Tiêu đề 4. P1C4. Newtonian-Mechanics C+Merged Ok-Ridoy 2.5.24 (Nashita) Ok.pdf ✅

wbDUwbqvb ejwe` ̈v  Engineering Question Bank 1 PZz_© Aa ̈vq wbDUwbqvb ejwe` ̈v Newtonian Mechanics weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| †Kv‡bv gva ̈‡g GKRb e ̈w3 2 m D”PZvq 3 m `xN© GKwU iwk‡Z cv_i †e‡a Avbyf~wgKfv‡e GKRb †jvK cÖwZ wgwb‡U 30 evi Nyi‡Q| e ̈w3 n‡Z `k©K mvwii `~iZ¡ b~ ̈bZg KZ n‡j `k©K wbivc‡` _vK‡e? [BUET 22-23] mgvavb: cv_i †Q‡o †`qvi gyû‡Z© †eM, v = r = 2N t × r  v = 2 × 30 60 × 3 = 9.42 ms–1  Abyf~wgK `~iZ¡, x = 2h g v    ∵  h = 1 2 gt2  x = 2 × 2 9.8 × 9.42 = 6.018 m r 6.018 m 3 m  e ̈w3 n‡Z b~ ̈bZg `~iZ¡ r = 3 2 + 6.0182 = 6.724 m `~i‡Z¡ _vK‡e n‡e| (Ans.) 2| Nbe ̄‘i D”PZv 8 cm Ges NbZ¡ 600 kg/m2 | G‡K 1000 kg/m3 Nb‡Z¡i Zi‡j Wzev‡j Gi WzešÍ •`N© ̈ KZ? G‡K m¤ú~Y© Wzwe‡q †Q‡o w`‡j Gi Z¡iY KZ? [BUET 22-23] mgvavb: AcmvwiZ Zi‡ji IRb = e ̄‘i fi  Vfg = mg  Axfg = vsg = Ahsg  x = hs f = 600 × 8 100 = 4.8 cm s = Nbe ̄‘i NbZ¡ f = Zi‡ji NbZ¡ GLb, Fnet = Vfg – mg = Vfg – Vsg  ma = Vg(f – s)  sVa = Vg(f – s)  a = 9.8(1000 – 600) 600 = 6.53 ms–2 (Dc‡ii w`‡K) (Ans.) 3| 3 kg f‡ii e ̄‘ we‡ùvwiZ n‡q 1 : 1 : 3 Abycv‡Z wef3 nq| nvjKv LÐ `ywU ci ̄úi j¤^fv‡e 120 ms–1 †e‡M wewÿß n‡j, Aewkó fiwUi †eM KZ? [BUET 21-22] mgvavb: (m1 + m2 + m3) u  = m1v1  + m2v2  + m3v3   3  0 = 3 5  120i  + 3 5  120j  + 3  3 5 v3   v3  = – 120i  – 120j  3 = – 40i  – 40j  |v |3  = 40 2 ms –1 (Ans.) hv x A‡ÿi mv‡_  = cos –1     – 40 40 2 = 135 †KvY Drcbœ K‡i| (Ans.) 4| GKwU †ij jvB‡bi euv‡Ki e ̈vmva© 250 m Ges †ij jvB‡bi cvZ؇qi ga ̈eZ©x `~iZ¡ 1 m| NÈvq 50 km †e‡M PjšÍ †ij Mvwoi †ÿ‡Î cÖ‡qvRbxq e ̈vswKs Gi Rb ̈ evB‡ii jvB‡bi cvZ‡K wfZ‡ii jvB‡bi cvZ A‡cÿv KZUzKz DuPz Ki‡Z n‡e? [BUET 21-22] mgvavb: tan = v 2 rg   = tan–1           50 3.6 2 250  9.8 = 4.5 sin = h x  h = 1  sin 4.5 = 0.078 m (Ans.) 5| 1.5 kg f‡ii GKwU g‡Wj †nwjKÞv‡ii Avw` MwZ‡eM 5j  ms–1 . 