Download presentation

Presentation is loading. Please wait.

Published byTommy Pinkins Modified over 2 years ago

1
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 1

2
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 2

3
Day 13 – June 6 – WBL 6.4-6.6 It is possible for an inelastic collision to occur in which the colliding objects stick together after the collision – the text mentions an example of two railcars that bump, couple together, and travel along while linked. This is termed a completely inelastic collision. Take care that you understand this definition. Many students assume that a completely inelastic collision is one in which the final kinetic energy is zero. This is not true; the final KE is determined by momentum conservation, as we shall see. 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 3 What we can say about a completely inelastic collision is that both objects have the same final velocity (since they’re joined together).

4
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 4

5
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 5

6
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 6 In this case, it is possible for the final speed to be zero. This simply requires that, pre-collision, the objects had equal and opposite linear momenta.

7
Day 13 – June 6 – WBL 6.4-6.6 Momentum and Energy in Elastic Collisions 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 7 For an elastic collision, both linear momentum and kinetic energy are conserved.

8
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 8

9
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 9

10
Day 13 – June 6 – WBL 6.4-6.6 6.4 Elastic and Inelastic Collision PC141 Intersession 2013Slide 10

11
Day 13 – June 6 – WBL 6.4-6.6 The short video shown here illustrates concepts of momentum conservation (in 2D) as it relates to the sport of curling. There’s some interesting information about friction as well. 6.4 Elastic and Inelastic Collisions PC141 Intersession 2013Slide 11 Videos are not embedded into the PPT file. You need an internet connection to view them.

12
Day 13 – June 6 – WBL 6.4-6.6 Problem #1: Inelastic Collision PC141 Intersession 2013Slide 12 WBL LP 6.11 Which of the following is not conserved in an inelastic collision… A linear momentum B mass C kinetic energy D total energy

13
Day 13 – June 6 – WBL 6.4-6.6 Problem #2: 2D Completely Inelastic Collision PC141 Intersession 2013Slide 13 WBL Ex 6.65

14
Day 13 – June 6 – WBL 6.4-6.6 Problem #3: Ballistic Pendulum PC141 Intersession 2013Slide 14 WBL Ex 6.55

15
Day 13 – June 6 – WBL 6.4-6.6 6.5 Center of Mass PC141 Intersession 2013Slide 15

16
Day 13 – June 6 – WBL 6.4-6.6 Our discussion relies on the concept of the center of mass: What this means is that, while the motion of any individual particle in a system may be complicated, the CM follows a simple path, dictated by the physics of chapter 4. The photo below provides an example. 6.5 Center of Mass PC141 Intersession 2013Slide 16 the center of mass (CM) is the point at which all of the mass of an object or system may be considered to be concentrated, for the purposes of describing its linear or translational motion only

17
Day 13 – June 6 – WBL 6.4-6.6 6.5 Center of Mass PC141 Intersession 2013Slide 17

18
Day 13 – June 6 – WBL 6.4-6.6 6.5 Center of Mass PC141 Intersession 2013Slide 18

19
Day 13 – June 6 – WBL 6.4-6.6 6.5 Center of Mass PC141 Intersession 2013Slide 19

20
Day 13 – June 6 – WBL 6.4-6.6 Finding the CM for an extended object (such as the wrench from a few slides ago) is much more complicated. In general, it requires integration, so it will not be covered in PC141. However, for extended objects with certain degrees of symmetry, the CM simply lies at the point / line / plane of symmetry. For instance, the CM of a meter-stick lies at exactly the 50-cm mark (and exactly halfway through the width and thickness of the stick, if you want to consider it as a 3D object). The CM of a 2D square or circle, or of a 3D sphere or cube, lies at the exact center of the object. The text describes the CM as the “balance point” of the system. Unfortunately, we don’t quite know enough physics to understand why something would or would not be balanced, so we’ll return to this later in the course. 6.5 Center of Mass PC141 Intersession 2013Slide 20

21
Day 13 – June 6 – WBL 6.4-6.6 For irregularly shaped objects, the CM can be found experimentally, using a suspension technique illustrated in the text. Finally, it is important to note that it is possible for the CM to exist at a point where no mass exists. This is illustrated below. 6.5 Center of Mass PC141 Intersession 2013Slide 21

22
Day 13 – June 6 – WBL 6.4-6.6 Problem #4: Center of Mass PC141 Intersession 2013Slide 22 WBL LP 6.15 The center of mass of an object… A …always lies at the center of the object B …is at the location of the most massive particle in the object C …always lies within the object D none of the preceding

23
Day 13 – June 6 – WBL 6.4-6.6 Problem #5: Sun-Earth System PC141 Intersession 2013Slide 23

24
Day 13 – June 6 – WBL 6.4-6.6 Problem #6: Three Point Masses PC141 Intersession 2013Slide 24 Three particles with masses m 1 = 1.2 kg, m 2 = 2.5 kg, and m 3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this system? Solution: In class

25
Day 13 – June 6 – WBL 6.4-6.6 Problem #7: Solid Plate PC141 Intersession 2013Slide 25 The figure shows a slab with dimensions d 1 = 11.0 cm, d 2 = 2.80 cm, d 3 = 13.0 cm. Half of the slab consists of aluminum (density = 2.70 g/cm 3 ) and the other half consists of iron (density = 7.85 g/cm 3 ). What are the coordinates of the slab’s center of mass? Solution: In class

26
Day 13 – June 6 – WBL 6.4-6.6 Problem #8: Elastic Collision PC141 Intersession 2013Slide 26 A body of mass 2.0 kg collides elastically with a stationary body and continues to move in the original direction but with one-fourth of its original speed. (a)What is the mass of the other body? (b)What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s? Solution: In class

27
Day 13 – June 6 – WBL 6.4-6.6 Linear momentum conservation (and Newton’s 3 rd law) dictate that when an object expels part of its mass in one direction, the remainder of the mass must gain velocity in the opposite direction. A good example of this is the recoil of a rifle: 6.6 Jet Propulsion and Rockets PC141 Intersession 2013Slide 27 (time permitting)

28
Day 13 – June 6 – WBL 6.4-6.6 This principle is put to good use during jet propulsion. Here, burning fuel produces exhaust products (gases) that are directed backwards at high speed (maximizing the speed of the exhaust is a formidable engineering problem…it literally requires the expertise of rocket scientists). The result is a forward-directed force on the aircraft, referred to as thrust. The basic mathematics involved with jet propulsion isn’t too tricky, but it does involve integration, so we’ll avoid it altogether. It’s just nice to know that something as complex as space travel is essentially governed by one month’s worth of introductory physics. 6.6 Jet Propulsion and Rockets PC141 Intersession 2013Slide 28 (time permitting)

Similar presentations

Presentation is loading. Please wait....

OK

Linear Momentum and Collisions

Linear Momentum and Collisions

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on mars one Ppt on current account deficit india Ppt on verbs for grade 5 Ppt on history of england Ppt on political parties and electoral process of the philippines Ppt on depth first search algorithm java Ppt on global warming in hindi language Ppt on computer graphics algorithms Ppt on perimeter and area of plane figures Ppt on thermal power generation