Title Motion of Particles Engg Practice Sheet Solution.pdf ✅

09 mgZ‡j e ̄‘KYvi MwZ Motion of Particles in a Plane WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. GKwU A ̈v¤^y‡jÝ 15 ms–1 †e‡M I 5 ms–2 Z¡i‡Y P‡j| 35 ms–1 mg‡e‡M Pjgvb GKwU ev‡mi 50 m wcQb †_‡K A ̈v¤^y‡jÝwU Pj‡j wgwjZ nevi Av‡M Av‡iv KZ `~iZ¡ AwZμg Ki‡e? [BUET 23-24] mgvavb: 15 ms–1 50 m A B C 5 ms–2 35 ms–1 awi, t mgq ci A ̈v¤^yj ̈vÝ I evm GK‡Î wgwjZ n‡e| SA = ut + 1 2 ft2 = 15t + 5 2 t 2 Ges SB = Vt = 35t GLb, SA = SB + 50  15t + 5 2 t 2 = 35t + 50  2.5t2 – 20t – 50 = 0 t = 10, – 2  t = 10 s [t  – 2]  A ̈v¤^yj ̈vÝ KZ©„K AwZμvšÍ `~iZ¡ SA = 15  10 + 5 2  102 = 400 m (Ans.) 2. wjI‡bj †gwm GKwU wd«-wKK †_‡K 20 ms–1 †e‡M kU wb‡j Zv 10 m DuPz GKwU μmevi †Kv‡bvg‡Z AwZμg K‡i| KZ †Kv‡Y ejwU‡K wKK Kiv n‡qwQj Ges Avbyf~wgK `~iZ¡ KZ wQ‡jv? [g = 10 ms– 2 ] [BUET 23-24] mgvavb: GLv‡b, m‡ev©”P D”PZv, H = 10 m  u 2 sin2 2g = 10  202  sin2 2  10 = 10 x u = 20 ms–1  10 m  sin = 1 2 [(+)ve gvb wb‡q]   = 45 (Ans.) Avevi, tan = 4H R  R = 4H tan  R = 4  10 tan45 = 40 m  x = R 2 = 20 m  μmevi AwZμg Kivi AwZμvšÍ Avbyf~wgK `~iZ¡ 20 m (Ans.) 3. GKwU ̧jwZ n‡Z `~‡M©i †`qvj eivei GKwU cv_i Avbyf~wg‡Ki mv‡_ 45 †Kv‡Y 22 m/s †e‡M wb‡ÿc Kiv n‡jv| `~M©wU 45 m `~‡i Ges †`qv‡ji D”PZv 20 m n‡j, cv_iwU wK †`qvj AwZμg Ki‡e? [Dˇii h_v_©Zv e ̈vL ̈v Ki] [BUET 22-23] mgvavb: y = xtan – gx2 2v2 cos2 = 45tan45 – 9.8  452 2  222  cos2 45 = 3.998 ≃ 4 < 20  AwZμg Ki‡Z cvi‡e bv| (Ans.) 4. GKwU cv_i †Kv‡bv wbw`©ó D”PZv †_‡K †d‡j †`qv n‡j Zv cZb mg‡qi †kl t †m‡K‡Û h wgUvi `~iZ¡ AwZμg K‡i| †`LvI †h, cZ‡bi mgq =     h gt + t 2 [BUET 22-23] mgvavb: H = 1 2 gT2 ....... (i) H – h = 1 2 g (T – t)2 ........ (ii)  h = 1 2 g (T2 – T 2 + 2Tt – t 2 )  2Tt – t 2 = 2h g  2Tt = 2h g + t2  T =     h gt + t 2 (Showed)
