Tiêu đề 2. Phy-1st-Vector.pdf ✅

†f±i  Practice Content 1 †f±i Vector wØZxq Aa ̈vq ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| 6 GKK I 8 GKK gv‡bi `ywU †f±i 60 †Kv‡Y †Kv‡bv KYvi Ici GKB mgq wμqv Ki‡Q| G‡`i jwäi gvb I w`K wbY©q K‡iv| [Easy] mgvavb: R A = 6 GKK 8 GKK = B  R 2 = A2 + B2 + 2.A.B.cos R = 6 2 + 82 + 2.6.8.cos60 = 12.16 (Ans.) tan = A sin B + A cos  = tan–1 6 sin60 8 + 6 cos = 25.28 (Ans.) 2| `ywU e‡ji m‡e©v”P I me©wb¤œ gvb h_vμ‡g 29 kg-wt I 5 kg-wt; hw` cÖ‡Z ̈KwU e‡ji gvb 3kg-wt K‡i evov‡bv nq, Z‡e bZzb ej؇qi jwäi gvb wbY©q Ki †hb ejØq ci ̄ú‡ii mv‡_ mg‡Kv‡Y _v‡K| [Easy] mgvavb: P + Q = 29 kg-wt P – Q = 5 kg-wt  P = 29 + 5 2 = 17 kg-wt  Q = 29 – 5 2 = 12 kg-wt  P = 20 kg-wt  Q = 15 kg-wt R = 202 + 15 2 = 25 kg-wt (Ans.)  = tan–1     15 20 = 36.869; 20 kg-wt Gi mv‡_ (Ans.) 3| hw` A, B I C we›`y ̧‡jvi Ae ̄’vb †f±iÎq h_vμ‡g  i + 2  j – 3  k , 3  i – 4  j + 5  k Ges 5  i – 10  j + 13  k nq Z‡e †`LvI †h, AB I BC †f±iØq mg‣iwLK ev collinear. [Medium] mgvavb:   AB =  B –  A = (3  i – 4  j + 5  k) – (  i + 2  j – 3  k) = 2  i – 6  j + 8  k   BC =  C –  B = (5  i – 10  j + 13  k) – (3  i – 4  j + 5  k) = 2  i – 6  j + 8  k   AC =  C –  A = (5  i – 10  j + 13  k) – (  i + 2  j – 3  k) = 4  i – 12  j + 16  k  |  AB | = 104  |  BC | = 104  |  AC | = 416 = 104 + 104 = |  AB | + |  BC | Avevi,  AB ×  BC =          i  j  k 2 – 6 8 2 – 6 8 = 0  AB I BC mg‣iwLK| (†`Lv‡bv n‡jv) 4|  A = (6  i – 8  j)m,  B = (– 8  i + 3  j)m Ges  C = (26  i + 19  j)m †f±iÎq ci ̄ú‡ii mv‡_ a  A + b  B +  C =  0 mgxKiY Øviv m¤úwK©Z| a I b Gi gvb wbY©q Ki| [Medium] mgvavb: a  A + b  B +  C =  0  a(6  i – 8  j)m + b(– 8  i + 3  j)m + (26  i + 19  j)m = 0   i(6a – 8b + 26) +  j(– 8a + 3b + 19) = 0 GLb, 6a – 8b = – 26 Ges – 8a + 3b = – 19  a = 5  b = 7 (Ans.) 5|  d1 +  d2 = 5d3,  d1 –  d2 = 3d3 Ges d3 = 2  i + 4  j n‡j,  d1 I  d2 wbY©q Ki| [Easy] mgvavb:  d1 +  d2 = 5(2  i + 4  j) = 10  i + 20  j  d1 –  d2 = 3(2  i + 4  j) = 6  i + 12  j  2  d1 = 16  i + 32  j   d1 = 8  i + 16  j (Ans.) Avevi, 2  d2 = 4  i + 8  j   d2 = 2  i + 4  j (Ans.) 6| hw` 3N, 4N I 12N gv‡bi wZbwU ej †Kv‡bv †Kv‡bv GKwU KYvi Dci Ggbfv‡e wμqvkxj nq †hb ej ̧‡jvi cÖ‡Z ̈KwU ci ̄ú‡ii mv‡_ mg‡Kv‡Y wμqvkxj, Z‡e jwä e‡ji gvb wbY©q Ki| [Medium]