2 sec mgZ¡i‡Y Pjvi ci Zvi MwZ‡eM nq (6i )  + 12j  ms–1 GB mg‡q cÖhy3 e‡ji gvb wbY©q Ki| [BUET 20-21] mgvavb: F  = ma   F  = m. v  – u  t  F  = 1.5 × (6i )  + 12j  – 5j  2 = 9 2 i  + 21 4 j   cÖhy3 e‡ji gvb, |F|  =     9 2 2 +     21 4 2 = 6.914 N (Ans.) 6| 150 g f‡ii GKwU wμ‡KU ej 12 ms–1 †e‡M MwZkxj n‡q GKwU e ̈vU Øviv AvNvZ Kivi d‡j ejwU 20 ms–1 †e‡M wd‡i Av‡m| ejwUi Dci wμqviZ e‡ji AvNv‡Zi mgqKvj 0.01 s, ejwUi Dci e ̈v‡Ui Mo ej wbY©q Ki| [BUET 19-20] mgvavb: F = m. |v – u| t  F = 0.150 × 20 – (–12) 0.01  F = 480 N (Ans.) 7| 0.56 kg fi wewkó GKwU wgUvi † ̄‹‡ji 20 cm wPwýZ `v‡Mi j¤^ A‡ÿi mv‡c‡ÿ wgUvi † ̄‹jwUi N~Y©b RoZv wbiƒcY Ki| † ̄‹jwU‡K cvZjv iW wn‡m‡e we‡ePbv Ki| [BUET 19-20]
2  Physics 1st Paper Chapter-4 mgvavb: I = I0 + Md2  I = 1 12 ML2 + M × 0.32  I = 0.097 kgm2 (Ans.) 20cm 20cm I I0 8| 2.0 kg f‡ii GKwU e ̄‘ w ̄’i Ae ̄’vq _vKv Av‡iKwU e ̄‘i mv‡_ w ̄’wZ ̄’vcK msNl© NUv‡jv Ges msN‡l©i ci cÖ_g e ̄‘wU Zvi Avw`‡e‡Mi GK-PZz_©vsk †eM wb‡q GKB w`‡K Pj‡Z _vKj| AvNvZcÖvß e ̄‘wUi fi KZ? [BUET 18-19] mgvavb: v1 =    m1  – m2 m1 + m2 u1 + 2m2 m1 + m2 u2  1 4 u1 = 2 – m2 2 + m2 × u1  m2 = 1.2 kg (Ans.) 9| GKwU 8 kg f‡ii PvKvi PμMwZi e ̈vmva© 25 cm n‡j Gi RoZvi åvgK KZ n‡e? PvKvwU‡Z 3 rads–2 †KŠwYK Z¡iY m„wó Ki‡Z KZ gv‡bi UK© cÖ‡qvM Ki‡Z n‡e? [BUET 17-18; BUTex 01-02] mgvavb: I = Mk2 = 8 × 0.252 = 0.5 kgm2   = I = 0.5 × 3 = 1.5 Nm (Ans.) 10| 4, 5 Ges 6 GKK f‡ii wZbwU KYvi ̄’vbv1⁄4 h_vμ‡g (4, 0, –1), (3, –2, 3) Ges (2, 1, 4) n‡j z A‡ÿi mv‡c‡ÿ Zv‡`i RoZvi åvgK I PμMwZi e ̈vmva© wbY©q Ki| [BUET 14-15] mgvavb: Iz 1 = mr 2 z = m(r ) 2 x + r 2 y  Iz 1 = 4 × (42 + 02 ) = 64 GKK Iz2 = mr 2 z = m(r ) 2 x + r 2 y  Iz2 = 5(32 + (–2)2 ) = 65 GKK Iz3 = mr 2 z = m(r ) 2 x + r 2 y = 6 × (22 + 12 ) = 30 GKK  K1 = Iz1 m1 = 64 4 = 4 GKK K2 = Iz2 m2 = 65 5 = 13 GKK K3 = Iz3 m3 = 30 6 = 5 GKK (Ans.) 11| 30 kg f‡ii GKwU †kj 48 ms–1 †e‡M Do‡Q| †kjwU we‡ùvwiZ n‡q `yB UzKiv n‡j 18 kg f‡ii UzKivwU w ̄’i n‡q hvq Ges evKx UzKivwU D‡V hvq| evKx As‡ki †eM KZ? [BUET 06-07] mgvavb: m1u1 = m2v2 + m3v3  30 × 48 = 18 × 0 + (30 – 18) × v3  v3 = 120 ms–1 (†k‡ji †e‡Mi w`‡K) 12| 7 metre DuPz n‡Z 2 kg f‡ii GKwU wcZ‡ji wb‡iU †MvjK GKwU bZ Z‡j Mov‡Z Mov‡Z f~wg‡Z G‡m c‡o| f~wg ̄úk© Kivi gyû‡Z© †MvjKwUi fi‡K‡›`ai MwZkw3 I †KŠwYK MwZkw3 KZ wQj? [g = 9.8 ms–2 ] [BUET 04-05] mgvavb: †gvU kw3 = mgh = 2 × 9.8 × 7 = 137.2 J  1 2 mv 2 + 1 2 I 2 = mgh  1 2 mv 2 + 1 2 × 2 5 mr 2 × v 2 r 2 = mgh  1 2 v 2 + 1 5 v 2 = 137.2 2  v 2 = 98(ms–1 ) 2  fi‡K‡›`ai MwZkw3 = 1 2 mv 2 = 1 2 × 2 × 98 = 98 J  †KŠwYK MwZkw3 = 137.2 – 98 = 39.2 J (Ans.) 13| 1000 kg f‡ii GKwU Mvwoi PvKvi mv‡_ iv ̄Ívi w ̄’wZ Nl©Y mnM 0.10 Ges Pj Nl©Y mnM 0.04| Avbyf~wg‡Ki w`‡K MvwowUi Dci b~ ̈bZg KZ ej cÖ‡qvM Ki‡j w ̄’i MvwowU Pj‡Z ïiæ Ki‡e| Avi AwZwi3 KZ ej cÖ‡qvM Ki‡j MvwowUi Z¡iY 2 ms–2 n‡e? [BUET 01-02] mgvavb: MvwowU MwZkxj nIqvi Dcμg n‡j mxgv ̄’ Nl©Y ej wμqv Ki‡e|  flim = smg = 0.1 × 1000 × 9.8 = 980 N MwZkxj Ae ̄’vq, F = ma + f = ma + mg  F = 1000 × 2 + 0.04 × 1000 × 9.8  F = 2392 N  AwZwi3 cÖhy3 ej = 2392 – 980 = 1412 N (Ans.) 14| 1000 kg f‡ii GKwU Mvwoi PvKv I iv ̄Ívi g‡a ̈ w ̄’wZ Nl©‡Yi mnM ev ̧Yv1⁄4 s = 0.8 n‡j MvwowU m‡e©v”P KZ Xvjy iv ̄Ívq wcQwj‡q bv c‡o †_‡g _vK‡Z cvi‡e? [BUET 00-01] mgvavb: mgcos  f R mgsin wcQ‡j covi Dcμg n‡j, mgsin = f = R  mgsin = mgcos   = tan–1 () = tan–1 (0.8) = 38.66 (Ans.) 15| gv‡Vi ga ̈ w`‡q Mvwo‡q hvIqv 0.5 kg f‡ii GKwU dzUej 50 m `~i‡Z¡ wM‡q †_‡g †Mj| dzUejwUi cÖv_wgK †eM 30 ms–1 n‡j Nl©Y e‡ji gvb KZ? [KUET 05-06] mgvavb: f = ma = m . u 2 2s  f = 0.5 × 302 2 × 50 = 4.5 N (Ans.) 16| GKwU 10 N ej 2 kg fi wewkó GKwU w ̄’i e ̄‘i Dci wμqv K‡i| hw` 5 †m‡KÛ ci e‡ji wμqv eÜ n‡q hvq Z‡e cÖ_g n‡Z 12 †m‡K‡Û e ̄‘wU KZ `~iZ¡ AwZμg Ki‡e? [KUET 04-05]
