Tiêu đề Motion of Particles Practice Sheet Solution HSC FRB-24.pdf ✅

2  Higher Math 2nd Paper Chapter-9 mgxKiY (i) G, v = 2S t ewm‡q cvB, t = 2S t     1 f1 + 1 f2  t = 2S t     f1 + f2 f1f2  S = t 2 f1f2 2(f1 + f2) (Proved) (M) GLv‡b, wb‡ÿcY †KvY,  = 45 Avgiv Rvwb, evqyk~b ̈ ̄’v‡b cÖwÿß KYvi MwZc‡_i mgxKiY, y = x tan     1 – x R CD †`Iqv‡ji D”PZv h n‡j, x = a Ges y = h Avevi, DÏxc‡Ki cÖvmwUi Avbyf~wgK cvjøv, R = a + b  h = atan     1 – a R = a tan45     1 – a a + b = a    a + b – a a + b  h = ab a + b  CD †`Iqv‡ji D”PZv, h = ab a + b (Proved) 2| `„k ̈Kí-1: w ̄’ive ̄’v n‡Z mij‡iLvq Pjgvb GKwU e ̄‘KYv cÖ_‡g y mgZ¡i‡Y Ges c‡i z mgg›`‡b P‡j| `„k ̈Kí-2: GKwU ͇̄¤¢i kxl© †_‡K 98 wg./†m‡KÛ †e‡M A e ̄‘‡K Lvov Dc‡ii w`‡K wb‡ÿc Kiv n‡jv| 2 †m‡KÛ c‡i GKB we›`y n‡Z Aci GKwU B e ̄‘‡K †Q‡o †`Iqv n‡jv| [ivRkvnx †evW©- Õ23] (K) 64 wgUvi DuPz `vjv‡bi Qv` †_‡K GKwU cv_i †Q‡o w`‡j f~wg‡Z co‡Z KZ mgq jvM‡e? (L) `„k ̈Kí-1 G KYvwU hw` t mg‡q d `~iZ¡ AwZμg K‡i Z‡e †`LvI †h, t = 2d     y + z yz (M) `„k ̈Kí-2 G e ̄‘ `ywU f~wg n‡Z KZ D”PZvq wgwjZ n‡e Zv wbY©q Ki| mgvavb: (K) GLv‡b, `vjv‡bi D”PZv, h = 64 m AwfKl©R Z¡iY, g = 9.8 ms–2 Avw`‡eM, u = 0 ms–1 GLb mgq t n‡j, Avgiv Rvwb, h = ut + 1 2 gt2  64 = 0  t + 1 2  9.8  t 2  t 2 = 64  2 9.8  t = 13.06  t = 3.61 †m‡KÛ (Ans.) (L) ̄’vb `yBwUi ga ̈eZx© †gvU `~iZ¡ = d awi, 1g As‡ki Rb ̈ `~iZ¡ = a, mgq = t1, mgZ¡iY = y Ges 2q As‡ki Rb ̈ `~iZ¡ = d – a, mgq = t2, mgg›`b = z mgq, t = t1 + t2 mgZ¡i‡Yi †ÿ‡Î: v = yt1  v y = t1 .......(i) Ges a = 1 2 yt 2 1 = 1 2 yt1.t1 = 1 2 vt1 .......(ii) mgg›`‡bi †ÿ‡Î: 0 = v – zt2  v = zt2  v z = t2 .......(iii) Ges d – a = vt2 – 1 2 zt 2 2 = vt2 – 1 2 vt2 = 1 2 vt2 ......(iv) (i) + (iii) K‡i cvB, t1 + t2 = v y + v z  t = v    1 y + 1 z ......(v) (ii) + (iv) K‡i cvB, d = 1 2 vt1 + 1 2 v2t2 = 1 2 v(t1 + t2)  d = 1 2 vt [⸪ t = t1 + t2]  v = 2d t ......(vi) mgxKiY (v) G v Gi gvb ewm‡q cvB, t = 2d t     y + z yz  t 2 2d = y + z yz  t 2 = 2d(y + z) yz  t = 2d     y + z yz (Showed) (M) B uA = 98ms–1 h A g‡b Kwi, B e ̄‘wU †d‡j †`Iqvi t †m‡KÛ c‡i ͇̄¤¢i kxl© n‡Z h wgUvi wb‡P A e ̄‘wU B e ̄‘i mv‡_ wgwjZ nq| A e ̄‘i †ÿ‡Î, h = – ut + 1 2 gt2  h = – 98(t + 2) + 1 2  9.8  (t + 2)2 .....(i)
