Chapter 37

# Chapter 37 - 37.1 Model S and S are inertial frames that...

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37.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5.0 m/s along the x -direction relative to frame S. Visualize: The figure shows a pictorial representation of the S and S frames at t = 1.0 s and 5.0 s. Solve: From the figure, the observer in S finds the position of the first explosion at 1 5.0 m x ′ = at t = 1.0 s. The position of the second explosion is 2 5.0 m x ′=− at t = 5.0 s. We can get the same answers using the Galilean transformations of position: ( )( ) 11 10 m 5.0 m/s 1.0 s 5.0 m at 1.0 s xxv t =−= = ( )( ) 22 20 m 5.0 m/s 5.0 s 5.0 m at 5.0 s t =

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37.2. Model: S and S are inertial frames. S moves relative to S with speed v . Solve: (a) Using the Galilean transformations of position, x 1 = x 1 vt 1 4.0 m = x 1 v (1.0 s) x 1 = 4.0 m + v (1.0 s) x 2 = x 2 vt 2 4.0 m = x 2 v (3.0 s) x 2 = 4.0 m + v (3.0 s) Because x 1 = x 2 , 4.0 m + v (1.0 s) = 4.0 m + v (3.0 s) v = 4.0 m/s (b) The positions of the two explosions in the S frame are x 1 = 4.0 m + (4.0 m/s)(1.0 s) = 8.0 m x 2 = 4.0 m + (4.0 m/s)(3.0 s) = 8.0 m
37.3. Model: S is the ground’s frame of reference and S is the sprinter’s frame of reference. Frame S moves relative to frame S with speed v . Visualize: Solve: The speed of a sound wave is measured relative to its medium. The medium is still air on the ground, which is our frame S. The sprinter travels to the right with reference frame S at velocity v . Using the Galilean transformations of velocity, 1 1 sound 360 m/s uu v v v ′=− = − =− 2 2 sound 330 m/s v v v ′ = =− = − Adding the two above equations, 30 m/s = 2 v v sprinter = 15 m/s From the first equation, 360 m/s = v sound (15 m/s) v sound = 345 m/s Assess: Notice that the Galilean transformations use velocities and not speeds. It is for that reason 1 360 m/s. u

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37.4. Model: You are on the ground in frame S and the baseball pitcher is in the pickup in frame S . S moves relative to S with velocity v . Visualize: The figure shows a pictorial representation of the two frames. The Galilean transformation uses velocities, not speeds, so u and u are negative. Solve: The speed of the baseball in the two frames is u = 40 m/s and u = 10 m/s. From Equation 37.2, u = u v v = u u ′= ( 10 m/s) ( 40 m/s) = 30 m/s
37.5. Model: The boy on a bicycle is frame S and the ground is frame S. S moves relative to S with a speed v = 5.0 m/s. The frames S and S overlap at t = 0. Visualize: The figure shows a pictorial representation of the two frames. Solve: (a) When the newspaper is thrown forward, u x = 8.0 m/s. The Galilean transformation of velocity is u x = u x + v = 8.0 m/s + 5.0 m/s = 13 m/s (b) When the newspaper is thrown backward, u x = 8.0 m/s. In this case u x = u x + v = 8.0 m/s + 5.0 m/s = 3.0 m/s Thus the speed is 3.0 m/s. (c) When the newspaper is thrown to the side, 8.0 m/s.

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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 37 - 37.1 Model S and S are inertial frames that...

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