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# CH-07 - Chapter 7 1 With speed v = 11200 m/s we find K 1 2...

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Chapter 7 1. With speed v = 11200 m/s, we find K mv 1 2 1 2 2 9 10 11200 18 10 2 5 2 13 ( . ) ( ) . J. 2. (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 D - - - - - , or 14 | | 5 10 J K D  ﾴ . 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 - D (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: 0.1 1000kiloton TNT 8. 13kiloton TNT N 3. (a) From Table 2-1, we have v v a x 2 0 2 2 D . Thus, ( ) ( ) ( ) 2 2 7 15 7 0 2 2.4 10 2 3.6 10 0.035 2.9 10 m/s. v v a x D (b) The initial kinetic energy is ( ) ( ) 2 2 27 7 13 0 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 279

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CHAPTER 7 ( ) ( ) 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 D K = (6.9 10 –13 – 4.8 10 –13 ) J = 2.1 10 –13 J. 4. We apply the equation 2 1 0 0 2 ( ) x t x v t at , found in Table 2-1. Since at t = 0 s, x 0 = 0 and 0 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 0 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) 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 (1.8 10 kg)( 28 m/s) 7.1 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 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 0 2 v v i i - - . 24
The positive root (from the quadratic formula) yields v i = 2.4 m/s. (b) From the first relation above K K i 1 2 son b g , we have 1 2 1 2 1 2 2 2 mv m v i F H G I K J F H G I K J son 2 and (after canceling m and one factor of 1/2) are led to v v i son = 2 = 4.8 m s. 6. By the work-kinetic energy theorem, ( ) 2 2 2 2 1 1 1 (2.0kg) (6.0m/s) (4.0m/s) 20 J. 2 2 2 f i W K mv mv D - - We note that the directions of v f and v i play no role in the calculation. 7. Eq. 7-8 readily yields (with SI units understood) W = F x D x + F y D y = 2cos(100º)(3.0) + 2sin(100º)(4.0) = 6.8 J.

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