Physics 111. ? (P1) Inclined plane & newton s. Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468

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1 ics Announcements day, ober 05, 2004 Ch 6: Last Example Ch 9: Impulse Momentum Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements Ch 6: Applying Newton s Laws Note the following: How does your weight change while you re This week s lab will be a workshop. Bring your Ranking Tasks workbook and your textbook. Ascending at v = const.? Descending at v = const.? Accelerating Down? Accelerating Up? Ch 6: Applying Newton s Laws Two blocks of mass 3.50 kg and 8.00 kg are connected by a massless string passing over a frictionless pulley. The inclines are also frictionless. Find (a the magnitude of the acceleration of each block and (b the tension in the string. Linear Momentum & Collisions =3.50 kg θ 1 =35 o θ 2 =35 o = 8.00 kg Problem Sheet #1 In the study of systems with multiple objects, particularly systems in which these objects interact with one another through collisions, it is useful to take advantage of the concept of momentum. We ll see that this new quantity is conserved in isolated systems, allowing us to analyze some rather complicated interactions.? (P1 Inclined plane & newton s 1

2 Ch 9: Momentum & Collisions Ch 9: Momentum & Collisions We ve applied Newton s Laws to better understand the behavior of systems - single object - multiple object - remote interactions Now, let s examine interactions between objects through another approach: What constitutes an impulse is probably best illustrated through a simple demonstration. When I drop the ball, the force of gravity exerts an impulse on the falling ball. When the ball hits the floor, a second force also exerts an impulse: the contact force of the floor. For a constant force, as with the falling ball, it s easy! I = F I = F t i t f Notice graphically that the impulse is the area under the curve of a force vs time graph. (Just like distance was the area under a velocity vs time graph! [ I ] = [ F][] [ I ] = N s = kg m s Graphical interpretation of Units: impulse Average Force I = ( F f + F i = 2 Let s look at a simple case of a force that increases linearly with time. F Non-constant force The answer is still just the area under the curve. I floor = ( F max + F min = F 2 floor Average Force Bouncing ball Let s assume that during the bounce, the floor exerts a force on the ball as given at left. Note: The force of gravity results in an impulse at the same time. 2

3 Ch 9: Momentum & Collisions 480 N F floor (N A kg rubber ball is released from a height of 2.00 m above the floor. The ball falls, bounces, and rebounds. How high does the ball make it on the rebound if the force function is given by the graph below? Problem Sheet #2 We can always find the a constant F such that the impulse delivered is the same as that delivered by a non-constant force. 5 ms 10 ms 15 ms Completely General F net = area under curve The constant value is always exactly the timeaveraged value of the nonconstant force! Having explored this quantity graphically and using results from calculus (not shown, we can say succinctly and generally that: I = F net The linear momentum of an object of mass m moving with velocity v is defined as the product of the mass and the velocity: Notice that momentum is a vector quantity, which means that it must be specified with both a magnitude and direction. Also notice that the direction of the momentum is necessarily parallel to the velocity. [ p] = kg (m/s [ p] OR = N s [ p] = [m][ v] The units suggest a relationship between force and momentum. Units: linear momentum 3

4 It accelerates. Its velocity changes. What happens when we apply a force to an object? Its momentum changes. The force imparts momentum. In fact, we can relate force to momentum changes directly using Newton s 2nd Law: F net = Δ p F net = m a = m Δ v This form is completely general, as it allows for changes in mass as well as velocity with time. m v f 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( = m v i + m a( m v f 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 Worksheet #1 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 #3 LAtimes.com v a CQ1: impulse? (P3 Bonds - motion 4

5 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 2i 1 p 1i = 1 v 1i v i v 2i 1 The total initial momentum of the system is p sys i = p 1i + p 2i = v 1i + 2 v 2i After these two balls hit, the situation changes... v 1f v 2f p 2i = v 2i p 1i = 1 v 1i p 2i = v 2i p 1f = v 1f p 2f = v 2f v 1f v 2f p 1f = v 1f The total final momentum of the system is p sys f = p 1f + p 2f = v 1f + v 2f And by conservation of momentum... p sys f = p sys i p 1f + p 2f = p 1i + p 2i v 1f + v 2f = v 1i + v 2i p 2f = v 2f v 1f + v 2f = v 1i + 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. 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 = = 2 1 Δ v left = Δ v left = F net,right F a right = = 1 2 Δ v right = F 1 2 Δ v right = F 1 2 F1 2 F1 2 5

6 Δ v left = Newton s 3 rd Law: Δ v right = F 1 2 = F 1 2 Δ p 1 = F 1 2 Δ p 2 = F 1 2 Δ p 1 = Δ p 2 F1 2 Δ p 1 = Δ p 2 Or stated slightly differently: p 1, f p 1,i = ( p 2, f p 2,i p 1, f + p 2, f = ( p 1,i + p 2,i p tot, f = ( p tot,i Worksheet Problems #2 CQ2: momentum 6