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Up until now, you have looked at momentum in only one dimension The law of conservation of momentum holds for all closed systems with no external forces It is valid regardless of the directions of the particles before or after they interact But what happens in two or three dimensions Figure 9-10 shows the result of billiard ball C striking stationary billiard ball D Consider the two billiard balls to be the system The original momentum of the moving ball is pCi and the momentum of the stationary ball is zero Therefore, the momentum of the system before the collision is equal to pCi After the collision, both billiard balls are moving and have momenta As long as the friction with the tabletop can be ignored, the system is closed and isolated Thus, the law of conservation of momentum can be used The initial momentum equals the vector sum of the final momenta, so pCi pCf pDf The equality of the momenta before and after the collision also means that the sum of the components of the vectors before and after the collision must be equal Suppose the x-axis is defined to be in the direction of the initial momentum, then the y-component of the initial momentum is equal to zero Therefore, the sum of the final y-components also must be zero: pCf, y pDf, y 0

The y-components are equal in magnitude but are in the opposite direction and, thus, have opposite signs The sum of the horizontal components also is equal: pCi pCf, x pDf, x

Figure 9-10 The law of conservation of momentum holds for all isolated, closed systems, regardless of the directions of objects before and after a collision C

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You have applied Newton s laws to a variety of equilibrium situations, but only to motions that were either horizontal or vertical How would you apply them in a situation like the one in Figure 5-13a, in which a skier glides down a slope Start by identifying the forces acting on the object, the skier, as shown in Figure 5-13b and sketching a free-body diagram The gravitational force on the skier is in the downward direction toward the center of Earth There is a normal force perpendicular to the hill, and the frictional forces opposing the skier s motion are parallel to the hill The resulting free-body diagram is shown in Figure 5-13c You can see that, other than the force of friction, only one force acts horizontally or vertically, and you know from experience that the acceleration of the skier will be along the slope How do you find the net force that causes the skier to accelerate

The Mac is a great computing platform, and Cocoa is the best tool to use to build your Mac applications. If you are coming to the Mac from another platform, you may find it a bit disorienting working with Cocoa and Objective-C. Even though you might find it hard to get your bearings at first, as you make your way through this book, we re confident that you ll soon start to wrap your head around the Cocoa way of doing things. With a little perseverance, it will all start to make sense. One thing you should keep in mind is that the exercises in this book are not simply a checklist that, when completed, grant you Cocoa developer guru status. Every step of the way, make sure you understand what you just did before moving on to the next project. Don t be afraid to make changes to the code. Experiment and observe the results. That s one of the best ways to wrap your head around the complexities of coding in an environment like Cocoa. Also, remember that we ve set up a message forum for readers to discuss the exercises and content of this book. So if you get stuck, or want to chat with us and other Cocoa developers about any Cocoa programming topics, head over to http://learncocoa.org and register for the discussion forum. That said, if you ve got your developer tools installed, turn the page. If not, get to it! Then let s go!

Figure 5-13 A skier slides down a slope (a) Identify the forces that are acting upon the skier (b) and draw a free-body diagram describing those forces (c) It is important to draw the direction of the normal and the friction forces correctly in order to properly analyze these types of situations

11:

Components of Weight for an Object on an Incline A crate weighing 562 N is resting on a plane inclined 300 above the horizontal Find the components of the weight forces that are parallel and perpendicular to the plane

Include a coordinate system with the positive x-axis pointing uphill Draw the free-body diagram showing Fg, the components Fgx and Fgy , and the angle Known:

Unknown:

Fgx and Fgy are negative because they point in directions opposite to the positive axes Fgx = Fg(sin ) = (562 N)(sin 300 ) Substitute Fg 562, 300 = 281 N Fgy Fg(cos ) (562 N)(cos 300 ) Substitute Fg 562, 300 487 N

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