Lecture 18 - Lecture 18 pls-18.1 Momentum Conservation and...

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Lecture 18 pls-18.1 Momentum Conservation and Collisions Media: 1. Two carts on track, weights 2. Newton’s cradle and video camera 1. Conservation of Momentum Let’s consider the center of mass (CM) of a two-object system where the two objects are on the x axis, such as two carts on a track . The CM component CM x of the two objects is 2 1 2 2 1 1 m m x m x m x CM + + = and the total mass of the system is 2 1 m m m CM + = Æ 2 2 1 1 x m x m x m CM CM + = . If we take the time derivative of this last equation, because dt dx v = , we get x x CMx CM v m v m v m 2 2 1 1 + = which can be written as x x CMx p p p 2 1 + = This equation says that the momentum associated with CM motion is equal to the sum of the momenta of the objects that make up the system. (This also applies to y components if motion is in 2D.) Clicker questions 1 - 3 .
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Lecture 18 pls-18.2 Let’s now assume that the net external force on the (two-cart) system is zero . Then N’s 2nd law Æ CM v r is constant. The mass CM m is also constant Æ the total momentum CM p r is also constant. That is, the total momentum CM CM CM m v p r r = of a system is conserved if the net external force is zero. This is known as the conservation of momentum . Even if the objects that make up a system are interacting via internal
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This note was uploaded on 03/26/2012 for the course PHY 1020 taught by Professor Kodera during the Fall '11 term at University of Florida.

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Lecture 18 - Lecture 18 pls-18.1 Momentum Conservation and...

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