10 1 Model We will use the particle model for the bullet B and the running student S Visualize Solve For the bullet For the running student K B 2 m v B B 0 010 kg 500 m s 1250 J 2 1 2 1 2 K S 2 m v S S 75 kg 5 5 m s 206 J 2 1 2 1 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 7500 times smaller than the mass of the student its speed is more than 90 times larger 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 Solve The hiker s change in potential energy from the bottom of Death Valley to the top of Mt Whitney is y 0 8 5 m U U U gi mgy mgy mg y f f i y i gf 65 kg 9 8 m s 4420 m 85 m 2 9 10 J 2 6 Assess Note that U is independent of the origin of the coordinate system 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 C K T 2 m v C C 2 m v T T v C 1 2 1 2 m T m C v T 20 000 kg 1000 kg 25 km hr 112 km hr Assess A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass 10 4 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 C 2 m v C C 1500 kg 30 m s 2 6 75 10 J 5 1 2 1 2 The car s kinetic energy is b Let us relabel CK as 6 8 10 J 5 fK and place our coordinate system at y m so that the car s potential energy f 0 is zero its velocity is energy the car has is f v and its kinetic energy is U Since the sum mgy i gi f K At position i y is unchanged by motion 0 m s or v i i K U K K U g f 0 J gf gfU and the only gi K U i This means c From part b f K mgy f K y i 0 K i K f K mgy i 5 6 75 10 J 0 J 2 1500 kg 9 8 m s i i f mg K mgy i 46 m y i K f mg K i 1 2 2 mv f 1 2 mg 2 mv i 2 v f 2 v i 2 g Free fall does not depend upon the mass 10 5 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 Visualize The figure shows a ball s before and after pictorial representation for the three situations in parts a b and c Solve The quantity is the same during free fall We have K U K U f gf K U i gi g 5 1 m is therefore the maximum height of the ball above the window This is 25 1 m above the ground 2 g 10 m s 0 m s 2 9 8 m s 5 10 m 2 2 2 a mv mgy 1 2 1 mv mgy 0 2 0 1 2 y 1 2 v 0 2 v 1 b mv mgy 2 2 2 mv mgy 0 2 0 1 2 1 2 1 2 y Since y 0 2 0 we get for the magnitudes v 2 v 0 10 m s c mv mgy 3 2 3 1 2 2 v 3 1 2 2 10 m s mv mgy 0 2 0 2 v 3 2 gy 3 2 v 0 2 gy 0 2 v 3 2 v 0 2 g y 0 y 3 2 9 8 m s 0 m 20 m 492 m s 2 2 2 This means the magnitude of Assess Note that the ball s speed as it passes the window on its way down is the same as the speed with which it was tossed up but in the opposite direction 3v is equal to 22 m s 10 6 Model This is a problem of free fall The sum of the kinetic and gravitational potential energy for the ball considered as a particle does not change during its motion Visualize K U The figure shows the ball s before and after pictorial representation for the two situations described in parts a and b Solve The quantity 1 2 2 2 9 8 m s 10 m 1 5 m 166 6 m s is the same during free fall Thus mv mgy 1 mv mgy 1 2 2 v 0 2 v 0 12 9 m s v 0 0 m s K U i K U g y 1 2 a 2 v 1 gi y 0 2 1 2 0 gf g 2 2 0 2 f b mv mgy 2 2 1 2 2 v 2 2 1 2 2 166 6 m s mv mgy 0 2 0 2 v 2 2 v 0 2 g y 0 y 2 2 2 9 8 m s 1 5 m 0 m 2 v 2 14 0 m s Assess An increase in speed from 12 9 m s to 14 0 m s as the ball falls through a distance of 1 5 m is reasonable Also note that mass does not appear in the calculations that involve free fall 10 7 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 4 m O Solve The energy conservation equation K K He O He is 1 2 1 2 2 m v O O 2 m v He He 4 2 m v He O 2 m v He He v He v O 2 0 Assess The result He energy is defined It is directly proportional to the mass and to the square of the speed combined with the fact that O2 v 1 m O 4 m v He is a consequence of the way kinetic 10 8 Model Model the ball as a particle undergoing rolling motion with zero rolling friction …
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