Physics 111. = Δ p. F net. p f. p i. = F net. m v i. = v i. v f. = m v i. + m a(δt) m v f. m v f. Δt = Δ p. I F net. = m a = m Δ v

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ics Announcements day, ober 07, 004 Ch 9: Momentum Conservation of Momentum Ch 7: Work Kinetic Energy Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements p =

1 ics Announcements day, ober 07, 004 Ch 9: Momentum Conservation of Momentum Ch 7: Work Kinetic Energy Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements p = mv In fact, we can relate force to momentum changes directly using Newton s nd Law: F net = m a = m Δ v Don t forget to read over the lab and prepare for the short quiz. F net = Δ p This form is completely general, as it allows for changes in mass as well as velocity with time. p = mv m By how much will the momentum change? That depends upon the amount of time over which the force is applied to the object. Here we ll derive the case for a constant force. = v i + a( = m v i + m a( m m v i = F net But that s just impulse. So we now have a relationship between momentum and impulse! We define impulse: I F net = Δ p p f p i = F net Impulse-Momentum Theorem 1

2 Worksheet #1 A 0.5 kg baseball arrives at the plate with a speed of 50 m/s. Barry Bonds swings and the ball leaves his bat at 40 m/s headed directly back toward the mound along the same path on which it arrived. Assuming that the ball is in contact with the bat for 10-3 s, what is the magnitude of the average force the bat exerts on the ball? Ignore gravity in this problem. Problem #1 LAtimes.com The total momentum of an isolated system of objects is conserved, regardless of the nature of the force between the objects in that system. Here, we speak of the linear momentum of the system. Thus, if our system consists of two objects... v i v i 1 p 1i = 1 v 1i v i v i 1 The total initial momentum of the system is p sys i = p 1i + p i = v 1i + v i After these two balls hit, the situation changes... v 1f v f p i = v i p 1i = 1 v 1i p i = v i p 1f = v 1f p f = v f collisions v 1f v f p 1f = v 1f The total final momentum of the system is p sys f = p 1f + p f = v 1f + v f And by conservation of momentum... p sys f = p sys i p 1f + p f = p 1i + p i v 1f + v f = v 1i + v i con t p f = v f v 1f + v f = v 1i + v i NOTE: our formulation above is derived for a system of two POINT MASS objects. With no external forces acting upon these objects, when the two objects collide, the total linear momentum before and after the collision are the same. Conservation of Momentum

3 Notice that this statement can also be derived through Newton s Laws. Let s say the collision lasts for a time. FBDs: F net,left F a left = = 1 Δ v left = Δ v left = F net,right F a right = = 1 Δ v right = F 1 Δ v right = F 1 F1 F1 Δ v left = Newton s 3 rd Law: Δ v right = F 1 = F 1 Δ p 1 = F 1 Δ p = F 1 Δ p 1 = Δ p F1 Δ p 1 = Δ p Or stated slightly differently: p 1, f p 1,i = ( p, f p,i p 1, f + p, f = ( p 1,i + p,i p tot, f = ( p tot,i Worksheet # Work & Kinetic Energy After developing an understanding of motion using velocity, acceleration, forces, momentum, and impulse (all vector quantities, we re now going to develop scalar tools that will help make some problems much simpler! CQ: momentum Ch 7: Work & Energy 3

4 We come to physics with at least a welldefined notion of what work is. However, in physics, the term has a VERY SPECIFIC meaning. System Motional energy (kinetic, K Stored energy (potential, U Mechanical energy= K + U Work is the mechanical transfer of energy between a system and its environment by pushes and pulls (forces at the boundary where the forces are applied. Environment Work: System & Environment Describe the following demonstrations: 1 Dropped ball Tugged cart Let s try to provide a context for understanding why work is such an important concept in physics. Worksheet #3 Worksheet #4 Remember to identify the system and forces. (W3 & W4 Work demos We come to physics with at least a welldefined notion of what work is. However, in physics, the term has a VERY SPECIFIC meaning. The work done by a constant force on an object is equal to the magnitude of the component of the force in the direction of motion of the object times the displacement of the object. Work definition in words Mathematically, this statement of work translates to W = F s Δs Where Δs is the distance the object travels as measured along the path it follows (i.e., its trajectory and F s is the component of the force in the direction of motion. W = F s Δs [W ] = [F s ][Δs] [W ] = N m Joule (J This statement is correct so long as the applied force F s is a constant. Work: math definition [W ] = kg m s Work: units 4

5 We re about to relate work to a new quantity by using an old expression... s f = s i + v i ( + 1 a( s f s i = Δs = v i ( + 1 a( Δs = (v i Δs = v i v i a + 1 a v v f i a / a (v i / a + 1 (v v f i + v i / a Δs = 1 (v v f i / a Now solve for Plug back in above... = v i + a( ( = v i a aδs = 1 (v v f i This last quantity looks something like work, right? All we need to do is multiply both sides by mass, and we have force times distance! ma s (Δs = F s,net Δs = 1 m(v v s, f s,i K = 1 mv Each term on the right-handside of this equation is called a kinetic energy term. K = 1 mv [K] = 1 [m][v ] And now we can rewrite work as a change in kinetic energy: W net = ΔK = K f K i = 1 mv f 1 mv i [K] = kg / s = Nm = J That s good! It has the same units as work! WHEW! Units: energy Although we ve derived this relationship between work and kinetic energy for the special case of a constant applied force (i.e., constant acceleration, it turns out to be generally true, regardless of whether or not the force is constant! The NET WORK done on an object is equal to its change in KINETIC ENERGY. generalization A man raises a pail of mass 5 kg at a constant velocity to a height of 0.5 m. How much work has he done? F 1-49 J J 3 0 J J J? (P Bucket Problem # 5

6 Let s think about this a little further We saw that a man raising a 5 kg bucket through 0.5 m at constant velocity does an amount of work W = 5 J But if the NET WORK done on the bucket equals the change in kinetic energy of the bucket, then W net = ΔK = 1 m(v v f i = 0 What s the deal? Worksheet #5 W = ± F s Δ s Δs is the magnitude of the displacement as measured along the path that the object actually travels (i.e., along its trajectory, where Δs is small or always in the same direction. F s is the component of the force in the direction of motion. + sign if F s and Δs are in the same direction. sign if F s and Δs are in opposite directions. Worksheet #6 CQ4: impulse & K 6

Physics 111. Thursday, October 07, Conservation of Momentum. Kinetic Energy

Physics 111. Thursday, October 07, Conservation of Momentum. Kinetic Energy ics Thursday, ober 07, 2004 Ch 9: Ch 7: Momentum Conservation of Momentum Work Kinetic Energy Announcements Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements