Physics I Class 07 Momentum and Impulse Rev. 10-Aug-03 GB 07-1 Momentum of an Object Definitions W e define m om entum for an object to be: p m v M omentum is a vector. It is in the same direction as velocity. SI units for m om entum : kg m /s.
Later in the sem ester, w e w ill talk about angular m omentum. Plain m omentum is also know n as linear or translational m om entum . 07-2 Change of Momentum C h a n g e o f m o m e n tu m is th e d iffe re n c e b e tw e e n th e fin a l v a lu e a n d th e in itia l v a lu e . B e w a r e : W e a re s u b tra c tin g v e c to rs ! p p p final fin a l
initial m v final v initial in itia l +3 +3 = 0 kg m/s +3
-3 = +6 kg m/s 07-3 Connecting Net Force and the Derivative of Momentum N e w to n s S e co n d L a w : F F net m a T h e tim e d e r iv a tiv e o f m o m e n tu m :
dp d dv (m v ) m m a dt dt dt W e c a n w r ite N e w to n s S e c o n d L a w in a n e w f o r m : dp F F net
dt 07-4 The Fundamental Theorem of Calculus R e c a ll fro m C la s s # 2 : M a th F a c t: B e c a u s e v e lo c ity is th e d e riv a tiv e o f d is p la c e m e n t, d is p la c e m e n t is th e a re a (in te g ra l) u n d e r th e g ra p h o f v v e rs u s t. T h e F u n d a m e n ta l T h e o r e m o f C a lc u lu s : If f is th e d e riv a tiv e o f g , th e n v v0 t0 t b
f ( t ) dt g (b ) g (a ) g a 07-5 Impulse and the ImpulseMomentum Theorem I m p u ls e is d e fin e d to b e th e tim e in te g ra l o f fo rc e . S I u n its = N s . L ik e fo r c e , it is a v e c to r . ( N e t o r to ta l fo r c e is im p lie d .) J F dt U s in g th e F u n d a m e n ta l T h e o re m o f C a lc u lu s :
J F dt p In P h y s ic s , th is is k n o w n a s th e I m p u ls e -M o m e n t u m T h e o r e m . 07-6 Example Problem Using Impulse-Momentum An object of mass 0.5 kg is subjected to a F force in the +X direction that varies as shown 5N in the graph from 0 to 7 seconds. Its initial X velocity is zero. What is its final X velocity? Doing this with F=ma would be hard. Doing 0 0 it with impulse-momentum is much easier. 7s t J = a r e a = 2 + 3 + 3 .5 + 3 + 2 + 3 + 1 .5 = 1 8 N s .
18 N s J p p final p initial p final p initial p final 0 p final 18 kg m / s v final p final m 36 m / s 07-7 Good-bye Average Force Favg A v e ra g e fo rc e c a n b e d e fin e d m a th e m a tic a lly , b u t it is a c o n c e p t to o e a s ily m is u s e d . W e w ill a v o id A v e ra g e F o rc e in P h y s ic s 1 . N ever use F av g = m a ! U s e th e I m p u ls e -M o m e n tu m T h e o r e m : J F dt p
07-8 Class #7 Take-Away Concepts 1. M o m e n t u m d e f i n e d f o r a n o b j e c t : 2. A n e w w a y to w rite N e w to n s S e c o n d L a w : p m v d p F F net
d t 3. I m p u ls e d e fin e d : 4. I m p u ls e -M o m e n tu m T h e o re m : J F dt J F dt p
07-9 Class #7 Problems of the Day ___1. For this question, take right as the positive direction. A 0.375 kg rubber ball traveling horizontally to the right at 10 m/s hits a wall and bounces back at 6 m/s to the left. The total impulse exerted by the wall on the ball is A) B) C) D) 6.0 N s. +6.0 N s. 1.5 N s. +1.5 N s. 07-10
Answer to Problem 1 for Class #7 The answer is A. J = p = p-final p-initial = m (v-final v-initial) v-final = 6 m/s. v-initial = +10 m/s. J = 0.375 * (6 10) = 0.375 * (16) = 6.0 N s. 07-11 Class #7 Problems of the Day ___2. Egg Toss Contest: You and a friend are tossing a raw egg back and forth. For a given speed of toss, you want to catch the egg without breaking it. To do this, you need to keep the force on the egg shell low. How can you do that? A) You cant. The egg experiences the same change in momentum no matter how you catch it, therefore it experiences the same force from your hand. B) Catch the egg with as stiff an arm as possible to bring it to rest quickly. That will decrease the time during which the force acts,
thus decreasing the force on the egg. C) Catch the egg with a lot of arm movement to stretch out the time of the catch, increasing the time during which the force acts. 07-12 Answer to Problem 2 for Class #7 The answer is C. The impulse on the egg will be the same no matter how you catch it. However, impulse is the integral (area under the curve) of force versus time. If you decrease the time over which the force acts, the force has to be greater to give the same area under the curve. So you want to catch the egg (bring it to rest) over a long time period in order to keep the force low. This is the same principle behind the automobile air bag. It brings a person to rest in a longer time than slamming into the steering wheel or dashboard. 07-13 Class #7
