Tiêu đề 4. P1C4 (Newtonian Mechanics) With Solve.pdf ✅

wbDUwbqvb ejwe` ̈v  Varsity Practice Sheet....................................................................................................................... 1 weMZ mv‡j DU-G Avmv cÖkœvejx 1. m GKwU 2m f‡ii `ywU AvqZvKvi ev· GKwU Nl©Ynxb Abyf‚wgK c„‡ô GKwU `wo Øviv mshy3| F gvÎvi GKwU m¤§yL ej Øviv fvix ev·wU‡K Wvbw`‡K Uvbv n‡”Q| d‡j, nvjKv ev·wU `wo Øviv Uvb Abyfe K‡i| `wowU‡Z Uvb KZ? [DU 23-24] F F 6 F 3 F 2 DËi: F 3 e ̈vL ̈v: anet = F 3m  T = manet = mF 3m = F 3 T F m 2m 2. 0.50 kg f‡ii GKwU KYv X Aÿ eivei x(t) = – 13.00 + 2.00t + 4.0022t2 – 3.005t3 mgxKiY Abyhvqx Pj‡Q, †hLv‡b x Gi GKK wgUvi Ges t Gi GKK †m‡KÛ| t = 2.0 s G KYvi Ici jwä ej KZ? [DU 22-23] – 28 N – 14 N 8 N 36 N DËi: – 14 N e ̈vL ̈v: †eM, v = dx dt = d dt (– 13.00 + 2.00t + 4.0022t2 – 3.005t3 ) = 2 + 8t – 9t2 Avevi, Z¡iY, a = dv dt = d dt (2 + 8t – 9t2 ) = 8 – 18t GLb, F = ma = 0.50  (8 – 18t)  F(t = 2s) = 0.50  (8 – 18  2) = – 14 N 3. w ̄’i Ae ̄’vq _vKv GKwU e ̄‘ we‡ùvwiZ n‡q m1 I m2 f‡ii `yBwU e ̄‘‡Z cwiYZ n‡q h_vμ‡g v1 I v2 †e‡M wecixZ w`‡K Pjgvb| v1 v2 Gi AbycvZ KZ? [DU 19-20, 17-18] m1 m2 – m1 m2 m2 m1 m2 m1 DËi: m2 m1 e ̈vL ̈v: m1u1 + m2u2 = m1v1 + m2v2  m1  0 + m2  0 = m1v1 + m2  (–v2)  0 = m1v1 – m2v2  m1v1 = m2v2  v1 v2 = m2 m1 4. 10 kg f‡ii GKwU e ̄‘i Dci 2F gv‡bi ej cÖ‡qvM Kivi d‡j e ̄‘wUi Z¡iY nq 60 ms–2 | M f‡ii GKwU e ̄‘i Dci 5F gv‡bi ej cÖ‡qvM Kivi d‡j hw` e ̄‘wUi Z¡iY 50 ms–2 nq, Z‡e fi M KZ? [DU 18-19] 3.3 kg 4.8 kg 21 kg 30 kg DËi: 30 kg e ̈vL ̈v: cÖ_‡g †ÿ‡Î, F1 = m1  a1  2F = 10  60  F = 300 N Avevi, wØZxq †ÿ‡Î, F2 = m2  a2  5F = M  50  5  300 = M  50  M = 30 kg 5. 5.0 N Gi GKwU Avbyf~wgK ej GKwU 0.50 kg f‡ii AvqZvKvi e ̄‘‡K GKwU Djø¤^ †`Iqv‡j av°v w`‡”Q| e ̄‘wU Avw`‡Z w ̄’i wQj| hw` w ̄’wZ I MZxq Nl©Y ̧Yv1⁄4 h_vμ‡g s = 0.6 Ges k = 0.8 nq, Z‡e ms–2 GK‡K e ̄‘wUi Z¡iY KZ? [DU 18-19] 1.8 2.0 6.0 8.0 DËi: 1.8 e ̈vL ̈v: Fk = kR = 0.8  5 = 4 N R = 5 N F = 5 N W Nl©‡Yi d‡j g›`b, a = 4 0.5 = 8 ms–2  MwZkxj Ae ̄’vq Z¡iY, a = g – a = 9.8 – 8 = 1.8 ms–2 6. †KŠwYK fi‡e‡Mi GKK †KvbwU? [DU 18-19] kgm2 s –1 kgms–2 kgms–1 km2 s –2 DËi: kgm2 s –1 e ̈vL ̈v: †KŠwYK fi‡eM = ˆiwLK fi‡eM  e ̈vmva© = mvr  GKK  kgm2 s –1
