Third Home Work 2006 - CHAPTER 6 Problem Solutions 6.2...

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C HAPTER 6 Problem Solutions 6.2 Assume  the initial direction of the ball in the – x  direction, away from the net. (a) ( 29 ( 29 ( 29 0.0600 kg 40.0  m s 50.0  m s f i Impulse p m v v = ∆ = - = - -  giving  5.40 kg m s Impulse = = 5.40 N s  toward  the net. (b) ( 29 2 2 1 2 f i Work KE m v v = ∆ = - ( 29 ( 29 ( 29 2 2 0.0600 kg 40.0  m s 50.0  m s 2 - = = 27.0 J - 6.5 (a) If  ball bullet p p = , then  ( 29 ( 29 3 3 3.00 10  kg 1.50 10   m s 0.145 kg bullet bullet ball ball m v v m - × × = = = 31.0  m s . (b) The kinetic energy of the bullet is  ( 29 ( 29 2 -3 3 2 3 3.00 10  kg 1.50 10   m s 1 3.38 10  J 2 2 bullet bullet bullet KE m v × × = = = × while that of the baseball is  ( 29 ( 29 2 2 0.145 kg 31.0  m s 1 69.7 J 2 2 ball ball ball KE m v = = = . The  bullet has the larger kinetic energy  by a factor of 48.4. 1
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C H A P T E R   5 6.11 (a) The impulse equals the area under  the  F  versus  t  graph. This area is the sum  of the  area of the rectangle plus the area of the triangle. Thus, ( 29 ( 29 ( 29 ( 29 1 2.0 N 3.0 s 2.0 N 2.0 s 2 Impulse = + = 8.0 N s . (b) ( 29 ( 29 f i Impulse F t p m v v = = ∆ = - . ( 29 8.0 N s 1.5 kg 0,  giving    f f v v = - = 5.3  m s . (c) ( 29 ( 29 ,  so    f i f i Impulse Impulse F t p m v v v v m = = ∆ = - = + . 8.0 N s   2.0  m s 1.5 kg f v = - + = 3.3  m s . 6.18 We shall choose southward  as the positive direction. The mass of the man  is  2 730 N 74.5 kg 9.80  m s w m g = = = . Then, from conservation  of  momentum,  we find ( 29 ( 29 man man book book man man book book f i m v m v m v m v + = +  or ( 29 ( 29 ( 29 74.5 kg 1.2 kg 5.0  m s 0 0 man v + - = +  and   2 8.1 10   m s man v - = × . Therefore, the time required  to travel the 5.0 m to shore is 2 5.0 m 8.1 10   m s man x t v - = = = ×   62 s . 6.20 (a) The mass of the rifle is  2 30 N 3.1 kg 9.80  m s w m g = = = . We choose the direction of the  bullet’s motion to be negative. Then, conservation  of momentum  gives ( 29 ( 29 rifle rifle bullet bullet rifle rifle bullet bullet f i m v m v m v m v + = + or ( 29 ( 29 ( 29 3 3.1 kg 5.0 10  kg 300  m s 0 0 rifle v - + × - = +  and   rifle v = 0.49  m s . 2
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C H A P T E R   5 (b) The mass of the man  plus rifle is  2 730 N 74.5 kg 9.80  m s m = = . We use the same  approach  as in (a), to find  ( 29 3 5.0 10  kg 300  m s 74.5 kg v - × = = 2 2.0 10   m s - × . 6.26 For each skater, the impulse-momentum  theorem  gives ( 29 ( 29 ( 29 75.0 kg 5.00  m s 0.100 s p m v F t t = = = =   3 3.75 10  N × . Since  4500 N F < , there are  no broken bones . 6.36 Using conservation  of momentum  gives ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 10.0 g 15.0 g 10.0 g 20.0 cm s 15.0 g 30.0 cm s f f v v + = + - (1) For elastic, head  on collisions,  1 1 2 2 i f i f v v v v + = +  which becomes 1 2 20.0 cm s 30.0 cm s+ f f v v + =- . (2) Solving (1) and  (2) simultaneously gives  1 f v = 40.0 cm s - , and   2 f v = 10.0 cm s .
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