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### PHY183-Lecture17pre

Course: PHY 183, Spring 2012
School: Michigan State University
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Word Count: 1694

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Momentum Linear Linear momentum is the product of mass (scalar) and velocity (vector) Physics for Scientists &amp; Engineers 1 p = mv Momentum vector and velocity vector are parallel to each other Magnitude of the linear momentum Spring Semester 2012 Lecture 17 Momentum and Collisions, Impulse, Conservation of Momentum p = mv February 14, 2012 1 Physics for Scientists&amp;Engineers 1 Momentum...

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Momentum Linear Linear momentum is the product of mass (scalar) and velocity (vector) Physics for Scientists & Engineers 1 p = mv Momentum vector and velocity vector are parallel to each other Magnitude of the linear momentum Spring Semester 2012 Lecture 17 Momentum and Collisions, Impulse, Conservation of Momentum p = mv February 14, 2012 1 Physics for Scientists&Engineers 1 Momentum and Force dv d p=m = ma = F dt dt K = 1 mv 2 2 Use p=mv and then obtain K= term is zero February 14, 2012 K= 3 Impulse tf and therefore: Then the impulse is simply: J = Fave t = m v This result seems rather trivial (the integral is still in the definition of the average force), but is very useful for practical purposes J = p ti February 14, 2012 Physics for Scientists&Engineers 1 4 i p x ,f dpx dt = dpx = px ,f px ,i px dt ti ti p x ,i tf tf pf dp F dt = dt = dp = pf pi p dt pi ti ti tf Impulse: J F dt Physics for Scientists&Engineers 1 Define the average force acting over a time interval t = tf - ti f Fdt = 1 f Fdt = 1 f Fdt Fave = i f t f t i i t i dt Obtain an expression for momentum change by going back to the relationship between momentum and force, and integrating both sides over time Fx dt = February 14, 2012 Impulse and Average Force Change in momentum (f = final, i = initial) p pf pi tf p2 2m We can reformulate all of mechanics we have studied thus far in terms of momentum instead of velocity dpy dp dpx ; Fy = ; Fz = z dt dt dt Physics for Scientists&Engineers 1 mv 2 m 2 v 2 p2 = = 2 2m 2m Find relationship between momentum and kinetic energy dp This Newtons Second Law for momentum F = dt In components: Fx = 2 We already know dv dm d d p = (mv ) = m + v dt dt dt dt If the mass is constant in time, Physics for Scientists&Engineers 1 Momentum and Kinetic Energy Take the time derivative of the definition of the momentum 2nd February 14, 2012 5 February 14, 2012 Physics for Scientists&Engineers 1 6 1 Example: A positive force and then a negative force acting on a mass Example: A positive force and then a negative force acting on a mass (2) A 0.442 kg object is at rest at the origin of a coordinate system. A 3.38 N force in the positive x direction acts on the object for 1.57 s. What is the velocity at the end of this interval? Use the relationship between impulse and linear momentum J = p F t = mv = m ( v f vi ) A 0.442 kg object is at rest at the origin of a coordinate system. A 3.38 N force in the positive x direction acts on the object for 1.57 s. What is the velocity at the end of this interval? ANS: +12.0m/s At the end of this interval, a constant force of 5.92 N is applied in the negative x direction for 3.65 s. What is the velocity at the end of the 3.65 s? F t =m ( v f vi ) F t F t v f vi = vf = + vi m m In this case, vi=+12.0 m/s F t 5.92 N 3.65s m m vf = + vf = + 12.0 = 36.9 m 0.442kg s s Starts from rest, so vi=0 F t 3.38 N 1.57 s m vf = = = +12.0 m 0.442kg s February 14, 2012 7 Physics for Scientists&Engineers 1 Baseball Home Run 9 Physics for Scientists&Engineers 1 ( 80.35 m/s)2 + ( 31.71 m/s )2 Start with