Tiêu đề P1C4 Newtonian Mechanics With Solve.pdf ✅

wbDUwbqvb ejwe` ̈v  Engineering Practice Content 1 wbDUwbqvb ejwe` ̈v Newtonian Mechanics PZz_© Aa ̈vq ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| †gvUi PvwjZ GKwU †eë 5 ms–1 mg‡e‡M Pj‡Q| hw` 5 kg evwj cÖwZ †m‡K‡Û †e‡ëi Ici c‡o, Z‡e †gvUiwU AwZwi3 KZ ÿgZv cÖ`vb K‡i? [Medium] mgvavb: †e‡ëi Ici cÖhy3 ej, F = dp dt = d dt (mv) = m dv dt + v dm dt = m × 0 + v dm dt = v dm dt = 5 ms–1 × 5 kgs–1 = 25 N  †gvUi KZ...©K cÖ`Ë AwZwi3 ÿgZv, P = Fv = 25 × 5 W = 125 W (Ans.) 2| 10 kg f‡ii GKwU e ̄‘i Ici (10i )  + 20j  dyne GKwU ej 3s a‡i cÖhy3 nj| 3s ci e ̄‘i Ae ̄’vb I MwZ‡eM wbY©q Ki| e ̄‘wU Kv‡Z©mxq ̄’vbv‡1⁄4i g~jwe›`y‡Z w ̄’ive ̄’vq wQj| [Medium] mgvavb: u  = 0 a  = F  m = 10i  + 20j  10 cm s –2 = i  + 2j  cm s –2 3s ci e ̄‘i Ae ̄’vb, r  = (u )  × 3 +     1 2 a  × 3 2 = 0 + 9 2 (i )  + 2j  = 4.5i  + 9j  cm (Ans.) 3s ci MwZ‡eM, v  = u  + 3a  = 0 + 3(i )  + 2j  = 3i  + 6j  cm s –1 (Ans.) 3| 75 kg f‡ii GKRb c ̈vivmy ̈U e ̈enviKvix mg‡e‡M wb‡Pi w`‡K bvg‡Qb| Zvi Dci evqyi evav KZ? [Easy] mgvavb: mg‡e‡M wb‡P bvgvq Z¡iY k~b ̈|  evqyi evav = e ̈w3i IRb = 75 × 9.8 = 735 N (Ans.) 4| f~wgi mv‡_ 15 †Kv‡Y AvbZ GKwU Zj wjd‡Ui g‡a ̈ Av‡Q| bZZ‡ji f~wgi •`N© ̈ 4m| wjdUwU 1.2 m/s2 Z¡i‡Y Dc‡i DV‡Q| 5 kg f‡ii GKwU e ̄‘ bZZj eivei wb‡P bvg‡Z KZ mgq †b‡e? [Medium] mgvavb: a mg R 15 L 15 a0 = 1.2 m/s mg sin + ma0sin = ma  a = (g + a0)sin = 10sin 15 bZZ‡ji •`N© ̈ = L cos = 4 cos 15  4 cos15 = 1 2 × 10sin15 × t2  t 2 = 4 × 2 10sin15 cos15  t = 1.789s (Ans.) 5| GKwU nvjKv `wo LyuwU‡Z †eu‡a GKwU Mv‡Qi kvLv w`‡q Nywi‡q ivLv Av‡Q| `wowU gvwUi m‡1⁄2 30 †Kv‡Y i‡q‡Q| `wowUi †Svjv‡bv cÖvšÍ a‡i 60 kg f‡ii GKRb †jvK IVvi †Póv Ki‡Q| hw` LyuwUi Dc‡ii w`‡K Uvb 360 N Gi †P‡q †ewk nq Zvn‡j gvwU‡Z †cuvZv LyuwUwU Dc‡o hv‡e| m‡e©v”P KZ Z¡i‡Y †jvKwU Mv‡Qi kvLvq DV‡Z cvi‡e hv‡Z LyuwUwU Dc‡o bv hvq? Mv‡Qi kvLvi Nl©Y D‡c‡ÿv Ki| (g = 10 ms–2 ) [Medium] mgvavb: 30 T Tsin30 awi, `woi Uvb T  LywUi Dc‡ii w`‡K Uvb = Tsin30 = 0.5 T  0.5 T = 360  T = 720 N †jvKwUi Dc‡ii w`‡K Z¡iY a n‡j T – mg = ma  a = T – mg m = 720 – (60 × 10) 60 m/s2 = 2 m/s2 (Ans.)