mgZ‡j e ̄‘KYvi MwZ  Final Revision Batch '24 3 B e ̄‘i †ÿ‡Î, h = 1 2 gt2 ......(ii) (i) I (ii) †_‡K, – 98(t + 2) + 1 2  9.8(t + 2)2 = 1 2 gt2  – 98t – 196 + 4.9(t2 + 2.t.2 + 4) = 1 2  9.8 t2  – 98t – 196 + 4.9t2 + 19.6t + 19.6 – 4.9t2 = 0  – 78.4t – 178.4 = 0  – 78.4 t = 178.4  t = 178.4 – 78.4  t = – 2.28 s wKš‘ t Gi gvb FYvZ¥K nIqv m¤¢e bq| A_©vr, A I B e ̄‘ `yBwU f~wg‡Z covi Av‡M wgwjZ n‡e bv| (Ans.) 3| (i) GKwU e ̄‘ mgZ¡i‡Y mij‡iLv eivei P‡j 25 Zg †m‡K‡Û 266 †m.wg. Ges 42 Zg †m‡K‡Û 402 †m.wg. `~iZ¡ AwZμg K‡i| (ii) †Kv‡bv we›`y O n‡Z cÖwÿß GKwU ej `ywU †`Iqvj AwZμg K‡i| O we›`y n‡Z Zv‡`i GKwUi Avbyf~wgK `~iZ¡ Q Ges Lvov `~iZ¡ P Ges O we›`y n‡Z AciwUi Avbyf~wgK `~iZ¡ P Ges Lvov `~iZ¡ Q| [h‡kvi †evW©- Õ23] (K) w ̄’i Ae ̄’v n‡Z GKwU KYv 10 †m.wg/†m.2 mgZ¡i‡Y †Kv‡bv wbw`©ó mij‡iLvq Pj‡Q| 5 †m‡KÛ c‡i e ̄‘wUi †eM KZ n‡e? (L) e ̄‘wUi Avw`‡eM wbY©q Ki| (M) †`LvI †h, O we›`yMvgx ejwUi Avbyf~wgK Z‡ji Dci cvjøv p 2 + pq + q2 p + q mgvavb: (K) GLv‡b, Avw`‡eM, u = 0 †m.wg./†m. mgZ¡iY, f = 10 †m.wg./†m.2 mgq, t = 5 †m. †eM v n‡j, Avgiv Rvwb, v = u + ft = 0 + 10  5 = 50 †m.wg./†m. (Ans.) (L) awi, e ̄‘i Avw`‡eM u Ges mgZ¡iY f Avgiv Rvwb, t Zg mg‡q AwZμvšÍ `~iZ¡, St = u + 1 2 f(2t – 1) 25 Zg mg‡q AwZμvšÍ `~iZ¡ S25 n‡j, S25 = u + 1 2 f(2  25 – 1)  S25 = u + 49 2 f  266 = u + 49 2 f ....(i) Ges 42 Zg †m‡K‡Û AwZμvšÍ `~iZ¡ S42 n‡j, S42 = u + 1 2 f(2  42 – 1) S42 = u + 83 2 f  402 = u + 83 2 f ......(ii) (ii) – (i) n‡Z cvB, 83 – 49 2 f = 402 – 266  34 2 f = 136  f = 136 17  f = 8 †m.wg./†m.2 (i) bs n‡Z, u + 49 2  8 = 266  u = 266 – 196  u = 70 †m.wg./†m. (Ans.) (M) DÏxc‡Ki Z_ ̈vbymv‡i, wPÎ A1⁄4b K‡i cvB, u  Q P Q P X awi, wb‡ÿcY †KvY =  Ges Avbyf~wgK cvjøv = R Avgiv Rvwb, evqyk~b ̈ ̄’v‡b cÖwÿß KYvi MwZc‡_i mgxKiY, y = xtan     1 – x R 1g †`Iqv‡ji †ÿ‡Î, P = Q tan     1 – Q R ......(i) 2q †`Iqv‡ji †ÿ‡Î, Q = P tan     1 – P R ......(ii) (ii)  (i) K‡i cvB, Q P = P Q 1 – P R 1 – Q R  Q 2 P 2 = R – P R – Q  P 2 R – P 3 = Q2 R – Q 3  R(P2 – Q 2 ) = P3 – Q 3  R = P 3 – Q 3 P 2 – Q 2 = (P – Q) (P 2 + PQ + Q2 ) (P + Q) (P – Q)  R = P 2 + PQ + Q2 P + Q (Showed) we: `a: cÖ‡kœ P I Q D‡jøL _vKvq, R = p + pq + q2 p + q Gi cwie‡Z© R = P 2 + PQ + Q2 P + Q