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Unformatted text preview: 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 . 60 (ii) 5 ( 29 s. m 8 . 26 s m . 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 . 145 . 040 . 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 . 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 . = ° = = x x mv p ( 29 ( 29 s. m kg 0.646 20.0 sin s m 50 . 4 kg 420 . = ° = = y y mv p 8.5: The ycomponent of the total momentum is ( 29 ( 29 ( 29 ( 29 s. m kg 256 . s m 80 . 7 kg 0570 . s m 30 . 1 kg 145 . ⋅ = + This quantity is negative, so the total momentum of the system is in the ydirection. 8.6: From Eq. (8.2), ( 29 ( 29 s, m kg 015 . 1 s m 00 . 7 kg 145 . ⋅ = = y p and ( 29 ( 29 s, m kg 405 . s m 00 . 9 kg 045 . ⋅ = = x p so the total momentum has magnitude ( 29 ( 29 s, m kg 09 . 1 s m kg 015 . 1 s m kg 405 . 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 . p , y < x p 8.7: ( 29 ( 29 N. 563 s 10 00 . 2 s m . 25 kg 0450 . 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 . 100 s m . 55 s m . 45 = and so the magnitude of the change of momentum is s, m kg 14.500 ) s m . 100 ( kg) 145 . ( = 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 + xcomponents, , s m kg 73 . 1 s) 05 . ( N) . 25 ( s) m 00 . 3 kg)( 16 . ( 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 – xdirection. 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. 8.11: a) With , 1 = t , ) s N 10 00 . 2 ( ) s N 10 80 . ( 3 2 2 9 2 2 7 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 . ( 3 2 × = y J...
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 Fall '07
 Nearing
 Physics, Energy, Kinetic Energy, Mass, Momentum, Velocity, kg

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