wbDUwbqvb ejwe` ̈v  Engineering Question Bank 3 mgvavb: Z¡iY, a = F m = 10 2 = 5 ms–2 cÖ_g 5 s G, s1 = 1 2 × at2 = 1 2 × 5 × 52  s1 = 62.5 m †kl‡eM, v = at = 5 × 5 = 25 ms –1  cieZ©x 7s G AwZμvšÍ `~iZ¡, s2 = vt = 25 × 7  s2 = 175 m  †gvU AwZμvšÍ `~iZ¡, s1 + s2 = 237.5 m (Ans.) 17| 6 kg I 4 kg f‡ii `yBwU e ̄‘ GKB mij‡iLv eivei wKš‘ wecixZ w`‡K Pjv Ae ̄’vq G‡K Aci‡K av°v w`j| av°vi c~‡e© Zv‡`i †eM h_vμ‡g 5 ms–1 (DËi w`‡K) I 2 ms–1 (`wÿY w`‡K) wQj| av°vi ci wØZxq e ̄‘wU 2.5 ms–1 †e‡M wcwQ‡q †Mj; cÖ_g e ̄‘wUi †eM KZ n‡e? [KUET 03-04] mgvavb: m1u1 – m2u2 = m1v1 + m2v2  6 × 5 – 4 × 2 = 6 × v1 + 4 × 2.5  v1 = 2 ms–1 (DËi w`‡K) 18| wZbwU w ̄’i e ̄‘, GKwU wis, GKwU wb‡iU wmwjÛvi Ges GKwU wb‡iU †MvjK GKB euvKv Z‡ji Dci w`‡q bv wcQwj‡q wb‡Pi w`‡K co‡Z _v‡K| wZbwU e ̄‘i e ̈vmva© GKB| †Kvb e ̄‘wU me‡P‡q †ewk †e‡M f~wg‡Z †cuŠQv‡e? [RUET 18-19] mgvavb: e ̄‘ wZbwUi RoZvi åvgK, I = kmr2 [k GKwU aaæeK] wis, wb‡iU wmwjÛvi I wb‡iU †Mvj‡Ki Rb ̈ k Gi gvb h_vμ‡g 1, 1 2 , 2 5  MwZkw3, 1 2 mv 2 + 1 2 × I 2 = mgh  1 2 v 2 + 1 2 × kr2 × v 2 r 2 = gh  1 2 v 2 + 1 2 kv2 = gh  v = 2 k + 1 gh  v  = 1 k + 1  h †h‡nZz kring > kcylinder > ksphere  vring < vcylinder < vsphere 19| GKwU iv ̄Ív 100 m e ̈vmv‡a© euvK wb‡q‡Q| H ̄’v‡b iv ̄ÍvwU PIov 5 m Ges Gi wfZ‡ii wKbviv n‡Z evB‡ii wKbviv 50 cm DuPz| m‡e©v”P KZ †e‡M H ̄’v‡b wbivc‡` euvK †bqv hv‡e? [RUET 15-16] mgvavb:  5m 0.5m tan = v 2 rg  v = tan     sin–1     0.5 5 × 100 × 9.8  v = 9.924 ms–1 (Ans.) 20| GKwU e ̄‘ w ̄’ive ̄’vq wQj| 15 N ej Gi Dci 4 s a‡i KvR K‡i Ges Zvici Avi †Kvb KvR Kij bv| e ̄‘wU Gici 9 sec G 54 m `~iZ¡ †Mj| e ̄‘wUi fi †ei Ki| [RUET 12-13] mgvavb: †kl‡eM v n‡j, s = vt  v = 54 9 = 6 ms–1 GLb, F = ma = m. v – u t  15 = m. 6 – 0 4  m = 10 kg (Ans.) 21| N›Uvq 40 gvBj †e‡M Pjgvb GKwU Mvwoi PvjK 59 MR `~‡i GKwU †QvU †Q‡j‡K †`L‡Z †cj| m‡1⁄2 m‡1⁄2 †m †ea‡K Pvc w`j| †Q‡jwUi 1 dzU Av‡M G‡m MvwowU †_‡K †Mj| MvwowU _vgv‡Z KZ mgq †j‡M‡Q Ges cÖhy3 e‡ji gvb KZ? Av‡ivnx m‡gZ Mvwoi IRb 1 Ub| [RUET 09-10] mgvavb: 59 MR = 59 × 3 ft = 177 ft 1 Ub = 2240 lb 40 gvBj/N›Uv = 40 mile 3600 s = 40 × 1600 m 3600 s = 40 × 1600 × 3.3 ft 3600 s = 176 3 ft/s s =     u + v 2 t  177 – 1 =       176 3 + 0 2 t  t = 6 s Avevi, F = m. |v – u| t  F = 2240 × 176 3 6 = 21902.22 poundal. (Ans.) 22| 100 kg f‡ii GKRb †jvK wjd‡U `uvwo‡q Av‡Q| wjdUwU hw` 2 ms–2 Z¡i‡Y Dc‡ii w`‡K DV‡Z _v‡K Zvn‡j †jvKwUi Dci EaŸ©gyLx cÖwZwμqv ej KZ? [RUET 06-07] mgvavb: EaŸ©Mvgx wjd‡Ui †ÿ‡Î, R = m(g + a) = 100 × (9.8 + 2)  R = 1180 N (Ans.) 23| 10 kg f‡ii GKwU e ̄‘i Dci KZ ej cÖ‡qvM Ki‡j e ̄‘ Lvov (i) Dc‡ii w`‡K 1.2 ms–2 (ii) wb‡Pi w`‡K 2.8 ms–2 Z¡i‡Y MwZkxj n‡e? [RUET 06-07] mgvavb: (i) F – mg = ma  F = m(g + a) = 10(9.8 + 1.2) = 10 × 11 = 110 N
4  Physics 1st Paper Chapter-4 (ii) mg – F = ma  F = m(g – a) = 10(9.8 – 2.8) = 10 × 7 = 70 N 24| w ̄’i cvwbi Dci fvmgvb GKwU †bŠKv n‡Z GKRb †jvK Avbyf~wgK w`‡K jvd w`‡q Zx‡i †cuŠQvj| evKx †jvKmn †bŠKvi fi 300 kg| jvd †`qv †jv‡Ki fi 60 kg| jv‡di †eM 20 ms–1 | GgZve ̄’vq †bŠKvq Aew ̄’Z 0.75 kg f‡ii GKwU w ̄’i ej‡K wKK gviv n‡jv| d‡j dzUejwU GKB w`‡K 18 ms–1 †eM cÖvß n‡jv| cv KZ...©K cÖhy3 e‡ji NvZ wbY©q Ki| [RUET 05-06] mgvavb: jvd †`qvi ci †bŠKvi †eM, vb = – m1v1 M  vb = – 60 × 20 300  vb = – 4 ms–1  dzUe‡ji fi‡e‡Mi cwieZ©b, p = m(v – u) = 0.75(18 – (–4)) = 16.5 kgms–1 (Ans.) 25| 200 ms–1 †e‡M AvMZ 0.2 kg f‡ii wμ‡KU ej‡K GKRb †L‡jvqvo K ̈vP a‡i 0.1 †m‡KÛ mg‡qi g‡a ̈ _vwg‡q w`j| †L‡jvqvo KZ...