecture 19, Pg 20 Recap of today’s lecture Chapter 9, Center of Mass Elastic Collisions Impulse Physics 151: Lecture 19, Pg 21