the change in velocity in the x-direction vx = ( 49.17 m/s ) ( cos 35.0 ) ( 40.23 m/s ) ( cos185) = 80.35 m/s Now in the y-direction vy = ( 49.17 m/s ) ( sin 35.0 ) ( 40.23 m/s ) ( sin185) = 31.71 m/s February 14, 2012 10 Why can one hit for more power with less tension on the tennis racket strings? Answer: relationship between impulse, force, and time = 86.38 m/s p = mv = ( 0.145 kg ) ( 86.38 m/s ) = 12.5 kg m/s PROBLEM 2 The ball-bat contact lasts 1.2 ms What is the average force exerted by the bat? SOLUTION 2 The force can be calculated in terms of the impulse p = J = Fave t weight of baseball team: p 12.5 kg m/s Fave = = = 10.4 kN 9 ( 90 kg ) ( 9.81 m/s2 ) = 8 kN t 0.0012 s Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 Impulse and Sports The change in momentum is then February 14, 2012 8 Lets make a sketch Baseball Home Run The change in velocity of the baseball is 2 2 v = vx + vy = Physics for Scientists&Engineers 1 Baseball Home Run A pitcher throws a fastball, which crosses the plate with a speed of 90.0 mph (= 40.23 m/s) and an angle of 5.0 relative to the horizontal The batter hits it for a home run, launching it with 110.0 mph (= 49.17 m/s) at an angle of 35.0 to the horizontal Mass of the baseball = 5.10 ounces (= 0.145 kg) PROBLEM 1: What is the magnitude of the impulse the baseball receives from the bat? SOLUTION 1: The impulse is equal to the momentum change that the baseball receives Need to calculate the change in velocity then multiply by the mass February 14, 2012 February 14, 2012 11 J = Fave t If the average force is fixed, increasing the contact time between racket and ball increases impulse given to ball February 14, 2012 Physics for Scientists&Engineers 1 12 2 Impulse and Car Safety Devices Collisions Car collision: very large impulse (cannot be changed) Typical collision time is very short J Since Fave = , this means a very large force acting on the t driver All passive safety devices belt, (seat air bag, crash crumple zones) are designed to extend the time interval and thus to reduce the average force Suppose two objects collide with each other They might rebound away from each other like two billiard balls Elastic collision They might stick together like a subcompact car colliding with an 18-wheeler Totally inelastic collision Looking at the momenta, p1 and p2, of two objects during a collision, we find that pf1 + pf2 = pi1 + pi 2 Conservation of total momentum The most important result of this chapter Our second conservation law February 14, 2012 Physics for Scientists&Engineers 1 13 Momentum Conservation Derivation Use definition of impulse and its relationship to momentum change we get tf F12 dt = p2 = pf 2 pi2 ( i means initial, f means final) Integration of this equation gives us tf tf tf 0 = ( F21 + F1 2 ) dt = F21 dt + F1 2 dt = pf1 pi1 + pf2 pi2 ti During a collision object 2 exerts a force on object 1 F21 So we get 2 1 ti dt = p1 = pf1 pi1 Physics for Scientists&Engineers 1 15 Momentum Conservation Derivation (3) F ext dt << p1 = p2 ti Physics for Scientists&Engineers 1 February 14, 2012 Physics for Scientists&Engineers 1 16 Totally Elastic Collisions in 1d Conversely, the forces between these two collision partners are very large, compare home run example February 14, 2012 ti pf1 + pf2 = pi1 + pi2 Only condition used for derivation is Newtons Third Law What about other, external forces? Neglected them in the derivation Why is this justified? Collision times are very short, and during these short times the impulse due to external forces can be neglected tf ti Collecting the initial momentum vectors on one side and final momentum vectors on the other side gives us ti February 14, 2012 14 Newtons Third Law tells us that these forces are equal and opposite to each other F12 = F21 F12 + F21 = 0 F 2 1 F Physics for Scientists&Engineers 1 Momentum Conservation Derivation (2) During a collision object 1 exerts a force on object 2 tf February 14, 2012 17 Video sequence shows collision between two carts on a (approximately) frictionless track. Equal time intervals Velocities before and after collision are marked by colored lines February 14, 2012 Physics for Scientists&Engineers 1 18 3 Totally Elastic Collisions in 1d (2) Totally Elastic Collisions in 1d (3) A totally elastic collision is an ideal case in which two conditions are fulfilled: Total momentum is conserved (valid for all collisions) We can also obtain expression for velocities instead of momenta, by using pi1,x=m1vi1,x, Expression for final velocities from initial ones pf1,x + pf2,x = pi1,x + pi2,x Total kinetic energy is conserved (only elastic collisions) 2 2 2 2 pf1,x pf2,x pi1,x pi2,x + = + 2 m1 2 m2 2 m1 2 m2 Result: can calculate momenta after collision m m2 2 m1 pi1,x + pi2,x pf1,x = 1 m1 + m2 m1 + m2 2 m2 m m1 pi1,x + 2 pi2,x pf2,x = m1 + m2 m1 + m2 February 14, 2012 vf2, x Go through Derivation 7.2 in the book Physics for Scientists&Engineers 1 19 Totally Elastic Collisions in 1d (4) m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x ) m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x Kinetic energy result m1 ( vf1,x vi1,x )( vf1,x + vi1,x ) = m2 ( vi2,x vf2,x )( vi2,x + vf2,x ) m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x ) Divide kinetic energy result by momentum result Total kinetic energy is conserved vf1,x + vi1,x = vi2,x + vf2,x 1 1 1 1 2 2 2 2 m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x 2 2 2 2 Rearrange to get Rearrange and simplify 2 2 2 2 2 2 2 2 m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x ) m1 ( vf1,x vi1,x )( vf1,x + vi1,x ) = m2 ( vi2,x vf2,x )( vi2,x + vf2,x ) Physics for Scientists&Engineers 1 20 Physics for Scientists&Engineers 1 Momentum result pf1,x + pf2,x = pi1,x + pi2,x February 14, 2012 February 14, 2012 Totally Elastic Collisions in 1d (5) Total momentum is conserved Rearrange m1 m2 2 m2 vi1, x + vi2.x m1 + m2 m1 + m2 2 m1 m m1 = vi1,x + 2 vi2, x m1 + m2 m1 + m2 vf1, x = 21 vf1, x vf2, x = vi1, x + vi2, x Collisions change the sign of the relative velocity! vf1, x vf2, x = (vi1, x vi2, x ) February 14, 2012 22 Physics for Scientists&Engineers 1 Example: P024 Example: P024 (2) y x p = mv v1, x = v1 sin 1 v1, y = v1 cos 1 v2, x = v2 sin 2 v2, y = v2 cos 2 p = p2 p1 px = mv2, x mv1, x = m ( v2 sin 2 v1 sin 1 ) p y = mv2, y mv1, y = m ( v2 cos 2 ( v1 cos 1 ) ) = m ( v2 cos 2 + v1 cos 1 ) px = ( 0.250 kg ) ( ( 2.20 m/s ) sin 57.0 ( 2.40 m/s ) sin 42.0 ) = 0.05979 kg m/s p y = ( 0.250 kg ) ( ( 2.20 m/s ) cos 57.0 + ( 2.40 m/s ) cos 42.0 ) = 0.7454 kg m/s p = February 14, 2012 Physics for Scientists&Engineers 1 23 February 14, 2012 ( p x ) + ( p x ) 2 2 = ( 0.05979 kg m/s ) + ( 0.7454 kg m/s ) 2 Physics for Scientists&Engineers 1 2 = 0.748 kg m/s 24 4 Example: P024 (3) p = p2 p1 p2 = p1 + p 2 px = 0.05979 kg m/s p y = 0.7454 kg m/s p y 1 0.7454 kg m/s = tan = 85.4 0.05979 kg m/s p x = tan 1 February 14, 2012 Physics for Scientists&Engineers 1 1 25 5
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Michigan State University - PHY - 183
Physics for Scientists &amp;Engineers 1Spring Semester 2012Lecture 18Special Cases for Elastic 1d Collisions, 2D CollisionsFebruary 15, 2012Physics for Scientists&amp;Engineers 11ElasticElastic Scattering Yesterday we introduced the following for the el
Michigan State University - PHY - 183
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Michigan State University - PHY - 183
Center of Mass and Center of Gravity In our previous discussions this semester we have alwaystreated the location of objects as a single point A car is located at x = 5.5 m Of course, the car is not ALL at x =5.5 m The natural choice for the location