Problems of the Day 3. The Physics Department decides that we need a more spectacular demonstration of projectile motion for class. We contact a supplier of physics education equipment and they send us the specifications for a long-range classroom projectile launcher, a more powerful version of the launcher we saw in class and in the VideoPoint activity. The mass of the ball fired by the new launcher is 10 g. The graph of net force versus time for launching the ball is shown below: force (N) 2 0 0 0.01 0.1 time (s) Calculate the speed of the ball as it leaves the launcher.
07-14 Answer to Problem 3 for Class #7 You could solve this problem using F = m a and calculus, but you cannot use the equations of motion for constant acceleration, because acceleration is not constant in this problem. Instead, use J = p. In this case, p-initial = 0, so we have J = m v. J is the area under the curve. Using the formula for a triangle, J = 0.1 N s. v = J/m = 0.1/0.01 = 10 m/s. (mass in kg) 07-15 Activity #7 Impulse and Momentum Objectives of the Activity: 1. 2. 3. Learn how to calibrate sensors to take accurate data.
Take data in a real experiment to study the relationship between impulse and momentum change. Learn how to use data analysis features of LoggerPro. 07-16 Class #7 Optional Material Special Theory of Relativity Albert Einstein (18791955) In Lecture 04, we mentioned that Albert Einstein showed that Newtons Second Law had to be modified in order to account for the observed behavior of electromagnetic waves (light) and the interaction of electromagnetic fields with matter. Einsteins Two Postulates of Special Relativity: 1. The laws of physics are the same in all inertial frames. 2. The speed of light, c, is constant in all inertial frames. 07-17
What is an Inertial Frame? An inertial frame (of reference) is a real or imaginary set of devices for measuring position and time that are in motion together according to Newtons First Law; in other words, these devices are not accelerating (or rotating). Neglecting gravity (well talk about that later in the semester), the track and motion detector that we use in our activities, along with the clock in your PC when you run LoggerPro, comprise an inertial reference frame with a special name: the laboratory reference frame (because this is the frame we use to make measurements). If we were to set the same equipment up in the Ferris Wheel we studied earlier, that would not be an inertial reference frame. Instead, we call that an accelerated frame. 07-18 Where Did Einsteins Postulates Come From? 1. The laws of physics are the same in all inertial frames. This idea goes back to Galileo. Imagine an experiment like our cart track
and hanging weight being performed in an airliner moving uniformly in one direction at a constant speed and altitude. (Assume no turbulence and the altitude is low enough so that the force of gravity is about the same as on the ground.) Our measurements that we take with LoggerPro should be the same as what we did in class if we set things up carefully. 2. The speed of light, c, is constant in all inertial frames. Nobody wanted to believe this prior to the Michelson-Morley experiment in 1887. In one of the most famous null results in the history of science, they showed by careful measurement that the speed of light is independent of the relative motion of the source and detector. This counter-intuitive result was not adequately explained until Einsteins Special Theory of Relativity. Do a web search and you will see that many people still refuse to believe it! 07-19 Einsteins Correction to Newtons Second Law Starting fromthe two postulates, Einstein showed that Newtons Second Lawwould be correct for all velocities if the definition of momentumwas modified:
dp F F net dt m0 p v 2 v 1 2 c Note that the square root termin the denominator allows the magnitude of the momentumto become arbitrarily large as the
speed approaches but never quite reaches the speed of light. At normal speeds much smaller than c, the correction is negligible. 07-20 http://www.amnh.org/exhibitions/einstein/?src=h_h 07-21
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