2 .........................................................................................................................................  Physics 1st Paper Chapter-4 7. 30 kg f‡ii GKwU w ̄’i e ̄‘i †eM 2 wgwb‡U e„w× K‡i 36 km/hr G DbœxZ Kivi Rb ̈ e ̄‘wUi Dci KZ ej cÖ‡qvM Ki‡Z n‡e? [DU 16-17] 2 N 2.5 N 300 N 5 N DËi: 2.5 N e ̈vL ̈v: cÖhy3 ej = fi‡e‡Mi cwieZ©‡bi nvi  F = mv t = 30  36 3.6 2  60 = 30  10 2  60 = 2.5 N 8. mgvb fi wewkó wZbwU LÐ A, B, C `woi Øviv wP‡Î cÖ`wk©Z iƒ‡c mshy3| LÐ C, F  ej Øviv Uvbv n‡j m¤ú~Y© e ̈e ̄’vwU Z¡wiZ nq| Nl©Y D‡cÿv Ki‡j LÐ B Gi Dci †gvU ej n‡jvÑ [DU 14-15] F  A B C 0 F  3 F  2 2F  3 DËi: 2F  3 e ̈vL ̈v: F  A B C B e ̄‘i Rb ̈ †jvW awi, cÖwZwU ev‡·i fi = m m¤ú~Y© wm‡÷‡gi Z¡iY, a = F  m + m + m = F  3m GLb, B e ̄‘‡Z †gvU ej, FB  = load(B)  a = 2m  F  3m = 2F  3 N weMZ mv‡j GST-G Avmv cÖkœvejx 1. GKRb mvB‡Kj Av‡ivnx NÈvq 24 km †e‡M 30 m e ̈vmv‡a©i GKwU e„ËvKvi c‡_ †gvo wb‡”Q| Zv‡K Dj‡¤^i mv‡_ KZ †Kv‡Y †n‡j _vK‡Z n‡e? [GST 23-24] tan–1 (0.15) tan–1 (0.015) tan–1 (0.115) tan–1 (0.215) DËi: tan–1 (0.15) e ̈vL ̈v: tan = v 2 rg   = tan–1           24 3.6 2 30 × 9.8   = tan–1     400 9 × 30 × 9.8   = tan–1 (0.15) 2. MvQ †_‡K 2 kg f‡ii GKwU bvwi‡Kj †mvRv wb‡Pi w`‡K co‡Q| evZv‡mi evav 8.6 N n‡j bvwi‡KjwUi Z¡iY KZ? [GST 23-24] 5.5 ms–2 4.5 ms–2 14.5 ms–2 –14.5 ms–2 DËi: 5.5 ms–2 e ̈vL ̈v: Fnet = ma  mg – 8.6 = 2a  2 × 9.8 – 8.6 = 2a  2a = 11  a = 5.5 ms–2 3. GKwU †`Iqvj Nwoi wgwb‡Ui KuvUvi ˆ`N© ̈ 18 cm n‡j Gi †KŠwYK †eM KZ? [GST 23-24] 1.74 × 10–3 rad/s 2.74 × 10–3 rad/s 1.47 × 10–3 rad/s 2.47 × 10–3 rad/s DËi: 1.74 × 10–3 rad/s e ̈vL ̈v:  = 2 T  2  3 3600  1 6  10–2  0.167  10–2 rad/s  1.67  10–3 rad/s Option me‡P‡q KvQvKvwQ| 4. †Kvb w ̄úas-Gi GK cÖv‡šÍ m f‡ii GKwU e ̄‘ Szjv‡j GwU 8 cm cÖmvwiZ nq| e ̄‘wU‡K Gici GKUz †U‡b †Q‡o w`‡j Gi ch©vqKvj KZ n‡e? [GST 23-24] 0.57 sec 0.56 sec 0.75 sec 0.65 sec DËi: 0.57 sec mgvavb: T = 2 e g = 2 0.08 9.8 s  T = 0.57 sec 5. GKRb dzUejv‡ii 0.4 kg f‡ii Ges 12 ms–1 †e‡M AvMZ GKwU dzUej‡K wKK gvivi d‡j †mUv wecixZ w`‡K 5 ms–1 †eM cÖvß n‡jv| wKK gvivi mgqKvj 0.02 s n‡j dzUejvi KZ...©K ejwUi Dci cÖhy3 ej KZ N? [GST 22-23] 340 300 240 140 DËi: 340