2  Higher Math 2nd Paper Chapter-9 5. GKwU †Uab w ̄’ive ̄’v †_‡K 750 m ch©šÍ P‡j _vg‡jv| 1g Ask 2 ms–2 mgZ¡i‡Y I cieZ©x Ask 4 ms–2 mgg›`‡b Pj‡j, †UabwUi m‡e©v”P †eM KZ? [BUET 20-21] mgvavb: A B  C u = 0 w v = 0 s1 s2 (1g Ask) (2q Ask) C we›`y‡Z G‡m w ms–1 (m‡e©v”P †eM) cÖvß nq| 1g Ask, w 2 = 02 + 2  2  s1  s1 = w 2 4 .......... (i) 2q Ask, 0 2 = w 2 – 2  4  s2  w 2 = 8s2  s2 = w 2 8 .......... (ii) (i) + (ii) K‡i, s1 + s2 = w 2 4 + w 2 8  750 = w2     3 8  w = 44.72 ms–1 (Ans.) 6. GK e ̈w3 Zvi 50 m `~‡i GKwU evm‡K w ̄’ive ̄’v n‡Z mylg Z¡i‡Y hvÎv ïiæ Ki‡Z †`‡L| H gyn~‡Z© †m mg‡e‡M ev‡mi w`‡K †`.o ïiæ K‡i Ges GK wgwb‡U evmwU‡K a‡i| †jvKwUi †eM I evmwUi Z¡iY wbY©q Ki| [BUET 19-20; MIST 17-18] mgvavb: awi, †jv‡Ki †eM = v, ev‡mi Z¡iY = f t = 1 min = 60 s ev‡mi AwZμvšÍ `~iZ¡, s1 = 1 2 ft2 = 1 2 f  60 2 = 1800f t mgq ci ev‡mi †eM, v= ft = 60f †jv‡Ki AwZμvšÍ †gvU `~iZ¡, vt = 50 + s1 = 50 + 1800f  v = 50 + 1800f 60 evm aivi gyn~‡Z© †jv‡Ki †eM I ev‡mi †eM mgvb n‡e|  v = v  60f = 50 + 1800f 60  3600f = 50 + 1800f  1800f = 50  f = 0.0278 ms–2 (Ans.)  †jv‡Ki †eM, v = v = 60  0.0278 = 1.67 ms–1 (Ans.) 7. w ̄’ive ̄’v †_‡K GKwU evm‡K 3 ms–2 mgZ¡i‡Y Pj‡Z †`‡L evmwU‡K aivi D‡Ï‡k ̈ GKRb †jvK ev‡mi †cQ‡b wKQz`~i †_‡K 12 m/sec mg‡e‡M †`.uov‡Z Avi¤¢ K‡i| evm †_‡K †jvKwU m‡e©v”P KZ `~‡i _vK‡j evmwU‡K ai‡Z cvi‡e? [BUET 17-18] mgvavb: A B  x C awi, B we›`y †_‡K evmwU 3 ms–2 Z¡i‡Y t mg‡q C we›`y‡Z †cuŠQvq| †jvKwU A we›`y †_‡K 12 ms–1 †e‡M †`Šuwo‡q t mg‡q C we›`y‡Z †cuŠQvq| A I B Gi `~iZ¡ x n‡j, 12t = x + 1 2  3t2  12t – 1 2 3t2 = x  – 3t2 + 24t = 2x  3t2 – 24t + 2x = 0 ......... (i) GLb, †jvKwU evm‡K ai‡Z n‡j t Gi gvb ev ̄Íe n‡Z n‡e|  (i) bs mgxKi‡Yi wbðvqK  0  (– 24)2 – 4 × 3 × 2x  0  24 × 24  4 × 3 × 2x  x  24  †jvKwU m‡e©v”P 24 wg. `~i †_‡K evmwU‡K ai‡Z cvi‡e| (Ans.) 8. `ywU †b.Kv cÖ‡Z ̈‡K 5 km/h †e‡M P‡j 3 km/h †e‡M cÖevwnZ 550 m PIov GKwU b`x cvwo w`‡Z Pvq| GKwU †b.Kv b~ ̈bZg c‡_ I AciwU b~ ̈bZg mg‡q b`xwU cvwo †`q| Zviv GKB mg‡q hvÎv Ki‡j Zv‡`i Aci cv‡o †c.uQv‡bvi mg‡qi cv_©K ̈ wbY©q Ki| [BUET 16-17] mgvavb: GLv‡b, †bŠKvi †eM, v = 5 km/h † ̄av‡Zi †eM, u = 3 km/h b`xi cÖ ̄’, d = 550 m = 0.55 km b~ ̈bZg c‡_ cvwo w`‡Z mgq, t1 = d v 2 – u 2 = 0.55 5 2 – 3 2 = 11 80 h b~ ̈bZg mg‡q cvwo w`‡Z mgq, t2 = d v = 0.55 5 = 11 100 h  mg‡qi cv_©K ̈, t = t1 – t2 =     11 80 – 11 100 h = 11 400 h = 33 20 wg. = 1 wg. 39 †m. (Ans.)
