Section07

# Section07 - Physics 204A Class Notes Section 7 Linear...

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Physics 204A Class Notes 7-1 Section 7 – Linear Momentum We are trying to understand why objects do what they do. The idea of force in general, and Newton’s Second Law in particular, have done a pretty good job of answering that question. However, it turns out the Newton never actually wrote F = ma. Instead, he defined a quantity which we now call “linear momentum” and expressed the Second Law in terms of it. Linear Momentum or just momentum, for short, turns out to be a very powerful idea for extending our understanding of why objects do what they do. Section Outline 1. The Definition of Momentum 2. The Impulse-Linear Momentum Theorem 3. The Second Law and the Center of Mass 4. The Law of Conservation of Linear Momentum 5. Linear Momentum and Collisions 1. The Definition of Momentum Newton actually wrote his Second Law in a different form than we have been using, The Original Second Law " r F = d r p dt , where we use, The Definition of Linear Momentum r p " m r v . For an object that can be treated like a single particle, such as a baseball of mass, m, moving at a velocity, r v , the original Second Law doesn’t seem to tell us anything new, " r F = d ( m r v ) dt = m d ( r v ) dt # " r F = m r a . However, it at least is consistent with the Second Law we have been using.

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Physics 204A Class Notes 7-2 Example 7.1: An 0.150kg baseball pitched at 30.0m/s is hit straight back at the pitcher at 50.0m/s. The ball is in contact with the bat for 1.00ms. Find (a)the initial momentum, (b)the final momentum and (c)the average force on the ball. (a)Using the definition of linear momentum, r p ! m r v " p o = mv o = (0.150)(30.0) " p o = 4.50 kg " m s . (b)Using the definition of linear momentum again, r p ! m r v " p = mv = (0.150)( # 50.0) " p = " 7.50 kg # m s . Note that momentum is a vector quantity and we have just found the x- component. (c)We can now apply the Second Law changing the d’s to ’s because we want the average force, " r F = d r p dt # F = \$ p \$ t = p % p o \$ t = % 7.50 % (4.50) 1.00 x 10 % 3 # F = ! 12.0kN . The force is to the left, in the negative x direction. Notice that we could have solved this problem using " r F = m r a = m # r v # t = m r v \$ r v o # t and gotten the same result. 2. The Impulse-Linear Momentum Theorem Even though it appears that we haven’t gained anything by developing the concept of momentum, we have. If you take the original Second Law and solve for the change in momentum. ! r F = d r p dt " d r p = r F dt " d r p r p o r p # = r F dt t o t # " r p \$ r p o = r F dt t o t # The quantity on the right hand side is defined to be the “impulse.” The Definition of Impulse r J ! r F dt t o t " The relationship becomes, r p ! r p o = r J , which is known as the “Impulse-Linear Momentum Theorem.” The Impulse-Linear Momentum Theorem ! r p = r J When a force acts on an object over a period of time, it changes the momentum of an object. Think about dropping a ball of soft clay. When is hits the ground, it comes to rest regardless of whether it hits a hard concrete floor or lands on carpet. Therefore, it has the same change in momentum in either case.
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