Page 8 - كتاب المعلم فيزياء اول ثانوي

‫م‪‬ا‪ ‬دا‪‬ل ال‪‬ض‪‬‬ ‫م‪‬ا‪2 ‬‬

‫�ض‪ ‬وا‪ ‬ﺍﲨﻊ ﺍﳌﺘﺠﻬﲔ ﺍﻟ ﹼﺘﺎﻟﻴﲔ ﺑﺎﺳـﺘﻌﲈﻝ ﻃﺮﻳﻘﺔ‬ ‫ال‪‬ري‪ ‬إال‪ ‬الم‪‬ز‪ ‬ﻳﺸﻴﺮ ﻣﺴﺘﻘﺒﻞ ﺟﻬﺎﺯ ﻧﻈﺎﻡ ﺗﺤﺪﻳﺪ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻌﺎﻟﻤﻲ ﺇﻟﻰ ﺃﻥ ﻣﻨﺰﻟﻚ ﻳﺒﻌﺪ ‪ 15.0 km‬ﻓﻲ ﺍﺗﺠﺎﻩ ﻳﺼﻨﻊ ‪40°‬‬

‫ﺍﳌﺮﻛﺒﺎﺕ‪ :‬ﻣﻘﺪﺍﺭ ﺍﳌ ﹼﺘﺠﻪ ‪ A‬ﻳﺴﺎﻭﻱ ‪ 4 m‬ﻭﻳ ﹼﺘﺠﻪ‬ ‫ﺷـﻤﺎﻝ ﺍﻟﻐﺮﺏ‪ ،‬ﻭﻟﻜﻦ ﺍﻟﻄﺮﻳﻖ ﺍﻟﻮﺣﻴﺪ ﺍﻟﻤﺘﺎﺡ ﺃﻣﺎﻣﻚ ﻟﻠﻮﺻﻮﻝ ﺇﻟﻰ ﺍﻟﻤﻨﺰﻝ ﻫﻮ ﻓﻲ ﺍﺗﺠﺎﻩ ﺍﻟﺸـﻤﺎﻝ‪ .‬ﻓﺈﺫﺍ ﺳـﻠﻜﺖ ﻫﺬﺍ ﺍﻟﻄﺮﻳﻖ‬
‫ﻧﺤﻮ ﺍﳉﻨﻮﺏ‪ ،‬ﻭﻣﻘﺪﺍﺭ ﺍﳌ ﹼﺘﺠﻪ ‪ B‬ﻳﺴﺎﻭﻱ ‪7.3 m‬‬
‫ﻭﺗﺤﺮﻛﺖ ﻣﺴﺎﻓﺔ ‪ ،5.0 km‬ﻓﻤﺎ ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﺘﻲ ﻳﺠﺐ ﺃﻥ ﺗﻘﻄﻌﻬﺎ ﺑﻌﺪ ﺫﻟﻚ ﺣﺘﻰ ﺗﺼﻞ ﺇﻟﻰ ﻣﻨﺰﻟﻚ؟ ﻭﻓﻲ ﺃﻱ ﺍﺗﺠﺎﻩ ﺗﺴﻴﺮ؟‬
‫ﻭﻳﺘﺠﻪ ﻧﺤﻮ ﺍﻟﺸﲈﻝ ﺍﻟﻐﺮﰊ‪.‬‬
‫‪ 1‬ليل ا‪�‬ضاألة و‪�‬ضمها‬
‫ا‪‬وا‪ ‬ﻣ ﹼﺜﻞ ﺍﻻﲡﺎﻩ ﻧﺤﻮ ﺍﻟـﴩﻕ ﻋﲆ ﺍﳌﺤﻮﺭ‬ ‫• ﺍﺭﺳﻢ ﻣﺘﺠﻪ ﺍﻟﻤﺤﺼﻠﺔ ‪ R‬ﻣﻦ ﻣﻮﻗﻌﻚ ﺍﻷﺻﻠﻲ ﺇﻟﻰ ﻣﻨﺰﻟﻚ‪.‬‬

‫‪ ،+x‬ﻭﺍﻻﲡﺎﻩ ﻧﺤﻮ ﺍﻟﺸﲈﻝ ﻋﲆ ﺍﳌﺤﻮﺭ ‪.+y‬‬ ‫• ﺍﺭﺳﻢ ﺍﻟﻤﺘﺠﻪ ﺍﻟﻤﻌﻠﻮﻡ ‪ ،A‬ﺛﻢ ﺍﺭﺳﻢ ﺍﻟﻤﺘﺠﻪ ﺍﻟﻤﺠﻬﻮﻝ ‪.B‬‬

‫ﺍﻟﻤﺠﻬﻮﻝ‬ ‫ﺍﻟﻤﻌﻠﻮﻡ‬

‫?=‪B‬‬ ‫‪ A = 5.0 km‬ﻧﺤﻮ ﺍﻟﺸﻤﺎﻝ‬

‫‪ R =15.0 km‬ﻓﻲ ﺍﺗﺠﺎﻩ ‪ 40°‬ﺷﻤﺎﻝ ﺍﻟﻐﺮﺏ‪.‬‬

‫‪θ =140°‬‬

‫‪ 2‬ا�ضتخرا‪ ‬الكمية ا‪‬جهولة‬

‫‪Ax = (4.0 m) cos 270° = 0‬‬ ‫‪Rx = R cos θ = (15.0 km) (cos 140°) = -11.5 km‬‬ ‫ﺟﺪ ﻣﺮﻛﺒﺎﺕ ﺍﻟﻤﺘﺠﻪ ‪R‬‬
‫‪Ry = R sin θ = (15.0 km) sin 140° = 9.64 km‬‬
‫‪Bx = (7.3 m) cos 135° = 5.16 m‬‬ ‫‪Ay = 5.0 km, Ax = 0.0 km‬‬ ‫ﻋﻮﺽ ﻋﻦ ﻛﻞ ﻣﻦ ‪R ، θ‬‬
‫ﺑﻤﺎ ﺃﻥ ‪ A‬ﻓﻲ ﺍﺗﺠﺎﻩ ﺍﻟﺸﻤﺎﻝ‪ ،‬ﻟﺬﺍ ﻓﺈﻥ‬
‫‪Ay = (4.0 m) sin 270° = -4.0 m‬‬ ‫ﻋﻮﺽ ﻋﻦ ﻛﻞ ﻣﻦ ