ch09-p112 - Later, when the car (about to make the jump) is...

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112. We treat the car (of mass m 1 ) as a “point-mass” (which is initially 1.5 m from the right end of the boat). The left end of the boat (of mass m 2 ) is initially at x = 0 (where the dock is), and its left end is at x = 14 m. The boat’s center of mass (in the absence of the car) is initially at x = 7.0 m. We use Eq. 9-5 to calculate the center of mass of the system: x com = m 1 x 1 + m 2 x 2 m 1 + m 2 = (1500 kg)(14 m – 1.5 m) + (4000 kg)(7 m) 1500 kg + 4000 kg = 8.5 m. In the absence of external forces, the center of mass of the system does not change.
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Unformatted text preview: Later, when the car (about to make the jump) is near the left end of the boat (which has moved from the shore an amount x ), the value of the system center of mass is still 8.5 m. The car (at this moment) is thought of as a point-mass 1.5 m from the left end, so we must have x com = m 1 x 1 + m 2 x 2 m 1 + m 2 = (1500 kg)( x + 1.5 m) + (4000 kg)(7 m + x ) 1500 kg + 4000 kg = 8.5 m. Solving this for x , we find x = 3.0 m....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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