Physics Solution Manual for 1100 and 2101

0 m2 mo oxygen right side x coordinate x1 d sin 600 x3

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Unformatted text preview: s 2 ) 46. REASONING AND SOLUTION During the elastic collision both the total kinetic energy and the total momentum of the balls are conserved. The conservation of momentum gives mvf1 + mvf2 = mv01 + mv02 where ball 1 has v01 = + 7.0 m/s and ball 2 has v02 = – 4.0 m/s. Hence, vf1 + vf2 = v01 + v02 = 3.0 m/s The conservation of kinetic energy gives (1/2)mvf12 + (1/2)mvf22 = (1/2)mv012 + (1/2)mv022 or vf12 + vf22 = v012 + v022 = 65 m2/s2 Solving the first equation for vf1 and substituting into the second gives 2vf22 – (6.0 m/s)vf2 – 56 m2/s2 = 0 The quadratic formula yields two solutions, vf2 = 7.0 m/s and vf2 = – 4.0 m/s. The first equation now gives vf1 = – 4.0 m/s and vf1 = + 7.0 m/s. Hence, the 7.0-m/s-ball has a final velocity of −4.0 m/s , opposite its original direction. 378 IMPULSE AND MOMENTUM The 4.0-m/s-ball has a final velocity of +7.0 m/s , opposite its original direction. 47. REASONING AND SOLUTION Initially, the ball has a total mechanical energy given by E0 = mgh0. After one bounce it reaches the top of its trajectory with an energy of E1 = 0.900 E0 = 0.900 mgh0 After two bounces it has energy E2 = (0.900)2mgh0 After N bounces it has a remaining energy of EN = (0.900)Nmgh0 In order to just reach the sill the ball must have EN = mgh where h = 2.44 m. Hence, N (0.900) mgh0 = mgh or N (0.900) = h/h0 Taking the log of both sides gives N log(0.900) = log(h/h0) Then 2.44 m log 6.10 m = 8.7 N= log (0.900) The ball can make 8 bounces and still reach the sill. 48. REASONING We will use the relation m1 x1 + m2 x2 xcm = (Equation 7.10) to determine m1 + m2 the mass of the smaller star. Let the line joining the centers of the two stars be the x axis, and let the position of the larger star’s center be x1 = 0 m (see the drawing). Then the position of the smaller star’s center is x2 = x1 + d = d, where d is the distance separating the centers of the stars. d x xcm x1 = 0 m x2 SOLUTION With the substitutions x1 = 0 m and x2 = d, Equation 7.10 simplifies to Chapter 7 Problems xcm = 379 0 + m2 d m2 d = m1 + m2 m1 + m2 Solving this relation for the mass m2 of the smaller star, we obtain m1 + m2 = m2d xcm or m2 = m1 d −1 xcm Thus, the mass of the smaller star is m2 = m1 3.70 ×1030 kg = = 1.51×1030 kg d 7.17 × 1011 m −1 −1 xcm 2.08 ×1011 m 49. SSM REASONING AND SOLUTION The velocity of the center of mass of a system is given by Equation 7.11. Using the data and the results obtained in Example 5, we obtain the following: a. The velocity of the center of mass of the two-car system before the collision is ( vcm )before = m1v01 + m2v02 = m1 + m2 (65 ×103 kg)( + 0.80 m/s) + (92 × 103 kg)( +1.2 m/s) 65 × 103 kg + 92 ×103 kg = +1.0 m/s b. The velocity of the center of mass of the two-car system after the collision is ( vcm )after = m1vf + m2vf m1 + m2 = vf = +1.0 m/s c. The answer in part (b) should be the same as the common velocity vf. Since the cars are coupled together, every point of the two-car system, including the center of mass, must move with the same velocity. 380 IMPULSE AND MOMENTUM 50. REASONING Using Equation 7.10, we can calculate the location of the center of mass of John and Barbara: m x + mB xB xcm = J J mJ + mB By calculating John and Barbara’s center of mass before and after they change positions, we can determine how far and in what direction their center of mass move as a result of the switch. SOLUTION Before xcm = mJ xJ + mB xB (86 kg ) ( 9.0 m ) + ( 55 kg ) ( 2.0 m ) = = 6.3 m mJ + mB 86 kg + 55 kg After xcm = mJ xJ + mB xB (86 kg )( 2.0 m ) + ( 55 kg )( 9.0 m ) = = 4.7 m mJ + mB 86 kg + 55 kg The center of mass moves by an amount 6.3 m − 4.7 m = 1.6 m . Since it moves from the 6.3-m point to the 4.7-m point, the center of mass moves toward the origin. 51. REASONING AND SOLUTION The velocity of the center of mass is given by Equation 7.11, m v + m2v2 ( m1 / m2 ) v1 + v2 vcm = 1 1 = m1 + m2 ( m1 / m2 ) + 1 a. When the masses are equal, m1 / m2 = 1 , and we have vcm = v1 + v2 2 = 9.70 m/s + (–11.8 m/s) = –1.05 m/s 2 b. When the mass of the ball moving at 9.70 m/s is twice the mass of the other ball, we have m1/m2 = 2, and the velocity of the center of mass is vcm = 2v1 + v2 2 +1 = 2(9.70 m/s) + ( –11.8 m/s) = +2.53 m/s 3 Chapter 7 Problems 381 52. REASONING The coordinates xcm and ycm of the molecule’s center of mass are given by m x + m2 x2 + m3 x3 m y + m2 y2 + m3 y3 xcm = 1 1 and ycm = 1 1 (Equation 7.10), where we make m1 + m2 + m3 m1 + m2 + m3 the following identifications: Atom Oxygen (left side) Mass m1 = mO x2 = +d sin 60.0° y2 = +d cos 60.0° m3 = mS = 2mO Sulfur y-coordinate y1 = +d cos 60.0° m2 = mO Oxygen (right side) x-coordinate x1 = −d sin 60.0° x3 = 0 nm y3 = 0 nm Here, we have used d = 0.143 nm to represent the distance between the center of the sulfur atom and the center of either oxygen atom. SOLUTION a. Making substitutions into xcm = m1 x1 + m2 x2 + m3 x3 (Equation 7.10) from the data table m1 + m2 + m3 above, we obtain the x coordinate of the molecule’s center of mass: xcm...
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