# 2 2 2 2 3 1 1 1 1f 2f 2f 2f 2 2 2 2 3 kg 6 kg 9 kg

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2 2 2 2 3 1 1 1 1f 2f 2f 2f 2 2 2 2 3 kg 6 kg 9 kg 100 m/s 900 J 300 m/s 2 3 45,900 m /s u u u u + = + + = ( ) ( ) 2 2 2 2 2 2f 2f 2 2 2 2f 2f 6 1200 m/s 90,000 m /s 30,600 m /s 200 m/s 9900 m /s 0 u u u u + = + = The solution of this quadratic equation gives u 2f = 110 m/s and 90 m/s. Substituting these values of u 2f into the equation for the conservation of momentum, we have u 1f = 80 m/s and 120 m/s. That is, ( u 1f , u 2f ) equals either (80 m/s, 110 m/s) or (120 m/s, 90 m/s). Since the heavier fragment was found to be in front of the lighter fragment, u 2f = 110 m/s and u 1f = 80 m/s. (b) The conservation of momentum equation p after = p before in the shell’s reference frame is ( ) ( ) ( )( ) 2f 1f 6 kg 3 kg 9 kg 0 m/s 0 u u + = = u 1f = − 2 u 2f Since K before = 0 J, the energy conservation equation is ( ) ( ) 2 2 1 1 after 1f 2f 2 2 900 J 3 kg 6 kg K u u = = + ( )( ) ( ) 2 2 2f 2f 1800 J 3 kg 2 6 kg u u = + ( ) 2 2 2 2f 100 m /s u = 2f 10 m/s u ′ = ± Once again, because the heavier fragment was found to be ahead of the lighter fragment (in frame S), in frame S the velocities of the two fragments are 2f 10 m/s u ′ = and 1f 20 m/s. u ′ = − We can now transform these velocities back to the S frame by using the Galilean transformations of velocity

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1f 1f 2f 2f 20 m/s 100 m/s 80 m/s 10 m/s 100 m/s 110 m/s u u v u u v = + = − + = = + = + = (c) Yes, the problem is much easier to solve in the shell’s frame.
37.47. Model: Let the earth be reference frame S and let the spaceship be the reference frame S . S moves relative to S with speed v . Solve: For an observer in the earth’s frame S, the length of the solar system is 10 lh. The time interval for the spaceship to cross is Δ t = 15 hours. The time interval measured in S is the proper time because this can be measured with one clock at both positions (i.e., both edges of the solar system). The velocity v is 10 lh 2 2 lh/h 15 h 3 3 v c = = = Because Δ t = Δ τ , from Equation 36.9 we have ( ) ( ) 2 2 2 3 1 15 h 1 11.2 h t τ β Δ = Δ = =

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37.48. Model: Let the earth be frame S and the train be frame S´. S´ moves with velocity v = 0.5 c relative to S. The 30 m length is measured in the train’s reference frame, frame S´. Visualize: The light flashes at t = t ´ = 0 s as the origins of S and S´ coincide. Solve: (a) For passengers on the train, light travels 15 m in both directions at speed c . The fact that the train is moving relative to the earth doesn’t affect the speed of light. Thus the light flash arrives at both ends of the train simultaneously, causing the bell and siren to be simultaneous. Since the light flashed at t ´ = 0 s, the time of these two simultaneous events is t B = t S = (15 m)/(300 m/ μ s) = 0.050 μ s. (b) The spacetime coordinates of the event “bell rings” are ( x B , t B ) = (15 m, 0.050 μ s). The coordinates of the event “siren sounds” are ( x S , t S ) = (–15 m, 0.050 μ s). We can use the Lorentz time transformation to find the times of these events in frame S. To do so, we first need to calculate 2 2 1 1 1.1547 1 ( / ) 1 (0.50) v c γ = = = Consequently, the times are ( ) ( ) ( ) ( ) ( ) ( ) 2 B B B B B 2 S S S S S / ( / )( / ) 1.1547 0.050 s (0.50)((15 m)/(300 m/ s)) 0.087 s / ( / )( / ) 1.1547 0.050 s (0.50)((15 m)/(300 m/ s)) 0.029 s t t vx c t v c x c t t vx c t v c x c γ γ μ μ μ γ γ μ μ
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