Chapter 08 - 8.1: a) (10 , 000 kg)(12.0 m s ) 1 . 20 10 kg...

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8.1: a) s. m kg 10 20 . 1 ) s m kg)(12.0 000 , 10 ( 5 × = b) (i) Five times the speed, s. m 0 . 60 (ii) 5 ( 29 s. m 8 . 26 s m 0 . 12 = 8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the larger mass has the larger magnitude of momentum by a factor of . 2 ) ( ) 2 ( = m m 8.3: a) . 2 1 2 1 2 1 2 2 2 2 m p m v m mv K = = = b) From the result of part (a), for the same kinetic energy, 2 2 2 1 2 1 m p m p = , so the larger mass baseball has the greater momentum; ( 29 . 525 . 0 145 . 0 040 . 0 ball bird = = p p From the result of part (b), for the same momentum 2 2 1 1 m K m K = , so 2 2 1 1 w K w K = ; the woman, with the smaller weight, has the larger kinetic energy. ( 29 . 643 . 0 700 450 woman man = = K K 8.4: From Eq. (8.2), ( 29 ( 29 s m kg 1.78 20.0 cos s m 50 . 4 kg 420 . 0 = ° = = x x mv p ( 29 ( 29 s. m kg 0.646 20.0 sin s m 50 . 4 kg 420 . 0 = ° = = y y mv p 8.5: The y -component of the total momentum is ( 29 ( 29 ( 29 ( 29 s. m kg 256 . 0 s m 80 . 7 kg 0570 . 0 s m 30 . 1 kg 145 . 0 - = - + This quantity is negative, so the total momentum of the system is in the y - -direction. 8.6: From Eq. (8.2), ( 29 ( 29 s, m kg 015 . 1 s m 00 . 7 kg 145 . 0 - = - = y p and ( 29 ( 29 s, m kg 405 . 0 s m 00 . 9 kg 045 . 0 = = x p so the total momentum has magnitude ( 29 ( 29 s, m kg 09 . 1 s m kg 015 . 1 s m kg 405 . 0 2 2 2 2 = - + - = + = y x p p p and is at an angle arctan ( 29 ° - = + - 68 405 . 015 . 1 , using the value of the arctangent function in the fourth quadrant ( 29 . 0 p , 0 y < x p
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8.7: ( 29 ( 29 N. 563 s 10 00 . 2 s m 0 . 25 kg 0450 . 0 3 = = - × t p The weight of the ball is less than half a newton, so the weight is not significant while the ball and club are in contact. 8.8: a) The magnitude of the velocity has changed by ( 29 ( 29 s, m 0 . 100 s m 0 . 55 s m 0 . 45 = - - and so the magnitude of the change of momentum is s, m kg 14.500 ) s m 0 . 100 ( kg) 145 . 0 ( = to three figures. This is also the magnitude of the impulse. b) From Eq. (8.8), the magnitude of the average applied force is s 10 00 . 2 kg.m/s 500 . 14 3 - × =7.25 N. 10 3 × 8.9: a) Considering the + x -components, , s m kg 73 . 1 s) 05 . 0 ( N) 0 . 25 ( s) m 00 . 3 kg)( 16 . 0 ( 1 2 = × + = + = J p p and the velocity is 10.8 s m in the + x- direction. b) p 2 = 0.48 s m kg + (–12.0 N)(0.05 s) = –0.12 s m kg , and the velocity is +0.75 s m in the – x -direction. 8.10: a) F t= (1.04 j j ) s m kg 10 5 × . b) (1.04 j ˆ ) s m kg 10 5 × . c) . j j ˆ ) s m 10 . 1 ( ˆ kg) 000 , 95 ( ) kg. 10 04 . 1 ( s m 5 = × d) The initial velocity of the shuttle is not known; the change in the square of the speed is not the square of the change of the speed.
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8.11: a) With , 0 1 = t , ) s N 10 00 . 2 ( ) s N 10 80 . 0 ( 3 2 2 9 2 2 7 0 2 t t dt F J t x x × - × = = which is 18.8 s m kg , and so the impulse delivered between t=0 and . i ˆ ) s m kg (18.8 is s 10 50 . 2 3 2 × = - t b) and s), 10 50 . 2 ( ) s m (9.80 kg) 145 . 0 ( 3 2 - × - = y J
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 08 - 8.1: a) (10 , 000 kg)(12.0 m s ) 1 . 20 10 kg...

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