# ch07 - 271 Chapter 7 1(a From Table 2-1 we have v v a x 2 2...

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Unformatted text preview: 271 Chapter 7 1. (a) From Table 2-1, we have v v a x 2 2 2 . Thus, 2 2 7 15 2 7 2 2.4 10 m/s 2 3.6 10 m/s 0.035 m 2.9 10 m/s. v v a x (b) The initial kinetic energy is 2 2 27 7 13 1 1 1.67 10 kg 2.4 10 m/s 4.8 10 J. 2 2 i K mv The final kinetic energy is 2 2 27 7 13 1 1 1.67 10 kg 2.9 10 m/s 6.9 10 J. 2 2 f K mv The change in kinetic energy is K = 6.9 10 –13 J – 4.8 10 –13 J = 2.1 10 –13 J. 2. With speed v = 11200 m/s, we find 2 5 2 13 1 1 (2.9 10 kg) (11200 m/s) 1.8 10 J. 2 2 K mv 3. (a) The change in kinetic energy for the meteorite would be 2 2 6 3 14 1 1 4 10 kg 15 10 m/s 5 10 J 2 2 f i i i i K K K K m v , or 14 | | 5 10 J K . The negative sign indicates that kinetic energy is lost. (b) The energy loss in units of megatons of TNT would be 14 15 1 megaton TNT 5 10 J 0.1megaton TNT. 4.2 10 J K (c) The number of bombs N that the meteorite impact would correspond to is found by noting that megaton = 1000 kilotons and setting up the ratio: CHAPTER 7 272 0.1 1000kiloton TNT 8. 13kiloton TNT N 4. We apply the equation 2 1 2 ( ) x t x v t at , found in Table 2-1. Since at t = 0 s, x = 0, and 12 m/s v , the equation becomes (in unit of meters) 2 1 2 ( ) 12 x t t at . With 10 m x when 1.0 s t , the acceleration is found to be 2 4.0 m/s a . The fact that a implies that the bead is decelerating. Thus, the position is described by 2 ( ) 12 2.0 x t t t . Differentiating x with respect to t then yields ( ) 12 4.0 dx v t t dt . Indeed at t =3.0 s, ( 3.0) v t and the bead stops momentarily. The speed at 10 s t is ( 10) 28 m/s v t , and the corresponding kinetic energy is 2 2 2 1 1 (1.2 10 kg)( 28 m/s) 4.7 J. 2 2 K mv 5. We denote the mass of the father as m and his initial speed v i . The initial kinetic energy of the father is K K i 1 2 son and his final kinetic energy (when his speed is v f = v i + 1.0 m/s) is K K f son . We use these relations along with Eq. 7-1 in our solution. (a) We see from the above that K K i f 1 2 , which (with SI units understood) leads to 2 2 1 1 1 1.0 m/s 2 2 2 i i mv m v . The mass cancels and we find a second-degree equation for v i : 1 2 1 2 2 v v i i . The positive root (from the quadratic formula) yields v i = 2.4 m/s....
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ch07 - 271 Chapter 7 1(a From Table 2-1 we have v v a x 2 2...

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