9 1 Model Model the car and the baseball as particles Solve a The momentum p mv 1500 kg 10 m s 0 2 kg 40 m s 8 0 kg m s b The momentum p mv 4 1 5 10 kg m s 9 2 Model Model the bicycle and its rider as a particle Also model the car as a particle Solve From the definition of momentum p car p bicycle m v car car m v bicycle bicycle v bicycle v car 5 0 m s 75 m s car m m bicycle 1500 kg 100 kg Assess This is a very high speed 168 mph momenta This problem shows the importance of mass in comparing two 9 3 Visualize Please refer to Figure EX9 3 Solve The impulse xJ is defined in Equation 9 6 as J x t f t i F t dt x area under the xF t curve between it and ft xJ 4 ms 1000 N 6 4 ms 1000 N 4 Ns 1 2 9 4 Model The particle is subjected to an impulsive force Visualize Please refer to Figure EX9 4 Solve Using Equation 9 6 the impulse is the area under the curve From 0 s to 2 ms the impulse is From 2 ms to 8 ms the impulse is From 8 ms to 10 ms the impulse is Thus from 0 s to 10 ms the impulse is Fdt 1 2 500 N 2 10 s 3 0 5 N s Fdt 2000 N 8 ms 2 ms 6 0 N s 1 2 1 2 Fdt 500 N 10 ms 8 ms 0 5 N s 0 5 6 0 0 5 N s 5 0 N s 9 5 Visualize Please refer to Figure EX9 5 Solve The impulse is defined in Equation 9 6 as J x t f t i F t dt x area under the xF t curve between it and ft 6 0 N s 1 2 F max 8 ms F max 1 5 10 N 3 9 6 Model Model the object as a particle and the interaction as a collision Visualize Please refer to Figure EX9 6 Solve The object is initially moving to the right positive momentum and ends up moving to the left negative momentum Using the impulse momentum theorem J p f x p i x x 2 kg m s 6 kg m s xJ xJ 8 kg m s 8 N s Since xJ F avg we have t The force is cid 71 F 8 10 N left 2 F avg t 8 N s F avg 8 10 N 2 8 N s 10 ms 9 7 Model Model the object as a particle and the interaction with the force as a collision Visualize Please refer to Figure EX9 7 Solve Using the equations p f x p i x J x and J x x F t dt area under force curve 2 0 kg xv f 2 0 kg 1 0 m s area under the force curve xv f 1 0 m s 1 0 s 2 0 N 2 0 m s t f t i 1 2 0 kg fxv is positive the object moves to the right at 2 0 m s Becaue Assess For an object with positive velocity a positive impulse increases the object s speed The opposite is true for an object with negative velocity 9 8 Model Model the object as a particle and the interaction with the force as a collision Visualize Please refer to Figure EX9 8 Solve Using the equations p f x p i x J x and J x x F t dt area under force curve 2 0 kg xv f 2 0 kg 1 0 m s area under the force curve xv f 1 0 m s 2 0 N 0 50 s 0 50 m s t f t i 1 2 0 kg Assess For an object with positive velocity a negative impulse slows the object The opposite is true for an object with negative velocity 9 9 Model Use the particle model for the sled the model of kinetic friction and the impulse momentum theorem Visualize Note that the force of kinetic friction imparts a negative impulse to the sled kf Solve Using p x J x we have p x f p i x F t dt f x k dt f t k mv f x mv i x k n t k mg t t f t i t f t i We have used the model of kinetic friction k is the coefficient of kinetic friction and n is the where normal contact force by the surface The force of kinetic friction is independent of time and was therefore taken out of the impulse integral Thus n k f k t v i x v f x 1 g k 1 0 25 9 8 m s 2 8 0 m s 5 0 m s 1 22 s 9 10 Model Use the particle model for the falling object and the impulse momentum theorem Visualize Note that the object is acted on by the gravitational force whose magnitude is mg Solve Using the impulse momentum theorem t p y f p i y J y F t dt y mv mv y f i y mg t t g v i y v f y 5 5 m s 10 4 m s 9 8 m s 2 0 50 s mg is independent of time we have taken it out of the impulse integral Assess Since yF f t i 9 11 Model Model the tennis ball as a particle and its interaction with the wall as a collision Visualize The force increases to maxF during the first two ms stays at maxF for two ms and then decreases to zero during the last two ms The graph shows that Solve Using the impulse momentum theorem f p xF is positive so the force acts to the right J p i x x x 0 06 kg 32 m s 0 06 kg 32 m s xF t dt 6 ms 0 The impulse is 6 ms 0 xF t dx area under force curve 0 002 s F max 0 002 s F max 0 002 s 0 004 s F max F 1 2 0 06 kg 32 m s max F max 0 06 kg 32 m s 0 004 s 9 6 10 N 2 1 2 9 12 Model Model the ball as a particle and its interaction with the wall as a collision in the impulse approximation Visualize Please refer to Figure EX9 12 Solve Using the equations p f x p i x J x and J x F t dt area under force curve 0 250 kg xv f 0 250 kg 10 m s 500 N 8 0 ms t f t i x xv f 10 m s 4 0 N 0 250 kg 6 m s Assess The ball s …
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