2  Physics 1st Paper Chapter-2 mgvavb: 3N R1 4N 12N R2 A D B C E O GLb, R1 = 3 2 + 42 N = 5N [OD eivei] Avevi, R2 = R1 2 + 122 N = 5 2 + 122 N = 13 N [OE eivei] (Ans.) 7| hw` mgvb gv‡bi `ywU e‡ji gvb F Ges Zv‡`i jwä gvb F 3 nq Z‡e ej `ywUi ga ̈eZ©x †KvY K‡Zv? [Easy] mgvavb: GLv‡b,     F 3 2 = F2 + F2 + 2F2 cos  1 9 = 2 + 2cos  2cos = 1 9 – 2  cos = 1 – 18 18   = cos–1     – 17 18 = 160.812 (Ans.) 8| GKwU wegvb 500km/h `aæwZ‡Z DËi w`‡K hvw”Q‡jv| GKRb hvÎx‡ekx nvBR ̈vKvi GKwU e›`yK wb‡q nVvr 1000km/ h `aæwZ‡Z Dc‡ii w`‡K ̧wj Quyo‡Z ïiæ Ki‡jv| c„w_ex‡Z Aew ̄’Z ch©‡eÿK †Uwj‡ ̄..v‡ci mvnv‡h ̈ m¤ú~Y© NUbv ch©‡eÿY Ki‡jv| ch©‡eÿ‡Ki mv‡c‡ÿ ̧wj Dj‡¤^i mv‡_ KZ †Kv‡Y MwZkxj n‡e? [Medium] mgvavb: awi, |  V1 | = 500 km/h |  V2 | = 1000 km/h Dj‡¤^i mv‡_ †KvY =   tan = v1 v2   = tan–1     v1 v2  V1   V2 D :   = tan–1     500 1000 = 26.565 (Ans.) 9| hw`  A +  B =  C nq, Z‡e †Kvb kZ©mv‡c‡ÿ A 2 + B2 = C2 n‡e? [Easy] mgvavb: awi,  A I  B Gi ga ̈eZ©x †KvY , GLv‡b,  C =  A +  B  C 2 = A2 + B2 + 2.A.B.cos A 2 + B2 = C2 n‡e hw`, 2AB cos = 0 nq  AB cos = 0   = 90  A 2 + B2 = C2 n‡e hw`  A I  B ci ̄ú‡ii j¤^ nq| (Ans.) 10| hw`  A +  B =  C nq, Z‡e †Kvb kZ©mv‡c‡ÿ A + B = C n‡e? [Easy] mgvavb: awi, ga ̈eZ©x †KvY =  GLb, C = A + B  C =  A +  B  C 2 = A2 + B2 + 2.A.B cos  (A + B)2 = A2 + B2 + 2.A.B.cos  cos = 1   = cos–1 (1) = 0 (Ans.) 11| †Kv‡bv we›`y‡Z  †Kv‡Y wμqviZ  P I  Q †f±‡ii jwä  R. hw`  Q Gi gvb wØ ̧Y Kiv nq Z‡e jwäi gvb wØ ̧Y nq| hw`  Q †K  – Q Øviv cÖwZ ̄’vwcZ Kiv nq Zvn‡jI jwäi gvb 2R nq|  P ,  Q I  R Gi gv‡bi AbycvZ wbY©q Ki| [Medium] mgvavb: awi,  P   Q =  1g †ÿ‡Î, R 2 = P2 + Q2 + 2.P.Q cos ............ (i) 2q †ÿ‡Î, 4R2 = P2 + 4Q2 + 4PQ cos .......... (ii) 3q †ÿ‡Î, 4R2 = P2 + Q2 – 2PQ cos ........... (iii) (i) + (iii)  5R2 = 2P2 + 2Q2  2P2 + 2Q2 – 5R2 = 0 .......... (iv) (ii) – 2 × (i)  2R2 = – P 2 + 2Q2  P 2 – 2Q2 + 2R2 = 0 .......... (v) (iv) I (v) n‡Z eRa ̧Y‡bi m~Îvbymv‡i, P 2 4 – 10 = Q 2 – 5 – 4 = R 2 – 4 – 2  P 2 – 6 = Q 2 – 9 = R 2 – 6  P : Q : R = 2 : 3 : 2 (Ans.) 12| `ywU mgvb †f±i‡K †hvM Ki‡j †Kv‡bv Ae ̄’vq Zv‡`i jwä (K) GKwU †f±‡ii gv‡bi 2 ̧Y n‡e Ges (L) GKwU †f±‡ii gv‡bi 3 ̧Y n‡e? [Easy] mgvavb: (K) awi, †f±iØq  P Ges ga ̈eZ©x †KvY  ( 2P)2 = P 2 + P2 + 2P2 cos  cos = 0   = 90 (Ans.) (L) ( 3P)2 = P 2 + P2 + 2P2 cos  2P 2 cos = P 2   = cos–1     1 2   = 60 (Ans.) 13|  2A Ges  A †f±iØq ci ̄ú‡ii m‡1⁄2 GKwU wbw`©ó †Kv‡Y AvbZ| cÖ_g †f±i‡K wØ ̧Y Ki‡j jwäi gvb wZb ̧Y nq| †f±i؇qi AšÍeZ©x †KvY KZ? [Easy]
