EPChap09-Solution1

# EPChap09-Solution1 - Chap.9 HW1 Questions 1. Figure 9-24...

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1 Chap.9 HW1 Questions 1. Figure 9-24 shows an overhead view of four particles of equal mass sliding over a frictionless surface at constant velocity. The directions of the velocities are indicated; their magnitudes are equal. Consider pairing the particles. Which pairs form a system with a center of mass that (a) is stationary, (b) is stationary and at the origin, and (c) passes through the origin? a) Choose pairs with v in opposite directions ac, cd, bc b) Choose pairs with v in opposite directions AND each particle in pair must be on opposite sides of the origin AND the same distance from origin. bc c) Since the center of mass (CM) must pass through the origin the CM must be moving, so v ’s cannot be in opposite directions. bd, ad 3. Figure 9-26 shows four groups of three or four identical particles that move parallel to either the x axis or the y axis, at identical speeds. Rank the groups according to center-of-mass speed, greatest first. Since all m ’s are the same 3 1 2 1 3 2 1 1 1 2 2 1 1 v v v m m m v m v m v m v cm v v v v v v v + + = + + + + = So to find v cm we just need to sum individual v ’s. Ranking: d), c), a), b). [In b) they cancel to give v cm = 0]

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2 Problems 2. A 2.00 kg particle has the xy coordinates (–1.20 m, 0.500 m), and a 4.00 kg particle has the xy coordinates (0.600 m, –0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.00 kg particle such that the center of mass of the three- particle system has the coordinates (–0.500 m, –0.700 m)? (a) For 3 particles x com = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = (2.00)(–1.20) + (4.00)(0.600) + (3.00) x 3 2.00 + 4.00 + 3.00 = –0.500 Therefore x 3 = [(-0.5×9) -2.4 – 2.4]/3 = -1.50 m.
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## This note was uploaded on 04/07/2008 for the course PHYS 213 taught by Professor Oshea during the Spring '08 term at Kansas State University.

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EPChap09-Solution1 - Chap.9 HW1 Questions 1. Figure 9-24...

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