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

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1 ics Thursday, ober 07, 2004 Ch 9: Ch 7: Momentum Conservation of Momentum Work Kinetic Energy

2 Announcements Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468

3 Announcements Don t forget to read over the lab and prepare for the short quiz.

4 Ch 9: Momentum and Collisions p = mv In fact, we can relate force to momentum changes directly using Newton s 2nd Law: F = ma = m Δ v net Δt F = Δ p net Δt This form is completely general, as it allows for changes in mass as well as velocity with time.

5 Ch 9: Momentum and Collisions p = mv 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 f = v i + a(δt) m v f = m v i + m a(δt) m v f m v i = F net Δt

6 Ch 9: Momentum and Collisions But that s just impulse. So we now have a relationship between momentum and impulse! p p = F Δt f i net We define impulse: I F Δt = Δp net Impulse-Momentum Theorem

7 Ch 9: Momentum and Collisions Worksheet #1

8 Thurs Ch 9: Momentum and Collisions Worksheet #1 Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart for 3 s and then the other for the same length of time, exerting equal force on each, the momentum of the light cart is 1. four times 2. twice 3. equal to 4. one-half 5. one-quarter the momentum of the heavy cart. PI, Mazur (1997)

9 Ch 9: Momentum and Collisions A 0.25 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. Motion Diagram: Problem #1 LAtimes.com v a

10 Ch 9: Momentum & Collisions Pictorial Representation: Knowns: v x,i = - 50 m/s v x,f = + 40 m/s Δt i = s m = kg Unknowns: <F x,net > a +x v yf, t f v xi, t i p x,i = mv x,i = (0.25 kg)( 50 m/s) = 12.5 kg m/s p x, f = mv x, f = (0.25 kg)(40 m/s) = 10 kg m/s

11 Ch 9: Momentum and Collisions Mathematical Representation: I = Δ p = p f p i = (10 ( 12.5)) kg m/s = 22.5 kg m/s I = F Δt net F net = I Δt (22.5 kg m/s) = 10 3 s = 22,500 N rightward

12 Ch 9: Momentum and Collisions 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 m 1i v 2i 1 m 2 p 1i = m 11 v 1i p 2i = m 2 v 2i

13 Ch 9: Momentum and Collisions v m 1i v 2i 1 m 2 p 1i = m 11 v 1i p 2i = m 2 v 2i The total initial momentum of the system is p sys i = p 1i + p 2i = m 1 v 1i + m 22 v 2i After these two balls hit, the situation changes... v 1f m 1 m 2 v 2f p 1f = m 1 v 1f p 2f = m 2 v 2f

14 Ch 9: Momentum and Collisions v 1f m 1 m 2 v 2f p 1f = m 1 v 1f p 2f = m 2 v 2f The total final momentum of the system is p sys f = p 1f + p 2f = m 1 v 1f + m 2 v 2f And by conservation of momentum... p sys f = p sys i p 1f + p 2f = p 1i + p 2i m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i

15 Ch 9: Momentum and Collisions m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i 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.

16 Ch 9: Momentum and Collisions Notice that this statement can also be derived through Newton s Laws. Let s say the collision lasts for a time Δt. FBDs: F 2 1 m 1 m 2 F1 2

17 a left = Δ v left Δt Ch 9: Momentum and Collisions F net,left m 1 = = F 2 1 m 1 F 2 1 m 1 a right = Δ v right Δt F net,right m 2 = = F 1 2 m 2 F 1 2 m 2 m 1 Δ v left = F 2 1 Δt m 2 Δ v right = F 1 2 Δt F 2 1 m 1 m 2 F1 2

18 Ch 9: Momentum and Collisions m 1 Δ v left = F 2 1 Δt Newton s 3 rd Law: F = F m 2 Δ v right = F 1 2 Δt Δ p 1 = F 1 2 Δt Δ p 2 = F 1 2 Δt Δ p 1 = Δ p 2 F 2 1 m 1 m 2 F1 2

19 Ch 9: Momentum and Collisions Δ p 1 = Δ p 2 Or stated slightly differently: p p = ( p p ) 1, f 1,i 2, f 2,i p + p = ( p + p ) 1, f 2, f 1,i 2,i p = p tot, f tot,i

20 Ch 9: Momentum and Collisions Worksheet #2

21 Thurs Ch 9: Momentum and Collisions Worksheet #2 Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart. 1. Increases 2. Remains the same 3. Decreases PI, Mazur (1997)

22 Ch 7: Work and Kinetic Energy 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!

23 Ch 7: Work and Kinetic Energy We come to physics with at least a welldefined notion of what work is. However, in physics, the term has a VERY SPECIFIC meaning. 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.

24 Ch 7: Work and Kinetic Energy System Motional energy (kinetic, K ) Stored energy (potential, U ) Mechanical energy= K + U Environment

25 Ch 7: Work and Kinetic Energy Describe the following demonstrations: 1) Dropped ball Let s try to provide a context for understanding why work is such an important concept in physics. Worksheet #3 2) Tugged cart Worksheet #4 Remember to identify the system and forces.

26 Ch 7: Work and Kinetic Energy 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.

27 Ch 7: Work and Kinetic Energy 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. This statement is correct so long as the applied force F s is a constant.

28 Ch 7: Work and Kinetic Energy W = F s Δs [W ] = [F s ][Δs] [W ] = N m Joule (J) [W ] = kg m2 s 2

29 Ch 7: Work and Kinetic Energy We re about to relate work to a new quantity by using an old expression... s f = s i + v i (Δt) a(δt)2 s f s i = Δs = v i (Δt) a(δt)2 Now solve for Δt v f = v i + a(δt) Plug back in above... (Δt) = v f v i a

30 Ch 7: Work and Kinetic Energy Δs = v i v f v i a a v v f i a 2 Δs = (v i v f / a) (v 2 i / a) + 1 (v 2 2v 2 f i v f + v 2 i ) / a Δs = 1 (v 2 v 2 2 f i ) / a aδs = 1 (v 2 v 2 2 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!

31 Ch 7: Work and Kinetic Energy ma s (Δs) = F s,net Δs = 1 2 m(v s, f 2 v 2 ) s,i Each term on the right-handside of this equation is called a kinetic energy term. K = 1 2 mv 2 And now we can rewrite work as a change in kinetic energy: W = ΔK = K K = 1 net f i 2 mv 2 1 f 2 mv i2

32 Ch 7: Work and Kinetic Energy K = 1 2 mv 2 [K] = 1 [m][v 2 ] 2 [K] = kg m 2 / s 2 = Nm = J That s good! It has the same units as work! WHEW!

33 Ch 7: Work and Kinetic 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.

34 Ch 7: Work and Kinetic Energy 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 2) J 3) 0 J 4) 24.5 J 5) 49.0 J Problem #2

35 Ch 7: Work and Kinetic Energy 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? Motion Diagram: v a= 0

36 Ch 7: Work and Kinetic Energy 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? FBD: F man = Force of man on bucket 3rd Law Pair???? W = weight NO!!

37 Ch 7: Work and Kinetic Energy 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? Pictorial Representation: Knowns: y i = 0; y f = 0.5 m a = 0; v f = v i F y,man = W = m g m = 5.00 kg Unknowns: W, v y f, t f, v f y i, t i, v i +y

38 Ch 7: Work and Kinetic Energy 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? Mathematical Representation: F man = W = (5.00 kg)(9.81 m/s 2 ) = 49.1 N Since the force is parallel to the displacement, the work done by the man is given by W man = F man,s Δs = (49.1 N) (0.5 m) = 24.5 J

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

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 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