©K cÖhy3 Mo ej KZ? [RUET 04-05] mgvavb: F = m. |v – u| t  F = 0.2 × 200 0.1 = 400 N (Ans.) 26| 25.2 wK‡jvwgUvi/N›Uv †e‡M Pjv GKRb mvB‡Kj Av‡ivnx 5 m e ̈vmv‡a©i GKwU e„ËvKvi †gvo NyiwQj| †Kvb `yN©Ubv Gov‡Z f~wgi mv‡_ KZUv †n‡j Zv‡K Pj‡Z n‡e? [RUET 04-05] mgvavb: Dj‡¤^i mv‡_ †KvY  n‡j, tan = v 2 rg =     25.2 3.6 2 5 × 9.8   = tan–1           25.2 3.6 2 5 × 9.8   = 45  Abyf~wg‡Ki mv‡_ †KvY = 90 – 45 = 45 (Ans.) 27| GKwU †ijjvB‡bi euv‡Ki e ̈vmva© 250 m Ges †ijjvB‡bi cvZ؇qi ga ̈eZ©x `~iZ¡ 1 m| N›Uvq 50 km †e‡M PjšÍ Mvwoi †ÿ‡Î cÖ‡qvRbxq e ̈vswKs-Gi Rb ̈ evB‡ii jvB‡bi cvZ‡K wfZ‡ii jvB‡bi cvZ A‡cÿv KZUzKz DuPz Ki‡Z n‡e? [CUET 13-14, 07-08; BUTex 11-12; RUET 05-06] mgvavb:  = tan–1     v 2 rg   = tan–1           50 3.6 2 250 × 9.8   = 4.5  1m h  sin = h 1  sin(4.5) = h  h = 7.87 cm (Ans.) 28| GKwU iv ̄Ív 60 m e ̈vmv‡a© euvK wb‡q‡Q| H ̄’v‡b iv ̄ÍvwU 6 m PIov Ges Gi wfZ‡i wKbviv n‡Z evB‡ii wKbviv 0.6 m DuPz| m‡e©v”P KZ †e‡M H ̄’v‡b wbivc` euvK †bIqv m¤¢e? [CUET 08-09] mgvavb: tan = v 2 rg  v = tan × rg  6m 0.6m  v = tan     sin–1     0.6 6 × 60 × 9.8  v = 7.68 ms–1 (Ans.) 29| mgZj iv ̄Ívq 500 m e ̈vmv‡a©i e„ËvKvi c‡_ 25 ms–1 †e‡M euvK wb‡Z †M‡j 900 kg f‡ii GKwU Mvwoi Uvqvi Ges iv ̄Ívi g‡a ̈ wμqvkxj b~ ̈bZg w ̄’wZ Nl©Y ̧Yv1⁄4 wbY©q Ki| [BUTex 22-23] mgvavb: cFlim = Fc  minmg = mv 2 r  min = v 2 rg  min = 252 500 × 9.8 = 0.127 (Ans.) 30| 2 Kg f‡ii GKwU e ̄‘i Ae ̄’vb †f±i r – = (i )  – 2j  + 2k  m Ges †eM v  = (2i )  – 4j  + 2k  ms–1 | e ̄‘wUi †KŠwYK fi‡e‡Mi gvb wbY©q Ki| [BUTex 21-22] mgvavb: L  = r  × p  = m(r )  × v   r  × v  =        i   1 2 j  –2 –4 k  2 2 = i  (–4 + 8) + j  (4 – 2) + k  (–4 + 4) = 4i  + 2j   L  = m(r )  × v  = 2(4i )  × 2j  = 8i  + 4j  †KŠwYK fi‡e‡Mi gvb, |L |  = 8 2 + 42 = 4 5 kgm2 s –1 (Ans.) 31| 4000 kg f‡ii GKwU wjdU 240 kg f‡ii GKwU ev· enb Ki‡Q| hLb wjd‡Ui Zv‡ii (Supporting cable) Dci EaŸ©gyLx Uvb 48000 N nq ZLb EaŸ©gyLx Z¡iY KZ? wjdUwU w ̄’i Ae ̄’vb †_‡K 3 s mg‡q KZ D”PZvq DV‡e? [BUTex 19-20] mgvavb: T = m(g + a)  48000 = 4240(9.8 + a)  a = 1.52 ms–2  3s mg‡q DV‡e, h = 1 2 at2