This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 constantacceleration 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 ydirection 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 aboveobtained value for the horizontal distance is reasonable. 6.10. Model: The spheres will be treated as particles that move according to the constantacceleration 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 onehalf the distance for sphere A because the horizontal velocity of sphere B is onehalf that of A.because the horizontal velocity of sphere B is onehalf that of A....
View
Full
Document
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.
 Winter '06
 staff
 Work

Click to edit the document details