mgZ‡j e ̄‘KYvi MwZ  Engineering Practice Sheet Solution 3 9. GKwU w ̄’i gm„Y cywji Dci w`‡q Szjv‡bv GKwU nvjKv iwki GK cÖv‡šÍ 51 kg IR‡bi GKwU e ̄‘ mshy3 Av‡Q| iwki Aci cÖvšÍ †e‡q mylg Z¡i‡Y GKwU evjK 4 s G 4.9 m Dc‡i D‡V| e ̄‘wU hw` w ̄’i Ae ̄’vq _v‡K Z‡e evjKwUi IRb wbY©q Ki| [g = 9.8 m/s2 ] [BUET 14-15] mgvavb: u = 0, t = 4 s, h = 4.9 m s = ut + 1 2 at2  4.9 = 0 + 1 2 × a × 42  a = 0.6125 ms–2 GLb, F = m (g + a)  51 × 9.8 = m (9.8 + 0.6125)  m = 48 kg-wt (Ans.) 10. GKwU †ijMvox GK †÷kb n‡Z †Q‡o 4 wgwbU ci 2 km `~‡i Aew ̄’Z Aci †÷k‡b _v‡g| MvwowU Zvi MwZc‡_i cÖ_gvsk x mgZ¡i‡Y Ges wØZxqvs‡k y mgg›`‡b Pj‡j 1 x + 1 y Gi gvb wbY©q Ki| [BUET 11-12] mgvavb: s1 = 0 + v 2 × t1  s1= v 2 t1  s1 + s2 = v 2 (t1 + t2) Ges s2 = v + 0 2 × t2  s2 = v 2 × t2  2 = v 2 × 4  v = 1 km/min Avevi, v = 0 + xt1  t1 = v x Ges 0 = v – yt2  t2 = v y  t1 + t2 = v     1 x + 1 y  4 = 1     1 x + 1 y  1 x + 1 y = 4 (Ans.) 11. GKwU ey‡jU GKwU Z3v †f` Ki‡Z Zvi †e‡Mi 1 10 Ask nvivq| Z3vi cÖwZ‡iva ÿgZv mylg n‡j _vgevi c~‡e© ey‡jUwU KZ ̧‡jv Z3v †f` Ki‡Z cvi‡e? [BUET 09-10] mgvavb: awi, ey‡jUwU u Avw`‡e‡M f g›`‡b x cyiæ‡Z¡i GKwU Z3v †f` Ki‡Z Zvi †e‡Mi 1 10 Ask nvivq|  Z3vwU †f` Kivi ci cÖvß †eM = u – u 10 = 9u 10 GLb,     9u 10 2 = u 2 – 2fx  2fx = u 2 – 81u2 100 = 19u2 100 awi, ey‡jUwU Abyiƒc n msL ̈K Z3v †f` Ki‡Z cv‡i| A_©vr, nx `~iZ¡ AwZμg K‡i †kl †eM k~b ̈ nq| 0 2 = u 2 – 2f × nx  2fx × n = u2  19u2 100 × n = u2  n = 100 19 = 5 5 19 wU  ey‡jUwU cici ̄’vwcZ 5 wU Z3v †f` Ki‡Z cv‡i| (Ans.) 12. GKwU RvnvR †Kv‡bv ̄’vb n‡Z 15 km/hr †e‡M DËi w`‡K hvÎv ïiæ Kij| hvÎvi ïiæ‡ZB Zvi c~e©w`‡K 10 km `~‡i Aci GKwU RvnvR †`L‡Z †cj| wØZxq RvnvRwU 20 km/hr †e‡M cwðg w`‡K hv‡”Q| KZÿY ci Zv‡`i gv‡S `~iZ¡ b~ ̈bZg n‡e? Zv‡`i gv‡S b~ ̈bZg `~iZ¡ KZ? [BUET 08-09] mgvavb: `: A B O c: c~: D D: x 15t 20t 10 – 20t 10 1g Rvnv‡Ri Ae ̄’vb O we›`y‡Z, 2q Rvnv‡Ri Ae ̄’vb D we›`y‡Z Ges t mgq ci 1g Rvnv‡Ri Ae ̄’vb B we›`y‡Z, 2q Rvnv‡Ri Ae ̄’vb A we›`y‡Z| awi, t mgq ci `~iZ¡ b~ ̈bZg n‡e,  OB = 15t Ges AD = 20t  OA = OD – AD = 10 – 20t ga ̈eZ©x `~iZ¡, x 2 = (15t)2 + (10 – 20t)2  x 2 = 225t2 + 100 – 400t + 400t2  d dt (x2 ) = 450t + 800t – 400 `~iZ¡ b~ ̈bZg n‡e hLb, d dt (x2 ) = 0  1250t – 400 = 0  t = 8 25 h (Ans.)  x 2 =     15  8 25 2 +     10 – 20  8 25 2 = 36  x = 6 km (Ans.)
