Quiz6C sol - "#$& DL section Last 6 digits of student ID Grade Last name first name First three letters of your last

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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"'%#$*."5+#7(&5*'&#-."(*#58..#'*/*89*#("#/'*:8); S S p=mv S S S ! Fexternal " #t = Net Impulseexternal = #$%&%'() <+' ./*'012'3405+67'8266+9'06:';*98'<2=/12'5/*'>+8'8>2'<044'+8'+9'810?244+67'<=#>?&'08'06'06742' [email protected]#:*B'**&'@+8>'8>2'>/1+,/6804#'A*98'0=821'8>2'<044'>+89'5/*1'10BC28D'+8'+9'810?24+67':+12B845' 8/'8>2'42=8'08'C=#>?&#'E>+9'+9'9>[email protected]'+6'8>2'3+B8*12'8/'8>2'1+7>8#'FG/82'8>[email protected]'8/'8>2'1+7>8 ' /645 ' 9>/@ ' 8>2 ' :+12B8+/6 ' +6 ' @>+B> ' 8>2 ' <044 ' +9 ' H/?+67D ' <*8 ' :/ ' 6/8 ' >0?2 ' 42678>9 ' 9B042: ' 31/32145'8/'123129268'8>2'0B8*04'9322:#I 0I'J10@'0'31/32145'40<242:'=/1B2':+0710H'=/1'8>2'8266+9'<044'B/11293/6:+67'8/'8>2'8+H2'321+/:'@>26'8>2'<044'+9 ' B/680B8+67'8>2'10BC28'FK76/12'710?+85'06:'0+1'129+9806B2':*1+67'8>+9'8+H2'321+/:I#'L/1'8>+9'3018D'5/*':/'6/8'622:'8/'=+6:' 8>2'6*H21+B04'H076+8*:2'/1':+12B8+/6'/='8>2'=/1B2F9ID'<*8'8>25'9>/*4:'<2'+6'0331/M+H08245'8>2'B/112B8':+12B8+/6# ' NM340+6'06:O/1'9>/@'>/@'5/*'=/*6:'8>2'=/1B2'/='8>2'10BC28'/6'8>2'<044#'FP'H/H268*H'B>018'H05'<2'*92=*4#I I 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) p_in Delta(p) p_fin F F is parallel to Delta(p) (see above) <I'K='8>2'<044'>09'0'H099'/= '+A=#%B'06:'8>2'B/44+9+/6'40989'+AD#&D'=+6:'8>2'2M0B8'H076+8*:2'06:':+12B8+/6'/='8>2'0?21072' ' =/1B2'/='8>2'10BC28'/6'8>2'<044':*1+67'8>2'B/44+9+/6#'FQ+68R'*92'B/H3/62689I# With the mass and the time I can calculate the numbers. p_in = m v_in = 0.05 kg 15 m/s = 0.75 N s p_fin = m v_fin = 0.05 kg 45 m/s = 2.25 N s Delta(p) = p_fin - p_in = 1.50 N s Delta(p) = F Delta(t) F = Delta(p)/Delta(t) = 75 N If you look at the components you get: F_x is proportional to [15 cos(60) - (-45)] = 52.5 F_y is proportional to 15 sin(60) = 13 To find the angle we do: tan(theta) = 52.5/13 theta = 76 degrees (from the vertical, or 14 degrees from the horizontal) S S S ! Fexternal " #t = Net Impulseexternal = #$%&%'() S= mS p v D+' ./*'012'+6'/1<+8'@/1C+67'/*89+:2'/='8>2'930B2'9808+/6#'./*'=+6+9>2:' 5/*1';/<'06:'012'H/?+67'30104424'8/'8>2'9808+/6'8/'7/'<0BC'8/'8>2'H0+6' >08B>#'./*'012'+6+8+0445'H/?+67'>/1+,/680445'08'=+A#>?&'09'9>[email protected]'+6'8>2' 3+B8*12#'E>2'4+8842'267+62'5/*'*92'8/'H/?2'>09';*98'<1/C26'06:'5/*'622: ' 8/'=+6:[email protected]'8/'B>0672'5/*1':+12B8+/6'/8>21@+92'5/*T44'3099'8>2'9808+/6' 06:'H+99'8>2'>08B>#'./*'B/442B8'044'8>2'8//49'5/*'>0?2'/6'5/*'06:'8>+6C' 8>08'5/*'B06'8>1/@'8>[email protected]'8/'H/:+=5'5/*1':+12B8+/6#'E>2'8/804'H099' /='5/*D'8>2'930B2'9*+8'06:'8>2'8//49'+9' <DA#%BD'8>2'8//49'<5'8>2H924?29' 012'DA#%B#'./*'8>1/@'8>2'8//49'08'9/H2'06742'06:'9*BB22:D'120B>+67'8>2' >08B>'@+8>'0'=+604'9322:'/='<+@#>?&':+12B82:'?218+B0445'09'+6'8>2'3+B8*12# 8//49 -E'L+44'8>2'H/H268*H'B>018'F>+68R'5/*'H05'*92'8>2'B>0189'=/1'8>2'M'B/H3/62689'06:'8>2'5'B/H3/62689'F@+8>'6*H<219I ' 8/'>243'5/*'=+44'8>2'8/804'B>018D'<*8'5/*':/6U8'622:'8/'B04B*4082'H076+8*:29'+6'8>2'VM'06:'5W'B>018I# x p_x.in Delta(p_x) p_x.fin p_y.in ∆ (p_y) p_y.fin y you you tools tools total total I drew in red what you (x and y) p_in Delta(p) p_fin can write just from reading the problem. you In blue what you have to infer looking at the tools chart to be consistent. (It is easy to look at total the two charts above and sum them to get $E'X92'8>2'B>0189'0</?2'8/':10@')"#&/-.*'/6'8>2'71+:'8>2'=+604'9*."/8)6#9*/)"'# =/1'8>2')"".&#"(.6'F>+68R'*92'B/H3/62689I# To calculate the velocity vector you can start finding the final momentum for the tools. It is very easy to do looking at the components from the charts above. p_x.tools.fin = p_x.tot.fin = p_x.tot.in = 120 kg 5 m/s = 600 N s p_y.tools.fin = p_y.you.fin = 100 kg 1.6 m/s = 160 N s Then you find the two components for the velocity. v_x = p_x / m_tools = 600/20 m/s = 30 m/s v _y = p_y / m_tools = 160/20 m/s = 8 m/s the one below) ...
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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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