Chapter10 - 10.1 Model We will use the particle model for...

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10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB). Visualize: Solve: For the bullet, K m v B B B 2 kg m/s J = = = 1 2 1 2 0 01 500 1250 2 ( . )( ) For the bowling ball, K m v BB BB BB kg m s J = = = 1 2 1 2 10 10 500 2 2 ( )( / ) Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 1000 times smaller than the mass of the bowling ball, its speed is 50 times larger.
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10.2. Model: Model the hiker as a particle. Visualize: The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley is y 0 8 5 = − . . m Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is U U U mgy mgy mg y y = = = = − − = × gf gi f i f i 2 kg ( m/s m m J ( ) ( ) . )[ ( )] . 65 9 8 4420 85 2 87 10 6 Assess: Note that U is independent of the origin of the coordinate system.
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10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize: Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal, K K m v m v v m m v C T C C T T C T C T kg kg km/hr km/hr = = = = = 1 2 1 2 20 000 1000 25 112 2 2 , ( ) Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
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10.4. Model: Model the oxygen and the helium atoms as particles. Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is four times heavier than the helium atom, so m m O He = 4 . Solve: The energy conservation equation K K O He = is 1 2 1 2 4 2 0 2 2 2 2 m v m v m v m v v v O O He He He O He He He O = = = ( ) . Assess: The result v v He O = 2 , combined with the fact that m m He O = 1 4 , is a consequence of the way kinetic energy is defined: it is directly proportional to the mass and to the square of the speed.
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10.5. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize: Solve: (a) The kinetic energy of the car is K m v C C C kg m/s J = = = × 1 2 1 2 1500 30 6 75 10 2 2 5 ( )( ) . (b) Let us relabel K C as K f and place our coordinate system at y f = 0 m so that the car’s potential energy U gf is zero, its velocity is v f , and its kinetic energy is K f . At position y i , v K i i m/s or J = = 0 0 , and the only energy the car has is U mgy gi i = . Since the sum K + U g is unchanged by motion, K U K U f gf i gi + = + . This means K mgy K mgy K K mgy y K K mg f f i i f i i i i 2 J - 0 J kg m /s m + = + + = + = = × = 0 6 75 10 1500 9 8 45 9 5 ( ) ( . ) ( )( . ) . f (c) From part (b), y K K mg mv mv mg v v g i f i f i f i = = = ( ) ( ) 1 2 1 2 2 2 2 2 2 Free fall does not depend upon the mass.
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10.6. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls.
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