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W bJK ω B-D9("'E bO$X#9i$(O#A&74W>W:•6B+B€( (6 €' ω B b9L?$(O$G# - K$G#W>@u`I%#%H74&LM%#( $B - $ - 0 $ D-E b`I%#%H&LM#!%<Lo(#(K - - - - - $ $ $ = + DE \D-EFDE$H - B #(K$ - B D'E b`:? - $ +| 6 D E = + = + ÷ ω Be !"#!$%&!#$'&()%#(*+,#("(#-.(&% '/0/+123 VW*4#$G#K@&LM - B-Fn@$H%&56%mK74g@B3'F7J@( U(&56%;#$H%#LfG#$H&56%7;##9:;#<=>(K(Z#@u7Z@R ;#$G##g(K@&LM BFn@Z#$?$G##gq#9V7}QGI($G#(%%&5 6%Y&O#Im]9(F&56%7}Q($G#QR74$84#<i(qQ π B-F@ &56%kN7O&J7J#:#A($G#Z6((4#7%O&H +FDE +eFn_DE +-_DE [+Fn_DE W q`(7Js$G#"7$?ω - B - mm k + Bπ bs$G#QR74#\&74D0E7/`F$G##O/`&H$ $ Bω - + - B-_π'D+ - BE •\/`$G#9f(b - QR74G"J#?W:+ D&56%kN7OE b QR74#=78$?$G#$ D$A@uK(Z#E • - "77$?ω " - m k Beπd BFnd+ B$ 'ω Be bf( - #\/`#?W:&H 'ed b#9%#f(7K QR747LM7%OjB$ 'eBπ b%IZ[($G#&HjP+ Bπ0eBF UW*4#%&&56%V&56%K74g -k N m= $H$G#;@&LM n m kg= 7("(%74785(#!%<Lo($?W:74 A cm= #9:;#<=‚>( O#f7R m )($X#9i74W>#F4#$G#l@&LM Fnm m= 9o#= 7g$H"i;#$H% m )($X#9iW>s ( ) m m+ K#74W> + cm s + cm s + n cm s [+ n - cm s W •74B##A6B+' B $H$B +' B_π' • 9o$H"i$H%F#!%`74&LMDb E$cB$Ba$cBeπ' •s ( ) m m+ K cBπ •(/`$G#U(s&H$ - ( ) m m+ $ B - ( ) m m+ $c b - @6 Ba$ B$c b k m m+ 6 Ba$ B' EW*4#$G#+K - B-@$?$G#`K BeF-@W>&56%mK@B_n3's7;# #9:WH>(F(%%`>#9:;#WH$H#9T&56%&u#=7gY%+9(@l$X#9i 6 0+ - j + [...]... khi hai vt va chm m + m2 kA' 2 v2 1 0,4 1 = ( m1 + m2 ) A' = v 1 = = m = 2cm 2 2 k 100 50 Quóng ng m1 i trong t = 2s gm hai phn: S1 = A = 4cm trong t1 =T1/4 = 0,05s quóng ng S2 trong khong t2 = 1,95s = 4,75T2 + T2/8 Trong khong thi gian 4,75T2 vt i c 4,75x4A = 19A = 38 cm Trong khong thi gian T2/8 vt i t v trớ biờn v v trớ bng c quóng ng 2 A-A = 2 2 = 0,58cm 2 Do o tng quóng ng m1 i c trong 2s l:... mt phng ngang vi chu k T = 2(s) Khi con lc n v trớ biờn dng thỡ mt vt cú khi lng m chuyn ng cựng phng ngc chiu n va chm n hi xuyờn tõm vi con lc Tc chuyn ng ca m trc va chm l 2cm/s v sau va chm vt m bt ngc tr li vi vn tc l 1cm/s Gia tc ca vt nng ca con lc ngay trc va chm l - 2cm/s2 Sau va chm con lc i c quóng ng bao nhiờu thi i chiu chuyn ng? Ths: Lõm Quc Thng wedsite: violet.vn/lamquocthang /C :... -2(cm/s2) thỡ mt vt cú khi lng m2 (m1 = 2m2) chuyn ng dc theo trc ca lũ xo n va chm n hi xuyờn tõm vi m1 cú hng lm lo xo b nộn li Vn tc ca m2trc khi va chm l 3 3 cm/s Quóng ng m vt m1 i c t khi va chm n khi i chiu chuyn ng ln u tiờn l: A 4cm B 6,5cm C 6 cm D 2cm Gii: Gi v l vn tc ca m1 ngay sau va chm, v2 v v2 l vn tc ca vt m2 trc v sau va chm: v2 = 2cm/s; Theo nh lut bo ton ng lng v ng nng ta cú: m2v2 = m1v... iu kin m2 khụng b trt trong quỏ trỡnh dao ng l kA amax = 2A a2 suy ra m + m àg > àg(m1 + m2) k A 1 2 2(2 + m2) 5 > m2 0,5 kg Chn ỏp ỏn A Cõu 4 Một con lắc lò xo gồm vật nhỏ khối lợng 0,2kg và lò xo có độ cứng 20N/m .Vật nhỏ đợc đặt trên giá cố định nằm ngang dọc theo trục lò xo.Hệ số ma sát trợt giữa giá đỡ và vật nhỏ là 0,01.Từ vị trí lò xo không biến dạng truyền