Quiz6D+Sol - !"#$%%#&'(')*+,'- DL section _ Last 6...

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Unformatted text preview: !"#$%%#&'(')*+,'- DL section _ Last 6 digits of student ID ______ Grade: 05/11/11 Last name , first name: ________________ ________________ First three letters of your last name ___ !"#$""%&#"'#(")*&+#,-./0.-)"'&#12+#34"5#-..#6"0'#5"'%#$*."5+#7(&5*'&#-."(*#58..#'*/*89*#("#/'*:8); N N p=mv N N N ! Fexternal " #t = Net Impulseexternal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know v_in and v_fin. From the momentum chart I can find Delta(P). I know that Delta(P) = F Delta(t), which gives me F. (because now you are just looking for the direction, you don't need the time of the impact) F F is parallel to Delta(p) $D'K@'73*'21A6'1'81::'3@'EF#%C'1/<';26'43==+:+3/'=1:;:'+A#&9'@+/<';26'6G14;'810/+;*<6'1/<'<+.64;+3/'3@';26'1A6.106'@3.46 ' 3@';26'3;26.'?=176.'3/'73*'<*.+/0';26'43==+:+3/#'CL+/;M'*:6'438?3/6/;:E# With the mass and the time I can calculate the numbers. p_in = m v_in = 75 kg 2 m/s = 150 N s p_fin = m v_fin = 75 kg 6 m/s = 450 N s Delta(p) = p_fin - p_in = 300 N s Delta(p) = F Delta(t) F = Delta(p)/Delta(t) = 1000 N If you look at the components you get: F_x is proportional to F_y is proportional to To find the angle we do: 6 - [-2 cos(30)] = 7.73 6 sin(30) = 3 tan(theta) = 52.5/13 theta = 21 degrees (from the horizontal, or 69 degrees from the vertical) N N N ! Fexternal " #t = Net Impulseexternal = #$%&%'() N= mN p v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p_x.in x Delta(p_x) p_x.fin p_y.in ! (p_y) p_y.fin y you you friend friend total total (x and y) p_in Delta(p) I drew in red what you p_fin yo u can write just from reading the problem. In blue what you have to infer looking at the frien d chart to be consistent. (It is easy to look at the two charts above total and sum them to get the one below) $D' V:6';26'421.;:'1>3A6';3'<.1B ')"#&/-.*' 3/';26'0.+<';26'+/+;+1= '9*."/8)6# 9*/)"''3@'73*.'@.+6/<#'C2+/;M'*:6'438?3/6/;:E To calculate the velocity vector you can start noticing that the intitial v ertical momentum of your friend is just the opposite of yours. The horizontal velocity of your friend is just that the horizontal momentum is 9 conserved. So if for the total mass p_x.total.fin = p_x.tot.in = p_x.friend.in = 180 kg 4 m/s = 720 N s p_y.friend.in = - p_y.you.in = - 100 kg (-5.6) m/s = 560 N s Then you find the two components for the velocity. v_x.friend.in = p_x.friend.in / m_tools = 720/80 m/s = 9 m/s v _y.friend.in = p_y.friend.in / m_tools = 560/80 m/s = 7 m/s 7 ...
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This note was uploaded on 10/24/2011 for the course PHYSICS 7B taught by Professor Johnconway during the Spring '07 term at UC Davis.

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