4  Higher Math 2nd Paper Chapter-9 13. GKwU KYv‡K 200 metre/sec †e‡M Lvov Dc‡ii w`‡K wb‡ÿc Kiv n‡jv| Gi 10 sec c‡i Aci GKwU KYv‡K GKB ̄’vb n‡Z Lvov Dc‡ii w`‡K wb‡ÿc Kiv n‡jv| hw` Zviv cÖ_g KYvi e„nËg D”PZvq wgwjZ nq, Z‡e wØZxq KYvi †eM wbY©q Ki| [g ≅ 9.8 metre/sec] [BUET 07-08; MIST 17-18] mgvavb: 1g KYvi e„nËg D”PZv, H = u 2 2g = 2002 2  9.8 = 2040.8 m Ges DÌvbKvj, t = u g = 200 9.8 = 20.4 s awi, 2q KYvi †eM = u2 kZ©g‡Z, H = u2(t – 10) – 1 2 g(t – 10)2  2040.8 = u2(20.4 – 10) – 1 2  9.8(20.4 – 10)2  u2 = 247.2 ms–1  2q KYvi †eM 247.2 ms–1 (Ans.) 14. † ̄avZnxb Ae ̄’vq GKwU †jvK 4 wgwb‡U mvuZvi †K‡U †mvRvmywRfv‡e 100 wgUvi cÖk ̄’ GKwU b`x AwZμg Ki‡Z cv‡i wKš‘ † ̄avZ _vK‡j Zvi mgq jv‡M 5 wgwbU| † ̄av‡Zi †eM wbY©q Ki| [BUET 03-04; RUET 12-13, 10-11, 06-07] mgvavb: b`xi cÖ ̄’, d = 100 m mgq, t = 4 wgwbU  †jv‡Ki †eM, v = d t = 100 4 = 25 m/min awi, † ̄av‡Zi †eM = u  † ̄avZ _vK‡j †mvRvmywR cvo n‡Z mgq, t = d v 2 – u 2  5 = 100 252 – u 2  u = 15 m/min (Ans.) 15. e›`yK †_‡K wbwÿß GKwU †Mvjv wb‡ÿcY we›`y †_‡K 50 yards `~‡i Ges 75 feet DuPz †`Iqv‡ji wVK Dci w`‡q Avbyf~wgKfv‡e AwZμg K‡i| †Mvjvi wb‡ÿcY MwZ‡eM I wb‡ÿcY †Kv‡Yi gvb wbY©q Ki| [BUET 02-03] mgvavb:  u 75 ft H = 75 ft ; R = 300 ft tan = 4H R = 4  75 150  2 = 1   = 45 (Ans.) Avevi, Rmax = u 2 g  u = Rmax  g = 300  32 = 97.98 ft/sec (Ans.) 16. GKwU e ̄‘ cÖ_g 25 ft/sec2 Z¡i‡Y Ges c‡i 5 ft/sec2 g›`‡b P‡j w ̄’i nq| w ̄’ive ̄’v †_‡K 192 ft `~iZ¡ AwZμg Ki‡j MwZc‡_ Zvi m‡e©v”P MwZ‡eM KZ wQj wbY©q Ki| [BUET 01-02] mgvavb: g‡b Kwi, e ̄‘wU s1 `~iZ¡ 25 ft/sec2 Z¡i‡Y AwZμg K‡i m‡ev©”P v †eM cÖvß nq| v 2 = 0 + 2  25  s1  s1 = v 2 50 ...... (i) 5 ft/sec2 g›`‡b s2 `~iZ¡ AwZμg K‡i w ̄’i nq|  0 = v2 – 2  5 s2  s2 = v 2 10 ...... (ii) (i) + (ii) K‡i, s1 + s2 = v 2     1 10 + 1 50  v 2 = 192  25 3  v = 1600 = 40 ft/s (Ans.) 17. GKRb †jvK AvovAvwofv‡e mvuZvi †K‡U b dzU cÖ ̄’wewkó GKwU † ̄avZwenxb b`x t1 †m‡K‡Û cvwo w`‡Z cv‡i Ges b`x‡Z † ̄avZ _vK‡j t †m‡K‡Û cvwo w`‡Z cv‡i| † ̄av‡Zi †eM wbY©q Ki| [BUET 00-01] mgvavb: b`xi cÖ ̄’ b, mgq t1  †jv‡Ki †eM, v = b t1 awi, † ̄av‡Zi †eM = u  † ̄avZ _vK‡j †mvRvmywR cvo n‡Z mgq, t = b v 2 – u 2  t 2 (v2 –u 2 ) = b2  v 2 – u 2 = b 2 t 2  u 2 = v 2 – b 2 t 2 = b 2 t 2 1 – b 2 t 2  u = b 1 t1 2 – 1 t 2 dzU/†m. (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 18. GKwU cÖ‡ÿcK 21 ms–1 †e‡M Ges Avbyf~wg‡Ki mv‡_ 30 †Kv‡Y k~‡b ̈ cÖ‡ÿc Kiv n‡jv| GwUi cvjøv, me©vwaK D”PZv Ges `yB †m‡KÛ ci Zvi Ae ̄’vb I †eM wbY©q Ki| [KUET 19-20] mgvavb: cvjøv, R = v 2 0 sin2 g = 212  sin(2 30) 9.8 = 38.97 m (Ans.) me©vwaK D”PZv, H = v 2 0 sin2  2g = 212  sin2 30 2 9.8 = 5.625 m (Ans.)