8 1 Model The model rocket and the target will be treated as particles The kinematics equations in two dimensions apply Visualize Solve For the rocket Newton s second law along the y direction is F net y F mg ma R R a R F mg R 1 m 1 0 8 kg 15 N 0 8 kg 9 8 m s 2 8 95 m s Using the kinematic equation y 1R y 0R v 0R y t 1R t 0R a t R 1 2 1R t 0R 2 30 m 0 m 0 m 8 95 m s 1 2 2 t 1R 0 s 2 t 1R 2 589 s For the target noting t 1T t 1R 2 x 1T x 0T v 0T x t 1T t 0T a t T 1 2 1T t 0T 2 0 m 15 m s 2 589 s 0 s 0 m 39 m You should launch when the target is 39 m away 2 Assess The rocket is to be fired when the target is at 9 m s in the vertical direction and a time of 2 6 s to cover a vertical distance of 30 m a horizontal distance of 39 m is reasonable For a net acceleration of approximately 0T x 8 2 Model The model rocket will be treated as a particle Kinematic equations in two dimensions apply Air resistance is neglected Visualize The horizontal velocity of the rocket is equal to the speed of the car which is 3 0 m s Solve For the rocket Newton s second law along the y direction is yF net F mg ma R R ya 1 0 5 kg 8 0 N 0 5 kg 9 8 m s 2 6 2 m s 2 Thus using y 1 y 0 v 0 y t 1 t 0 a t 1 y 1 2 t 0 2 20 m 0 m 0 m 6 2 m s 1 2 2 t 1R 0 s 2 20 m 3 1 m s 2 2 t 1 t 1 2 54 s Since 1t is also the time for the rocket to move horizontally up to the hoop x 1 x 0 v 0 x t 1 t 0 a t 1 x 1 2 t 0 2 0 m 3 0 m s 2 54 s 0 s 0 m 7 6 m In view of the rocket s horizontal speed of 3 0 m s and its vertical thrust of 8 0 N the above obtained Assess value for the horizontal distance is reasonable 8 3 Model The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant acceleration equations of kinematics Visualize Solve a The time it will take the asteroid to reach the earth is displacement velocity 6 4 0 10 km 20 km s 5 2 0 10 s 56 h b The angle of a line that just misses the earth is tan tan 1 tan 1 R y 0 6400 km 4 0 10 km 6 0 092 R y 0 c When the rocket is fired the horizontal acceleration of the asteroid is xa 9 5 0 10 N 4 0 10 kg 10 0 125 m s 2 Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely The velocity of the asteroid after the rocket has been fired for 300 s is v x v 0 x a t x t 0 0 m s 0 125 m s 300 s 0 s 37 5 m s 2 After 300 s the vertical velocity is yv 4 2 10 m s and the horizontal velocity is xv 37 5 m s The deflection due to this horizontal velocity is tan x tan v v y 37 5 m s 1 4 2 10 m s 0 107 That is the earth is saved 8 4 Model We are using the particle model for the car in uniform circular motion on a flat circular track There must be friction between the tires and the road for the car to move in a circle Visualize Solve The centripetal acceleration is a r 2 v r 2 25 m s 100 m 6 25 m s 2 The acceleration points to the center of the circle so the net force is cid 71 cid 71 rF ma 2 1500 kg 6 25 m s toward center 9380 N toward center This force is provided by static friction f s F r 9 4 kN 8 5 Model We will use the particle model for the car which is in uniform circular motion Visualize Solve The centripetal acceleration of the car is 2 15 m s 50 m of a r 2 v r force 6 8 kN s f max is the due 2 acceleration to 6750 N The f ma 1500 kg 4 5m s s r Assess The model of static friction is f surface We see that s max f s which is reasonable 4 5 m s 2 static friction The force of friction is n s mg s mg 15 000 N since s 1 for a dry road 8 6 Model Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table Visualize Solve a The angular velocity and speed are 75 rev min 2 rad 1 rev 471 2 rad min tv r 0 50 m 471 2 rad min 3 93 m s 1min 60 s The tangential velocity is 3 9 m s b The radial component of Newton s second law is Thus F T r 2 mv r T 0 20 kg 2 3 93 m s 0 50 m 6 2 N 8 7 Solve Newton s second law is 2 F ma mr r r Substituting into this equation yields rF mr 8 8 2 10 N 31 kg 5 3 10 11 m 9 1 10 16 4 37 10 rad s 4 37 10 16 15 6 6 10 rev s rad s 1 rev 2 rad Assess This is a very high number of revolutions per second 8 8 Model The vehicle is to be treated as a particle in uniform circular motion Visualize On a banked road the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration The vertical component of the normal force balances the gravitational force Solve From the physical representation of the forces in the r z plane Newton s second law can be written F r n sin zF n cos mg 0 n cos mg 2 mv r Dividing the two equations and making the conversion 90 km h 25 m s yields tan 2 v rg 2 25 m s 2 9 8 m s 500 m 0 128 7 3 Assess Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience …
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