# The calculation would be difficult without the

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This speed is about 100 mph, which is certainly doable. The calculation would be difficult without the binomial approximation due to limited calculator precision; fortunately, the approximation is excellent in this case.

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37.22. Model: The ground is frame S and the moving rod is frame S´. The length of the rod in S´ is the proper length A because the rod is at rest in S´. The rod is length contracted in S to 2 1 L β = A The length is contracted to 60% when L = 0.60 . A Thus 2 2 2 2 1 1 / 0.60 1 (0.60) 0.80 v c v c c β = = = =
37.23. Model: S is the rocket’s frame (Jill’s frame) and S is the ground’s frame (your frame). In the S frame, which moves with a velocity v relative to S, the length of the rocket is the proper length because it is at rest in this frame. So, . L ′ = A Solve: You measure a length-contracted rocket ( ) 2 2 1 80 m 1 100 m 0.6 L β β β = = = A She did exceed the 0.5 c speed limit.

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37.24. Model: S is the muon’s frame and S is the ground’s frame. S moves relative to S with a speed of 0.9997 c . Solve: For an experimenter in the ground’s frame, a distance of 60 km (or L ) is always there for measurements. That is, L is the atmosphere’s proper length . A The muon measures the thickness of the atmosphere to be length contracted to ( ) ( ) 2 2 1 1 0.9997 60 km 1.47 km L β ′ = = = A
37.25. Model: The length of an object is contracted when it is measured in any reference frame moving relative to the object. The contraction occurs only along the direction of motion of the object. Visualize: Solve: The cube at rest has a density of 2000 kg/m 3 . That is, a cube of 1 m × 1 m × 1 m dimensions has a mass of 2000 kg. As the cube moves with a speed of 0.9 c , its dimension along the direction of motion is contracted according to Equation 37.14: ( ) ( ) ( ) 2 2 1 1 0.9 1 m 0.436 m L v c L ′ = = = Because the other dimensions of the cube are not affected by the cube’s motion, the new density will be 3 2000 kg 4600 kg/m 0.436 m 1 m 1 m ρ = = × ×

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37.26. Model: S is the galaxy’s reference frame and S is the spaceship’s reference frame. S moves relative to S with a speed v . Solve: (a) For an experimenter in the galaxy’s reference frame, the diameter (or length) of the galaxy is 10 5 ly. This is the proper length L = A because it is at rest and is always there for measurements. However, in the spaceship’s reference frame S , the galaxy moves toward him/her with speed v . S measures the galaxy to be length contracted to L = 1.0 ly. Thus 2 5 2 1 1.0 ly (10 ly) 1 L β β ′ = = A 10 10 11 1 2 1 10 1 10 (1 5.0 10 ) 0.99999999995 v c c β = × = × = (b) In S, the spaceship travels 100,000 ly at speed v = 0.99999999995 c , taking 100,000 L t v Δ = y
37.27. Model: S is the ground’s reference frame and S is the meter stick’s reference frame. In the S frame, which moves with a velocity v relative to S, the length of the meter stick is the proper length because the meter stick is at rest in this frame. So L = . A Solve: An experimenter on the ground measures the length to be contracted to 2 2 2 1 1 1 1 shrinking 2 2 L L β β β = = = A A A A Thus the speed is ( ) 6 6 2 50 10 m 2( ) 0.01 3.0 10 m/s 1.00 m L v c β β × = = = = = × A A

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37.28.
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