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Circular Motion Circular motion is motion along the perimeter of a circle Circular motion is surprisingly common CD, DVD, Blu-ray, Indy-car racing, carousel, ferris wheel, etc.Physics for Scientists &amp;Engineers 1Spring Semester 2012Lecture 23Circul
Michigan State University - PHY - 183
ClickerClicker Quiz A horizontal table rotates counterclockwise withconstant angular velocity. The components of thelinear acceleration are at (tangential) and ac(centripetal). Select the correct statement withrespect to the components of the accele
Michigan State University - PHY - 183
Example: Hammer Throw Throw the hammer, a 12 cm diameter ballattached to a grip by a steel cable, a maximumdistance The hammers total length is 121.5 cm, and itstotal weight is 7.26 kg (4 kg for women) The athlete has to accomplish the throw whilen
Michigan State University - PHY - 183
Rotation We have been treating bodies by looking at themotion of the center of mass We introduced the concept of motion of the centerof mass + rotation about the center of mass We have covered circular motion of the center ofmass Now we will study
Michigan State University - PHY - 183
Rolling without Slipping Lets assume that we have a round object withradius R rolling without slipping This object has special relationships between itslinear and angular quantitiesPhysics for Scientists &amp;Engineers 1 Displacement and angular displa
Michigan State University - PHY - 183
Torque So far in our discussion of forces, we have shownthat forces can cause linear motion of objects We described the motion of these objects in termsof the motion of the center of mass of the object However, we have not addressed one generalquest
Michigan State University - PHY - 183
Newtons Second Law for Rotation Newtons Second Law for Rotation: = IPhysics for Scientists &amp;Engineers 1 Yesterday we saw that taller trees will fall more slowly thanshorter trees. The angular acceleration is proportional to 1/L Consider how easy i
Michigan State University - PHY - 183
Equilibrium Conditions In chapter 4 we explored the conditions formechanical equilibrium We found that the necessary condition formechanical equilibrium is the absence of a netexternal force In that case Newtons First Law stipulates that anobject i
Michigan State University - PHY - 183
Stability For a skyscraper or a bridge,designers and builders need toworry about the ability of thethe structure to remainstanding under the influence ofexternal forces For example, consider thebridge carrying Interstate 35Wacross the Mississippi
Michigan State University - PHY - 183
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Michigan State University - PHY - 183
Final ExamYesterdays Exam An Adobe print error caused some exams to printoutincorrectly. Your exam score on the correction exam will be used as yourexam 8 score. Our Common Comprehensive Final Exam is onThursday, May 3rd from 8:00-10:00PM Location
Michigan State University - PHY - 183
Final Exam Our Common Comprehensive Final Exam is on Thursday, May3rd from 8:00-10:00PM Location Chemistry (CEM) room 138Physics for Scientists &amp;Engineers 1 Two 8.5 inch by 11 inch sheets of notes (both sides) Calculator Alternate Final Exam Univ
Michigan State University - PHY - 183
Final Exam Our Common Comprehensive Final Exam is on Thursday, May3rd from 8:00-10:00PM Location Chemistry (CEM) room 138Physics for Scientists &amp;Engineers 1 Two 8.5 inch by 11 inch sheets of notes (both sides) Calculator Alternate Final Exam Univ
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
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