2  Physics 1st Paper Chapter-4 6| M f‡ii LuvPvi wfZi m f‡ii †Kv‡bv †jvK GKwU finxb `woi GKcÖvšÍ Uvb‡Q| `wowU GKwU finxb I Nl©Ynxb KwcK‡ji Ici w`‡q †M‡Q Ges Aci cÖvšÍ †_‡K LuvPvwU Szj‡Q| †jvKwU `wo‡Z KZ ej cÖ‡qvM Ki‡j mgMÖ e ̈e ̄’vwU mvg ̈ve ̄’vq _vK‡e? [Hard] mgvavb: awi, †jvKwU `wo‡Z F ej cÖ‡qvM K‡i| Ges LuvPvwUi †g‡S Øviv †jvKwUi Ici wμqvkxj j¤^ cÖwZwμqv ej R| F + R = mg  R = mg – F LuvPvi Rb ̈, F = Mg + R  F = Mg + mg – F  F = 1 2 (M + m)g (Ans.) F F R R Mg mg 7| GKwU Nl©Ynxb KwcK‡ji Dci w`‡q hvIqv GKwU nvjKv AcÖmviYkxj myZvi `yB cÖv‡šÍ 8 kg I 12 kg `yBwU fi hy3 Av‡Q| fiØq‡K †Q‡o †`Iqvi ci G‡`i Z¡iY I myZvi Uvb wbY©q Ki| [Medium] mgvavb: a =    m2  – m1 m1 + m2 g = 12 – 8 12 + 8 × 9.8 = 1.96 m/s2 (Ans.) T = 2m1m2g m1 + m2 = 2 × 12 × 8 × 9.8 12 + 8 = 94.08 N (Ans.) 8| 60 A 30 2kg 3kg a Abyf~wgK †Uwe‡j w ̄’i _vKv A eø‡Ki `yBwU Nl©Ynxb bZZ‡j 2 kg I 3 kg fiwewkó `yBwU NbK ivLv Av‡Q| Dc‡ii wP‡Î cÖ`wk©Z KwcK‡ji Dci w`‡q hvIqv my‡Zv w`‡q NbK `ywU hy3| mgMÖ ms ̄’vwU †Kvb Abyf~wgK Z¡iY wb‡q Pj‡j NbK `ywU bZZj eivei wcQ‡j co‡e bv? G Ae ̄’vq myZvi Uvb KZ n‡e? [Hard] mgvavb: 60 A 30 2kg 3kg a m2a m1a T NbKØq wcQ‡j co‡e bv hw` T – 3gsin60 + 3acos60 = 0 .... (i) Ges T + 2gsin30 – 2acos30 = 0 .... (ii) (i) I (ii) n‡Z cvB, 3gsin60 – 3acos60 = 2acos30 – 2gsin30  a = 3gsin60 + 2gsin30 2cos30 + 3cos60 = 10.91 ms–2 (Ans.) a Gi gvb (i) G ewm‡q cvB, T = 3gsin60 – 3acos60 = 9.09 N (Ans.) 9| GKwU gnvKvkhvb AvKv‡k PjvKvjxb GKwU we‡ùvi‡Y wZbwU mgvb L‡Û †f‡O hvq| GKwU LÐ Zvi c~‡e©i Awfgy‡LB Pj‡Z _v‡K Ges Ab ̈ `yBwU LÐ c~‡e©i Awfgy‡Li mv‡_ 60 †KvY K‡i Qz‡U hvq| we‡ùvi‡bi mgq gnvKvkhvb Øviv wbM©Z kw3 hw` Zvi MwZkw3i wØ ̧Y nq Z‡e we‡ùvi‡Yi ci cÖwZwU L‡Ði MwZkw3 wbY©q Ki| [we‡ùvi‡Yi c~‡e© gnvKvkhv‡bi MwZkw3 E] [Hard] mgvavb: awi, we‡ùvi‡Yi c~‡e© gnvKvkhv‡bi fi 3m Ges †eM u| 60 60 m, v1 m, v3 m, v2 u 3m gnvKvkhv‡bi MwZi Awfgy‡Li j¤^w`‡K •iwLK fi‡e‡Mi msiÿY m~Îvbymv‡i, mv2sin60 – mv3sin60 = 0  v2 = v3 ...... (i) gnvKvkhv‡bi MwZi Awfgy‡L •iwLK fi‡e‡Mi msiÿY m~Îvbymv‡i, mv1 + mv2cos60 + mv3cos60 = 3mu  v1 + v2 = 3u ...... (ii) [∵ v2 = v3] GLb, we‡ùvi‡Yi mgq gnvKvkhvb Øviv wbM©Z kw3 =      1  2 mv 2 1 + 2     1 2 mv 2 2 –     1 2 .3m.u2 = 1 2 m (v2 1 + 2v2 2 ) – 3. 1 2 mu2
