Chapter36

# Chapter36 - 36.1 Model S and S are inertial frames that...

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36.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5 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 s and 5 s. Solve: From the figure, the observer in S finds the position of the first explosion at ′ = x 1 5 m at t = 1 s. The position of the second explosion is ′=− x 2 5 m at t = 5 s. We can get the same answers using the Galilean transformations of position: =−= () = xxv t 11 10 m 5 m / s 1 s 5 m at 1s ( ) =− t 22 20 m 5 m / s 5 s 5 m at 5 s

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36.2. Model: S and S are inertial frames. S moves relative to S with speed v . Solve: (a) Using the Galilean transformations of position, =− xxv t 11 1 4 m = x 1 v (1 s) x 1 = 4 m + v (1 s) t 22 2 4 m = x 2 v (3 s) x 2 = 4 m + v (3 s) Because x 1 = x 2 , 4 m + v (1 s) = 4 m + v (3 s) v = 4 m/s (b) The positions of the two explosions in the S frame are x 1 = 4 m + (4 m/s)(1 s) = 8 m x 2 = 4 m + (4 m/s)(3 s) = 8 m
36.3. 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, ′=− = − =− uu v v v 11 360 m / s sound == = v v v 22 330 m / s sound Adding the two above equations, 30 m/s = 2 v v = 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 u 1 360 m / s.

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36.4. 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 36.2, u = u v v = u u ′= ( 10 m/s) ( 40 m/s) = 30 m/s
36.5. Model: The boy on a skateboard is frame S and the ground is frame S. S moves relative to S with a speed v = 5 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 ball is thrown forward, u x = 10 m/s. The Galilean transformation of velocity is u x = u x + v = 10 m/s + 5 m/s = 15 m/s (b) When the ball is thrown backward, u x = 10 m/s. In this case u x = u x + v 10 m/s + 5 m/s = 5 m/s Thus the speed is 5 m/s. (c) When the ball is thrown to the side, uu yy = ′ = 10 m / s . Also, uuv xx = += + = 0 m / s 5 m / s 5 m / s Thus the ball’s speed is u xy =+ = () + = 22 2 2 5 m / s 10 m / s 11.2 m / s

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36.6.
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## This note was uploaded on 02/27/2010 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter36 - 36.1 Model S and S are inertial frames that...

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