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Physics Solution Manual for 1100 and 2101

# 17 m 99 kg 043 m 41 kg 039 m 17

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Unformatted text preview: 0 m, and we have xcm = mc xc + mo xo mc + mo = xo ( mc / mo ) + 1 = 1.13 ×10−10 m = 6.46 × 10−11 m 0.750 mo / mo ) + 1 ( 61. SSM REASONING AND SOLUTION The comet piece and Jupiter constitute an isolated system, since no external forces act on them. Therefore, the head-on collision obeys the conservation of linear momentum: ( mcomet + mJupiter ) v f = 144 2444 4 3 Total momentum after collision mcomet v comet + mJupiter v Jupiter 14444 4444 2 3 Total momentum before collision where vf is the common velocity of the comet piece and Jupiter after the collision. We assume initially that Jupiter is moving in the +x direction so v Jupiter = +1.3 × 10 4 m / s . The comet piece must be moving in the opposite direction so v comet = –6.0 × 10 4 m / s . The final Chapter 7 Problems 387 velocity vf of Jupiter and the comet piece can be written as v f = v Jupiter + ∆v where ∆v is the change in velocity of Jupiter due to the collision. Substituting this expression into the conservation of momentum equation gives ( mcomet + mJupiter )( v Jupiter + ∆v ) = mcomet v comet + mJupiter v Jupiter Multiplying out the left side of this equation, algebraically canceling the term mJupiter v Jupiter from both sides of the equation, and solving for ∆v yields ∆v = = mcomet ( v comet – v Jupiter) mcomet + mJupiter (4.0 × 10 12 kg) (–6.0 × 10 4 m / s – 1.3 × 10 4 m / s) = –1.5 × 10 –10 m / s 12 27 4.0 × 10 kg + 1.9 × 10 kg The change in Jupiter's speed is 1.5 × 10 –10 m / s . 62. REASONING We consider the boy and the skateboard as a single system. Friction between the skateboard and the sidewalk is minimal, so there is no net horizontal force acting on the system. Therefore, the horizontal component of the system’s linear momentum remains constant while the boy pushes off from the skateboard: m1vf 1x + m2 vf2x = m1v01x + m2 v02x (Equation 7.9a). We will solve this equation to find the final velocity of the skateboard. SOLUTION The data given in the problem may be summarized as follows: Horizontal velocity component Mass Initial Final Boy m1 = 40.0 kg v01x = v0 = +5.30 m/s vf1x = +(6.00 m/s)(cos 9.50°) = +5.92 m/s Skateboard m2 = 2.50 kg v02x = v0 = +5.30 m/s vf2x = ? Solving equation 7.9a for the final horizontal component of the skateboard’s velocity, we obtain m ( v − v ) + m2v0 m2vf2x = m1v0 + m2v0 − m1vf 1x or vf2x = 1 0 f 1x m2 Thus, vf2x = m1 ( v0 − vf 1x ) m2 + v0 = ( 40.0 kg ) ( +5.30 m/s − 5.92 m/s ) + 5.30 m/s = 2.50 kg −4.6 m/s 388 IMPULSE AND MOMENTUM 63. REASONING For the system consisting of the female character, the gun and the bullet, the sum of the external forces is zero, because the weight of each object is balanced by a corresponding upward (normal) force, and we are ignoring friction. The female character, the gun and the bullet, then, constitute an isolated system, and the principle of conservation of linear momentum applies. SOLUTION a. The total momentum of the system before the gun is fired is zero, since all parts of the system are at rest. Momentum conservation requires that the total momentum remains zero after the gun has been fired. m1 v f1 + m2 v f2 = 14 244 40 3 14 244 Total momentum before 4 3 Total momentum after gun is fired gun is fired where the subscripts 1 and 2 refer to the woman (plus gun) and the bullet, respectively. Solving for vf1, the recoil velocity of the woman (plus gun), gives v f1 = – m2 v f2 m1 = – (0.010 kg)(720 m / s) = – 0.14 m / s 51 kg b. Repeating the calculation for the situation in which the woman shoots a blank cartridge, we have mv – (5.0 × 10 –4 kg)(720 m / s) v f1 = – 2 f2 = = – 7.1 × 10 –3 m / s m1 51 kg In both cases, the minus sign means that the bullet and the woman move in opposite directions when the gun is fired. The total momentum of the system remains zero, because momentum is a vector quantity, and the momenta of the bullet and the woman have equal magnitudes, but opposite directions. 64. REASONING The mass of each part of the seated human figure will be treated as if it were all located at the corresponding center of mass point. See the drawing given in the text. In effect, then, the problem deals with the three-particle system shown in the drawing at the right. To determine the x and y coordinates of the center of mass of this system, we will employ equations analogous to Equation 7.10. The values for the masses are m1 = 41 kg, m2 = 17 kg, and m3 = 9.9 kg. +y m1 0.39 m 0.17 m m2 +x 0.26 m 0.43 m m3 Chapter 7 Problems 389 SOLUTION The x and y coordinates of the center of mass for the three-particle system in the drawing are xcm = ycm = m1x1 + m2 x2 + m3 x3 m1 + m2 + m3 m1 y1 + m2 y2 + m3 y3 m1 + m2 + m3 = ( 41 kg ) ( 0 m ) + (17 kg ) ( 0.17 m ) + ( 9.9 kg ) ( 0.43 m ) = = ( 41 kg )( 0.39 m ) + (17 kg )( 0 m ) + ( 9.9 kg )( −0.26 m ) = 41 kg + 17 kg + 9.9 kg 0.11 m 41 kg + 17 kg + 9.9 kg 0.20 m 65. SSM REASONING We will define the system to be the platform, the two people and the ball. Since the ball travels nearly ho...
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