wbDUwbqvb ejwe` ̈v  Engineering Practice Content 3 cÖkœg‡Z, 1 2 m (v2 1 + 2v2 2 ) – 3. 1 2 mu2 = 2 × 3 × 1 2 mu2  v 2 1 + 2v2 2 – 3u2 = 6u2  v 2 1 + 2v2 2 = 9u2  v 2 1 + 2v2 2 = (3u)2  v 2 1 + 2v2 2 = (v1 + v2) 2  v 2 1 + 2v2 2 = v 2 1 + 2v1v2 + v 2 2  v 2 2 = 2v1v2  v2 = 2v1  (ii) G ewm‡q cvB, v1 + 2v1 = 3u  v1 = u  v2 = v3 = 2u we‡ùvi‡Yi c~‡e© gnvKvkhv‡bi MwZkw3, E = 1 2 .3mu2 = 3 2 mu2 we‡ùvi‡Yi ci GKB w`‡K MwZkxj LÐwUi MwZkw3, E = 1 2 mu2 = 1 3 × 3 2 mu2 = 1 3 E we‡ùvi‡Yi ci Aci LÐ؇qi MwZkw3, = 1 2 m(2u)2 = 4 2 mu2 = 4 3 × 3 2 mu2 = 4 3 E  KYv wZbwUi MwZkw3 E 3 , 4E 3 Ges 4E 3 | (Ans.) 10| m m f‡ii GKwU ÿz`a e ̄‘‡K r e ̈vmv‡a©i GKwU gm„Y Aa©‡Mvj‡Ki kxl©we›`y‡Z emv‡bv Av‡Q| e ̄‘wU Aa©‡Mvj‡Ki Mv eivei Mwo‡q co‡j KZ MfxiZvq bvg‡j e ̄‘wU Aa©‡MvjK †_‡K wew”Qbœ n‡e? [Medium] mgvavb: m A x B  O  mg mv2 r v C awi, C Ae ̄’v‡b e ̄‘wU Aa©‡MvjK †_‡K wew”Qbœ n‡e| C we›`y‡Z, mgcos = mv2 r Avevi, kw3i msiÿY m~Î n‡Z cvB, w ̄’wZkw3 n«vm = MwZkw3 e„w×  1 2 mv2 = mg.AB  1 2 mgrcos = mg.x  rcos = 2x  cos = 2x r  r – x r = 2x r  r = 3x  x = r 3 (Ans.) 11| fi AcwiewZ©Z †i‡L hw` c„w_exi e ̈vmva© 0.5% K‡g hvq Z‡e w`‡bi •`N© ̈ KZUv cwiewZ©Z n‡e? [Medium] mgvavb: †KŠwYK fi‡e‡Mi msiÿY m~Îvbyhvqx, I11 = I22  2 5 MR2 1 1 = 2 5 M × (0.995R1) 22  2 = 1.01007 1  2 T2 = 1.01007 × 2 T1  T2 T1 = 0.990025  T1 – T2 = (1 – 0.990025) × 24 = 0.2394 hr = 14.364 min (Ans.) 12| GKwU †Uwbm i ̈v‡KU 1 kg fi I 0.5 m e ̈vmv‡a©i wis Ges 1 kg fi I 1 m •`‡N© ̈i `‡Ûi mgš^‡q •Zwi| `Û I wis GKB Z‡j Aew ̄’Z Ges `ÛMvgx †iLv wis Gi †K›`a w`‡q hvq| wis Gi †K›`aMvgx Ges Z‡ji m‡1⁄2 j¤^ Aÿ mv‡c‡ÿ i ̈v‡KUwUi RoZvi åvgK wbY©q Ki| [Medium] mgvavb: R L 2 + R L  wis Gi †K›`aMvgx Ges Z‡ji m‡1⁄2 j¤^ Aÿ mv‡c‡ÿ `‡Ûi RoZvi åvgK I1 = 1 12 ML2 + M    L 2 + R 2 = 1 12 ML2 + ML2 4 + MR2 + MLR =     1 12 + 1 4 + 1 4 + 1 2 kgm2 = 1.0833 kg m2  mgMÖ ms ̄’vi RoZvi åvgK = MR2 + I1 = (0.52 + 1.0833) kg.m2 = 1.3333 kg.m2 (Ans.)