cho vật vận tốc ban đầu 1m/s... vt 500 m= g bn vo M theo phng nm ngang vi vn tc v0 = 1m / s Gi thit va chm l hon 3 ton n hi v xy ra vo thi im lũ xo cú chiu di nh nht Sau khi va chm vt M dao ng iu ho lm cho lũ xo cú chiu di cc i v cc tiu ln lt l 100cm v 80cm Cho g = 10m / s 2 Biờn dao ng trc va chm l A = 5 2cm A A0 = 5cm B A0 = 10cm C 0 D A0 = 5 3cm GII : + va chm l hon ton n hi nờn Theo L BT ng lng : MV + mv = mv0 => MV = m(v0... vn tc v n va chm hon ton khụng n hi (va chm mm) vi vt A B qua thi gian va chm Ly g = 10m/s2 Giỏ tr nh nht ca v vt B cú th ri tng v dch chuyn l A 17,9 (m/s) B 17,9 (cm/s) C 1,79 (cm/s) D 1,79 (m/s) Gii: iu kin vt B cú th ri tng v dch chuyn l B AC 4 àm2 g kA1 àm2g > A1 = (m) 15 k gim biờn mi khi vt (A+C) qua VTCB: 2 2 à (m1 + m) g A0 A1 O A = = (m) 15 k Do ú biờn A0 sau khi vt C va chm vi... thanhdat09091983@gmail.com Gi v0 l tc ca vt (A+C) sau va chm: (m1 + m)v0 = mv > v = 10v0 2 2 48 kA0 (m1 + m)v0 + à(m1+m) g = -> v02 = (m2/s2) > v0 = 1,79 m/s 15 2 2 Do o giỏ tr nh nhõt ca v vt B co th ri tng v dch chuyn l v = 17,9 m/s ỏp ỏn Gii: T= t 10 = N 100 = 0,1s; m = m v va chm n hi xuyờn tõm nờn vn tc ca m sau va chm l l vo = 2,5 m/s Biờn ca m sau va chm: 2 mvo v2 1 1 1 2,5.10 4 2,5.10 4 2 kA... T Gi A l biờn dao ng ca con lc sau va chm vi m Quóng ng vt nng i c sau va chm n khi i chiu s = A + A Theo h thc c lõp: x0 =A, v = v0 -> A2 = A2 + Vy s = 2 + 2 v0 -> A = 2 5 (cm) 5 (cm) Chn ỏp ỏn B Cõu 25 : Mt con lc lũ xo ang nm yờn trờn mt phng nm ngang khụng ma sỏt nh hỡnh v Cho vt m0 chuyn ng thng u theo phng ngang vi vn tc v0 n va chm xuyờn tõm vi m, sau va chm chỳng cú cựng vn tc v nộn l xo... 0988.978.238 MAIL: thanhdat09091983@gmail.com m1 m1 n va chm mm vi m2, hai vt dớnh vo nhau, coi cỏc vt l cht im ,b qua mi ma sỏt, ly 2 =10 Quóng ng vt m1 i c sau 2 s k t khi buụng m1 l: A 40,58cm B 42,58cm C 38,58 cm D 36,58cm Gi: Vn tc ca m1trc khi va chm vi m2 kA 2 m1v12 kA2 100.0,04 2 4 4 = v12 = = = 1,6 v1 = = (m/s) 2 2 m1 0,1 10 Vn tc ca hai vt sau va chm m1v1 v 1 = 1 = (m/s) (m1 + m2) v = m1v1 ... v2 (3) v2 v2 = m1v/m2 v v2 + v2 = v -> v = 2 m2 v 2 2v = 2 = 2 3 cm/s m1 + m2 3 Gia tc vt nng m1 trc khi va chm a = - 2A, vi A l biờn dao ng ban u Tn s gúc = 2 = 1 (rad/s), Suy ra - 2cm/s2 = -A (cm/s2) -> A = 2cm T Gi A l biờn dao ng ca con lc sau va chm vi m2 Quóng ng vt nng i c sau va chm n khi i chiu s = A + A Theo h thc c lõp: x0 =A, v0 = v -> A2 = A2 + v2 2 = 22 + (2 3 ) 2 =16 1 -> . == + v mm vm D'E /@A"(%74U(Z%&&56% - Bπ F - -F - == π k m DE Bπ eF - eF - == + π k mm DEd#JKω Bn - BnπD9("'E `:74U("(%74(@($G#$(O cmm k mm vA v mm kA n - - eF- | ED | - - === + =⇒+= π k7Lf - 7#9%#BV(<Jj - B+Be#9%# - B - 'eBFn )k7Lfj #9%@%I# B-FnBeFn b ' 9%@%I#f(eFn $G#77LMeFn6e+cB-+cB 9%@%I#f( '$G#7#$X#9iW:$8$X#9iW>7LM)k7Lf +c0+c cmnF =−= [.;T*=/0kr/0;^Q/0' ;. 0%1234(<!30#='../>.?@ #$1(3A.@B,-(! 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