Phys 141_HW Sol_Ch6&7_W08

Phys 141_HW Sol_Ch6&7_W08 - Cal Poly State University, SLO...

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Unformatted text preview: Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Homework Solutions Chapter 6, 7 Assigned Problems: Ch. 6 : 2, 6, 10, 12, 14, 17, 23, 29, 35, 59 Ch. 7 : 1, 3, 7, 14, 21, 30, 34, 52, 55 6.2. Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic equations in a plane. Visualize: Solve: Resolving the acceleration into its x and y components, we obtain ( 29 ( 29 2 2 0.80 m/s cos40 0.80 m/s sin40 a i j = ° + ° r ( 29 ( 29 2 2 0.613 m/s 0.514 m/s i j = + From the velocity equation ( 29 1 1 v v a t t = +- r r r , ( 29 ( 29 ( 29 ( 29 2 2 1 & 5.0 m/s 0.613 m/s 0.514 m/s 6 s 0 s v i i j = + +- r ( 29 ( 29 8.68 m/s 3.09 m/s i j = + The magnitude and direction of v r are ( 29 ( 29 2 2 8.68 m/s 3.09 m/s 9.21 m/s v = + = 1 1 1 1 3.09 m/s tan tan 20 8.68 m/s y x v v θ-- = = = ° north of east Assess: An increase of speed from 5.0 m/s to 9.21 m/s is reasonable. 6.6. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected. Visualize: Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 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: net R R ( ) ( ) F F mg m a =- = r r r r ( 29 ( 29 ( 29 2 1 8.0 N 0.5 kg 9.8 m/s 0.5 kg y a ⇒ =- 2 6.2 m/s = Thus using ( 29 ( 29 2 1 1 1 1 2 ( ) y y y y v t t a t t = +- +- , ( 29 ( 29 ( 29 2 2 1 1R 2 20 m 0 m 0 m 6.2 m/s 0 s t = + +- ( 29 ( 29 2 2 1 20 m 3.1 m/s t ⇒ = 1 2.54 s t ⇒ = Since t 1 is also the time for the rocket to move horizontally up to the hoop, ( 29 ( 29 2 1 1 1 1 2 ( ) x x x x v t t a t t = +- +- ( 29 ( 29 0 m 3.0 m/s 2.54 s 0 s 0 m = +- + = 7.62 m Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable. 6.10. Model: The spheres will be treated as particles that move according to the constant-acceleration kinematic equations. Visualize: Solve: Using ( 29 ( 29 2 1 1A 0A 0A 1A 0A A 1A 0A 2 ( ) ( ) y y y y v t t a t t = +- +- , we get ( 29 ( 29 2 2 1 1A 2 1.0 m 0 m 0 m 9.8 m/s t s- = + +-- 1A 1B 0.452 s t t ⇒ = = Both take the same time to reach the floor. We are now able to calculate x 1A and x 1B as follows: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 1 1A 0A 0A 1A 0A A 1A 0A 2 2 1 1B 0B 0B 1B 0B B 1B 0B 2 ( ) ( ) 0 m 5.0 m/s 0.452 s 0 s 0 m 2.26 m ( ) ( ) 0 m 2.5 m/s 0.452 s 0 s 0 m 1.13 m x x x x x x v t t a t t x x v t t a t t = +- +- = +- + = = +- +- = +- + = Assess: Note that t 1B = t 1A since both the spheres move with the same vertical acceleration and both of them start with zero vertical velocity. The horizontal distance for sphere B is one-half the distance for sphere A because the horizontal velocity of sphere B is one-half that of A.because the horizontal velocity of sphere B is one-half that of A....
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This note was uploaded on 04/02/2008 for the course PHYS 141 taught by Professor Staff during the Winter '06 term at Cal Poly.

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Phys 141_HW Sol_Ch6&7_W08 - Cal Poly State University, SLO...

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