†f±i  Practice Content 3 mgvavb: GLv‡b, R 2 = (2A)2 + A2 + 4A 2 cos  R 2 = A2 (5 + 4cos) .......... (i) Avevi, (3R)2 = (4A)2 + A2 + 2.4A 2 cos  9R2 = A2 (17 + 8 cos) ........... (ii) (i)  (ii)  1 9 = 5 + 4 cos 17 + 8 cos  17 + 8cos = 45 + 36 cos  28 cos = – 28   = cos–1 (– 1) = 180 (Ans.) 14| wPÎ-1 cÖ`wk©Z †f±i  A I  B Gi mgwó (  A +  B) I AšÍi (  A –  B ) wbY©q Ki| [Easy] y x 30  B(15N)  A(6N) mgvavb: |  A +  B| = A 2 + B2 + 2A.B cos = 6 2 + 152 + 2.6.15 cos 150 = 10.25 N  tan1 = A sin B + A cos  tan1 = 6 sin150 15 + 6cos150  1 = 17  x A‡ÿi mv‡_ (30 + 17) = 47 y x  A  B  A+  B 1 Avevi, |  A –  B | = A 2 + B2 – 2AB cos = 6 2 + 152 – 2.6.15 cos = 20.42 N  tan2 = 15 sin150 6 – 15 cos150  2 = 21.55  x A‡ÿi mv‡_ 201.55 (Ans.) y x  A –  B  A–  B 2 15| †f±i ivwk A = 3 Ges B = 4, hw` A Gi w`K B Gi Dj‡¤^i w`K nq Zvn‡j Zv‡`i jw¤^ †f±i R KZ n‡e? A Gi mv‡c‡ÿ R Gi w`K wK n‡e? [Easy] mgvavb: R = A 2 + B2 = 3 2 + 42 = 5 GKK tan = B sin A + B cos = 4 sin90 3 + 4 cos90   = tan–14 3   = 53.13 (Ans.) 16| DËi Ges c~e© w`‡K h_vμ‡g 3 ms–1 Ges 4 ms–1 gv‡bi `ywU †eM †f±i‡K  v1 Ges  v2 Øviv m~wPZ Kiv n‡jv| G‡`i we‡qvMdj  v1 –  v2 wbY©q Ki| GwU †Kvb w`‡K KZ †Kv‡Y _vK‡e? [Easy] mgvavb:  v1 –  v2 †f±‡ii gvb 5 ms–1 , Gi w`K DËi w`‡Ki mv‡_ 53.13 †Kv‡Y DËi-cwðg w`‡K |  v1 –  v2 | = 3 2 + (4) 2 = 5 ms–1 tan = 4 3   = 53.13 v1 143.13  DËi c~e©  – v2 v2 Dˇii mv‡_ 53.13 †Kv‡Y DËi-cwð‡g| (Ans.) 17| x-A‡ÿi abvZ¥K w`‡Ki mv‡_ 170 I 50 †Kv‡Y `ywU ej †Kvb we›`y e ̄‘i Dci wμqvkxj| Dfq e‡ji gvb 100 N n‡j jwä e‡ji gvb I w`K wbY©q K‡iv| [Medium] mgvavb: y x 100N 100N 120 60 50  R = 1002 × 2 + 2 × 1002 × cos120 = 100 N  = 120 2 = 60  x A‡ÿi mv‡_ = 60 + 50 = 110 (Ans.) GLv‡b,  = (170 – 50) = 120 18| (  A +  B) Ges (  A –  B) Gi ga ̈eZ©x †KvY KZ n‡j Zv‡`i jwäi gvb 3A2 + B2 n‡e? [Easy] mgvavb: (A + B) 2 + (A – B) 2 + 2(A + B) (A – B) × cos = 3A2 + B2  2(A2 + B2 ) + 2(A2 – B 2 ) cos = 3A2 + B2  2(A2 – B 2 )cos = A2 – B 2  cos = 1 2 [∵ A  B ZvB A 2 – B 2  0]   = 60 (Ans.) 19| GKwU D‡ovRvnvR 110 m/s †e‡M DËi w`‡K hvÎv ïiæ Kij| H mgq evZvm 40 m/s †e‡M cwðg n‡Z c~e© w`‡K eBwQjv| D‡ovRvnvRwU f‚wg mv‡c‡ÿ KZ †e‡M †Kvb w`‡K Pj‡e? [Easy]
