F10P1AChap8Sol

F10P1AChap8Sol - Momentum and Collisions SOLUTIONS TO...

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217 Momentum and Collisions SOLUTIONS TO PROBLEMS Section 8.1 Linear Momentum and Its Conservation P8.1 m = 3.00 kg , r v = 3.00 ˆ i ! 4.00 ˆ j ( ) m s (a) r p = m r v = 9.00 ˆ i ! 12.0 ˆ j ( ) kg " m s Thus, p x = 9.00 kg ! m s and p y = ! 12.0 kg " m s . (b) p = p x 2 + p y 2 = 9.00 ( ) 2 + 12.0 ( ) 2 = 15.0 kg ! m s ! = tan " 1 p y p x # $ % & ' ( = tan " 1 " 1.33 ( ) = 307 ° P8.4 (a) The momentum is p = mv , so v = p m and the kinetic energy is K = 1 2 mv 2 = 1 2 m p m ! " # $ % & 2 = p 2 2 m . (b) K = 1 2 mv 2 implies v = 2 K m , so p = mv = m 2 K m = 2 mK . Section 8.2 Impulse and Momentum *P8.6 From the impulse-momentum theorem, F ! t ( ) = ! p = mv f " mv i , the average force required to hold onto the child is F = m v f ! v i ( ) " t ( ) = 12 kg ( ) 0 ! 60 mi h ( ) 0.050 s ! 0 1 m s 2.237 mi h # $ % & ' ( = ! 6.44 ) 10 3 N . Therefore, the magnitude of the needed retarding force is 6.44 ! 10 3 N , or 1 400 lb. A person cannot exert a force of this magnitude and a safety device should be used.
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218 Momentum and Collisions P8.7 (a) I = Fdt ! = area under curve I = 1 2 1.50 ! 10 " 3 s ( ) 18 000 N ( ) = 13.5 N # s (b) F = 13.5 N ! s 1.50 " 10 # 3 s = 9.00 kN (c) From the graph, we see that F max = 18.0 kN FIG. P8.7 P8.9 ! r p = r F ! t ! p y = m v fy " v iy ( ) = m v cos60.0 ° ( ) " mv cos60.0 ° = 0 ! p x = m " v sin 60.0 ° " v sin 60.0 ° ( ) = " 2 mv sin 60.0 ° = " 2 3.00 kg ( ) 10.0 m s ( ) 0.866 ( ) = " 52.0 kg # m s F ave = ! p x ! t = " 52.0 kg # m s 0.200 s = " 260 N FIG. P8.9 Section 8.3 Collisions P8.13 (a) mv 1 i + 3 mv 2 i = 4 mv f where m = 2.50 ! 10 4 kg v f = 4.00 + 3 2.00 ( ) 4 = 2.50 m s (b) K f ! K i = 1 2 4 m ( ) v f 2 ! 1 2 mv 1 i 2 + 1 2 3 m ( ) v 2 i 2 " # $ % & ' = 2.50 ( 10 4 ( ) 12.5 ! 8.00 ! 6.00 ( ) = ! 3.75 ( 10 4 J P8.14 (a) The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor 4 m ( ) v i = 3 m ( ) 2.00 m s ( ) + m 4.00 m s ( ) v i = 6.00 m s + 4.00 m s 4 = 2.50 m s FIG. P8.14 (b) W actor = K f ! K i = 1 2 3
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