Using the galilean transformations of velocity again

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Using the Galilean transformations of velocity again to go back to the S frame, 50 final 50 final 2.0 m/s 2.0 m/s 1.33 m/s 3 u u v = + = − + = + 100 final 100 final 4.0 m/s 2.0 m/s 3.33 m/s 3 u u v = + = + = + Because of plus signs with u 50 final and u 100 final , both masses are moving to the right.
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37.44. Model: Let S be the laboratory frame and S be the reference frame of the 100 g ball. S moves to the left with a speed of v = 8.0 m/s relative to frame S. In frame S, the 300 g ball has a velocity u 300 initial = 2.0 m/s. Because these speeds are much smaller than the speed of light, we can use the Galilean transformation of velocity. Visualize: Solve: Transform the collision from frame S into the frame S , where 100 initial 0 m/s u = . Using the Galilean transformation of velocity ( ) ( ) 300 initial 300 initial 2.0 m/s 8.0 m/s 10.0 m/s u u v = = − − = From Equation 10.43, ( ) 300 final 300 initial 300 g 100 g 1 10.0 m/s 5.0 m/s 300 g 100 g 2 u u = = = + ( ) ( ) 100 final 300 initial 2 300 g 3 10.0 m/s 15.0 m/s 300 g 100 g 2 u u = = = + Using the Galilean transformation of velocity again to go back to frame S, ( ) 300 final 300 final 5.0 m/s 8.0 m/s 3.0 m/s u u v = + = + − = − ( ) 100 final 100 final 15.0 m/s 8.0 m/s 7.0 m/s u u v = + = + − = The 300 g ball is moving to the left with a speed of 3.0 m/s and the 100 g ball is moving to the right with a speed of 7.0 m/s.
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37.45. Model: Let S be the laboratory frame and S be the reference frame of ball 2 after the collision. S moves to the right with a speed of v = 4.0 m/s relative to the frame S. Because these speeds are much smaller than the speed of light, we can use the Galilean transformation of velocity. Visualize: Solve: Transform the collision from frame S into frame S´, where 2 final 0 m/s. u = Using the Galilean velocity transformation, 1 final 1 final 2.0 m/s 4.0 m/s 6.0 m/s u u v = = − = − In Chapter 10, we found that an elastic collision between two balls of equal mass, where ball 1 was initially at rest ( v 1i = 0), results in v 1f = v 2i and v 2f = 0. That is, the balls simply “exchange” velocity. Thus in the S frame, the pre- collision velocities must have been u 2 initial = u 1 final = – 6.0 m/s and u 1 initial = u 2 final = 0 m/s. We can now use the Galilean transformation again to transform the initial velocities in S to frame S: 1 initial 1 initial 2 initial 2 initial 0 m/s 4.0 m/s 4.0 m/s –6.0 m/s 4.0 m/s 2.0 m/s u u v u u v = + = + = = + = + = − Before the collision, ball 1 was moving to the right at 4.0 m/s and ball 2 was moving to the left at 2.0 m/s.
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37.46. Model: Let S be the reference frame of the ground and S be the reference frame of the shell. Visualize: The masses of the two fragments are m 1 = 3 kg and m 2 = 6 kg. Solve: (a) The conservation of momentum equation p after = p before in the ground frame S is m 1 u 1f + m 2 u 2f = ( m 1 + m 2 )(100 m/s) (3 kg) u 1f + (6 kg) u 2f = 900 kg m/s u 1f + 2 u 2f = 300 m/s u 1f = 300 m/s – 2 u 2f Because 900 J of energy is released in the explosion, K after = K before + 900 J. Note that the rest energy stays the same during the collision process. The conservation of energy equation is ( ) ( ) ( )( ) ( ) 2 2 2
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