wbDUwbqvb ejwe` ̈v  Varsity Practice Sheet...................................................................................................................... 3 e ̈vL ̈v: wbDU‡bi wØZxq m~Î g‡Z, cÖhy3 ej = fi‡e‡Mi cwieZ©‡bi nvi  F = m (v – u) t = 0.4 (– 5 – 12) 0.02 = 20  (–17) = – 340 N  e‡ji gvb = 340 N 6. mgvšÍivj mgev‡q hy3 PviwU Zvgvi Zv‡ii mvnv‡h ̈ 6 kg f‡ii GKwU e ̄‘‡K Lvov Dc‡ii w`‡K 2 ms–1 mg‡e‡M Uvbv n‡j cÖwZwU Zv‡i KZ N Uvb co‡e? [GST 22-23] 10.7 12.7 14.7 17.8 DËi: 14.7 e ̈vL ̈v: mg‡e‡M Uvbv n‡j, a = 0 ms–2 cÖhy3 Uvbej, F = m (g + a) = 6  (9.8 + 0) = 58.8 N  cÖwZ Zv‡i Uvbej, T = F 4 = 58.8 4 = 14.7 N 7. †Kvb e ̄‘i Dci 15 N ej 3 s a‡i wμqv Ki‡j e ̄‘wUi fi‡e‡Mi cwieZ©b KZ kgms–1 ? [GST-A 21-22] 25 30 45 50 DËi: 45 e ̈vL ̈v: fi‡e‡Mi cwieZ©b, mv = Ft = 15  3 = 45 Ns = 45 kgms–1 8. MvQ †_‡K 0.5 kg f‡ii GKwU Avg Lvov wb‡Pi w`‡K co‡Q| evZv‡mi evav hw` 2.4 N nq, Zvn‡j AvgwUi Z¡iY KZ ms–2 ? [GST 21-22] 5.0 5.5 6.0 6.5 DËi: 5.0 e ̈vL ̈v: evZv‡mi evav = 2.4 N  evZv‡mi d‡j g›`b, a = 2.4 0.50 = 4.8 ms–2  AvgwUi wbU Z¡iY = g – a = 9.8 – 4.8 = 5 ms–2 9. GKwU PvKvi e ̈vmva© 1 m| GwU wgwb‡U 15 evi Nyi‡j Gi cÖv‡šÍi ˆiwLK †eM KZ ms–1 ? [GST-A 21-22]   2 2  3 DËi:  2 e ̈vL ̈v: ˆiwLK †eM, v = r = 2N T r = 2  15 60  1 =  2 ms–1 10. †K›`axq ej F Gi cÖfv‡e r e ̈vmv‡a©i e„ËvKvi c‡_ N~Y©biZ GKwU KYvi Dci wμqvkxj U‡K©i gvbÑ [GST 21-22] rF – rF 0  DËi: 0 e ̈vL ̈v: F  r  Avgiv Rvwb,  = r   F  = rF sin = rF sin180 = 0 11. 5 kg f‡ii GKwU e ̄‘ 1.2 ms–1 †e‡M GKwU †`qv‡j j¤^fv‡e av°v †L‡q 0.8 ms–1 †e‡M wecixZ w`‡K wd‡i Avm‡j e‡ji NvZ KZ Ns n‡e? [GST 20-21] 4 5 6 10 DËi: 10 e ̈vL ̈v: e‡ji NvZ = fi‡e‡Mi cwieZ©b = m (v – u) = 5 (– 0.8 – 1.2) = – 10 kgms–1  e‡ji NvZ = 10 kgms–1 = 10 Ns 12. f~-mgZ‡ji mv‡_ 30 †Kv‡Y AvbZ c‡_ GKwU 2 kg f‡ii e ̄‘‡K 3 ms–2 Z¡i‡Y DVv‡Z n‡j e ̄‘wUi Dci KZ wbDUb (N) ej cÖ‡qvM Ki‡Z n‡e? [GST 20-21] 6.8 11.8 12.8 15.8 DËi: 15.8 e ̈vL ̈v: 30 3 ms –2 g g sin30  3 ms–2 mgZ¡i‡Y Zzj‡Z ej, F = m (g sin + a) = 2  (9.8 sin30 + 3) = 15.8 N
4 .........................................................................................................................................  Physics 1st Paper Chapter-4 weMZ mv‡j Agri-G Avmv cÖkœvejx 1. ˆiwLK Z¡iY I †KŠwYK Z¡i‡Yi g‡a ̈ m¤úK© †KvbwU? [Agri. Guccho 21-22]  = ra  = a 2 r r = a  ar = 1 DËi: r = a  e ̈vL ̈v: ˆiwLK Z¡iY = †KŠwYK Z¡iY  e ̈vmva©  a = r  r = a  2. GKwU cvLv cÖwZ wgwb‡U 30 evi Nyi‡Q| Gi †KŠwYK †eM KZ? [Agri. Guccho 21-22]  rads–1 2 rads–1 15 rads–1 60 rads–1 DËi:  rads–1 e ̈vL ̈v: †KŠwYK †eM,  = 2N T = 2  30 60 =  rads–1 3. GKwU Nwoi †m‡KÛ, wgwbU, NÈvi KuvUvi †KŠwYK †e‡Mi AbycvZÑ [Agri. Guccho 21-22] 720 : 60 : 1 1 : 60 : 720 1 : 12 : 720 720 : 12 : 1 DËi: 720 : 12 : 1 e ̈vL ̈v: S : M : H = 2 60 : 2 60  60 : 2 12  60  60 = 1 : 1 60 : 1 12  60 = 1 : 1 60 : 1 720 = 720 : 12 : 1 4. GKwU Kv‡Vi LЇK Avbyf~wg‡Ki mv‡_ 60 †Kv‡Y 200 N ej Øviv Uvbv n‡”Q| e ̄‘wUi Dci Avbyf~wg‡Ki w`‡K Kvh©Kix ej KZ? [Agri. Guccho 20-21; DU 13-14] 100 N 200 N 125 N Zero DËi: 100 N e ̈vL ̈v:  = 60 Fsin Fcos F = 200N  Avbyf‚wgK w`‡K Kvh©Kix ej = F cos = 200  cos60 = 100 N 5. wbDU‡bi MwZi Z...Zxq m~Îvbymv‡i wμqv-cÖwZwμqvi ga ̈eZ©x †KvY KZ wWwMÖ? [Agri. Guccho 21-22; KU 19-20; JU 18-19] 0 90 180 360 DËi: 180 e ̈vL ̈v: Òwμqv I cÖwZwμqv ej mgvb I wecixZgyLxÓÑ wbDU‡bi Z...Zxq m~Î| wecixZgyLx nIqv‡Z ga ̈eZ©x †KvY 180| 6. 5 kg f‡ii GKwU ivB‡dj †_‡K 20 g f‡ii GKwU ey‡jU 1000 ms–1 MwZ‡Z Qz‡U hvq| wcQb †_‡K ivB‡d‡ji av°vi †eM KZ? [Agri. Guccho 20-21] 4000 ms–1 4 ms–1 400 ms–1 40 ms–1 DËi: 4 ms–1 e ̈vL ̈v: m1u1 + m2u2 = m1v1 + m2v2  0 =     20 1000  1000 + 5v2  v2 = – 20 5 = – 4 ms–1  cðvr‡eM = 4 ms–1 weMZ mv‡j JU-G Avmv cÖkœvejx 1. NvZ Gi GKK Kx‡mi Abyiƒc? [JU 22-23] ej fi Z¡iY fi‡eM DËi: fi‡eM e ̈vL ̈v: NvZ ej‡Z Ôe‡ji NvZÕ‡K †evSv‡bv n‡q‡Q| e‡ji NvZ = fi‡e‡Mi cwieZ©b = m (v – u) = mv  GKK = kgms–1 2. c‡_i Nl©Y ej 10 N n‡j, 2 kg f‡ii GKwU e ̄‘‡K 10 ms–2 Z¡i‡Y MwZkxj Ki‡Z KZ ej `iKvi? [JU 22-23] 30 N 35 N 40 N 25 N DËi: 30 N e ̈vL ̈v: MwZkxj Ki‡Z †gvU ej, F = Fk + Fnet = 10 + (2  10) = 10 + 20 = 30 N 3. 30 kg f‡ii e ̄‘i Dci KZ ej cÖ‡qvM Ki‡j 1 wgwb‡U Gi †eM 36 kmh–1 e„w× cv‡e? [JU 22-23] 5 N 9.8 N 4.9 N 10 N DËi: 5 N e ̈vL ̈v: e ̄‘i fi‡e‡Mi cwieZ©‡bi nvi Zvi Dci cÖhy3 e‡ji mgvbycvwZK|  F = mv t = 30  36 3.6 60 = 30  10 60 = 5 N