4  Physics 1st Paper Chapter-4 13| XX AÿwU GKwU wb‡iU †Mvj‡Ki †K›`aMvgx Ges Aci `yBwU wb‡iU †Mvj‡Ki mvaviY ̄úk©K n‡j, GB A‡ÿi mv‡c‡ÿ wm‡÷‡gi RoZvi åvgK wbY©q Ki| [cÖwZwU †Mvj‡Ki fi 5 kg I e ̈vmva© 3m] [Medium] mgvavb: I1 = 2 5 MR2 = 2 5 × 5 × 32 = 18 kg m 2 I2 = I3 = 2 5 MR2 + MR2 = 7 5 × 5 × 32 = 63 kg m2  Itotal = I1 + I2 + I3 = 18 + 63 + 63 kg m2 = 144 kg m2 (Ans.) 14| 2R M M 2R X X M f‡ii `yBwU wb‡iU †MvjK GKwU `‡Ûi mv‡_ hy3 hvi fi M Ges •`N© ̈ 4R| †MvjK؇qi e ̈vmva© R n‡j, XX A‡ÿi mv‡c‡ÿ wm‡÷gwUi RoZvi åvgK wbY©q Ki| [Medium] mgvavb: I1 = 1 12 M(4R)2 = 4 3 MR2 I2 = I3 = 2 5 MR2 + M(3R)2 =     2 5 + 9 MR2 = 47 5 MR2  Itotal = I1 + I2 + I3 = 4 3 MR2 + 47 × 2 5 MR2 =     4 3 + 94 5 MR2 = 20 + 282 15 MR2 = 302 15 MR2 (Ans.) 15| X X 1.75 kg fiwewkó eM©vKvi cv‡Zi cÖwZ evûi •`N© ̈ 5.03 m| cvZwUi K‡Y©i mgvšÍivj XX A‡ÿi mv‡c‡ÿ RoZvi åvgK KZ? [Medium] mgvavb: eM©vKvi cv‡Zi †K›`aMvgx Awfj¤^ eivei A‡ÿi mv‡c‡ÿ RoZvi åvgK, Iz = Ix + Iy = 1 12 ma2 + 1 12 ma2 = 1 6 ma2 Avevi, e‡M©i KY©Øq ci ̄úi j¤^ nIqvq, 2I = 1 6 ma2 [†hLv‡b I = KY© mv‡c‡ÿ RoZvi åvgK]  I = 1 12 ma2  XX Aÿ mv‡c‡ÿ RoZvi åvgK = 1 12 ma2 + m     a 2 2 =     1 12 + 1 2 ma2 = 7 12 × 1.75 × 5.032 kgm2 = 25.828 kgm2 (Ans.) 16| r e ̈vmv‡a©i GKwU †QvU †MvjK R e ̈vmv‡a©i GKwU e„nr Aa©‡MvjvKvi cv‡Îi wKbviv †_‡K w ̄’ive ̄’v n‡Z cv‡Îi Mv eivei Mwo‡q co‡Q| †MvjKwU hLb cv‡Îi Zj‡`‡k Av‡m ZLb Gi MwZkw3 wbY©q Ki| [†QvU †MvjKwUi fi m] [Medium] mgvavb: r R B A awi, †QvU †MvjKwUi cÖv_wgK Ae ̄’vb A we›`y| B we›`y‡Z †eM, v = 2g R B we›`y‡Z MwZkw3 = •iwLK MwZkw3 + †KŠwYK MwZkw3 = 1 2 mv2 + 1 2 I 2 = 1 2 mv2 + 1 2 × 2 5 mr2 2 = 1 2 mv2 + 1 5 mv2 = 7 10 mv2 = 7 10 m.2gR = 7 5 mgR (Ans.)