4  Physics 1st Paper Chapter-2 mgvavb:  R = 1102 + 402 ms– 1 = 117.05 ms–1 Ges tan = 40 110   = tan–1 4 11 = 19.98  DËi 110 ms –1 40 ms–1 c~e© A_©vr, DËi w`‡Ki mv‡_ 19.98 K‡i DËi-c~e© w`‡K| (Ans.) 20| †f±i  B †K  A Gi mv‡_ †hvM Ki‡j 6  i +  j Ges †f±i  A n‡Z we‡qvM Ki‡j – 4  i + 7  j nq| †f±i  A I  B Ges Zv‡`i gvb wbY©q Ki| [Easy] mgvavb: A = 17 B = 34  A +  B = 6  i +  j ............. (i)  A –  B = – 4  i + 7  j ............. (ii) (i) + (ii)  2  A = 2  i + 8  j   A =  i + 4  j |  A| = 1 + 16 = 17 GKK (i) – (ii)  2  B = 10  i – 6  j   B = 5  i – 3  j  |  B | = 25 + 9 = 34 GKK 21| 75 †Kv‡Y wμqviZ `ywU e‡ji jwä 12N Ges Zv GKwUi mv‡_ 45 †KvY Drcbœ K‡i| ej `ywUi gvb wbY©q Ki| [Medium] mgvavb: 12 N P Q 30 45 awi, P e‡ji mv‡_  = 45  Q e‡ji mv‡_ = 30  P sin30 = Q sin45 = 12 sin75  P = 12 sin 30 sin 75 = 6.21 N  Q = 12 sin45 sin75 = 8.78 N 22| hw`  r1 = 2  i –  j +  k,  r2 =  i + 3  j – 2  k,  r3 = – 2  i +  j – 3  k Ges  r4 = 3  i + 2  j + 5  k nq Z‡e, a, b I c Gi gvb wbY©q Ki †hb  r4 = a  r1 + b  r2 + c  r3 nq| [Medium] mgvavb: a = – 2 b = 1 c = – 3 3  i + 2  j + 5  k = a(2  i –  j +  k) + b (  i + 3  j – 2  k) + c(– 2  i +  j – 3  k)  3  i + 2  j + 5  k =  i(2a + b – 2c) +  j(– a + 3b + c) + (a – 2b – 3c)  k mnM mgxK...Z K‡i cvB, 2a + b – 2c = 3 .......... (i) – a + 3b + c = 2 ......... (ii) a – 2b – 3c = 5 ........... (iii)  a = – 2, b = 1 c = – 3 (Ans.) 23| GKwU wÎfz‡Ri wZbwU †K.wYK we›`yi ̄’vbv1⁄4 h_vμ‡g A(3, – 2, 1), B(1, – 3, 5), C(2, 1, – 4). [Easy] K. BC evûi •`N© ̈ wbY©q K‡iv| mgvavb:  BC = (2 – 1)  i + (1 + 3)  j + (– 4 – 5)  k =  i + 4  j – 9  k  |  BC | = 1 2 + 42 + 92 = 7 2 GKK L. wÎfzRwU mg‡KvYx wK-bv g~j ̈vqb c~e©K gZvgZ `vI| mgvavb:  AB = – 2  i –  j + 4  k  AC = –  i + 3  j – 5  k   AB .  AC = 2 – 3 – 20 = – 21  0   AB .  BC = – 2 – 4 – 36 = – 42  0   BC .  AC = – 1 + 12 + 45 = 56  0  ABC wÎfzR, mg‡KvYx wÎfzR bq| 24| evqy f‚wgi mgvšÍivj DËi w`‡K 5 km/h †e‡M cÖevwnZ n‡”Q| wb‡¤œv3 w`Kmg~n Gi Dcvsk KZ? [Easy] (K) c~e© w`K : (L) cwðg w`K (M) Lvov Dc‡ii w`K mgvavb: DËi Dci cwðg c~e© (K) c~e© w`‡K, Ac = 5 cos90 = 0 km/h (L) cwðg w`‡K, Aw = 5cos90 = 0 km/h (M) Lvov Dc‡ii w`‡K, Au = 5 cos90 = 0 km/h 25| evqy `wÿY c~e© w`‡Ki gvSvgvwS †Kv‡bv w`K n‡Z DËi- cwðg w`‡K cÖevwnZ n‡”Q| evqyi †e‡Mi DËigyLx I cwðggyLx Ask‡Ki gvb h_vμ‡g 3 km/h I 4 km/h. [Easy] (K) evqyi cÖK...Z †eM K‡Zv?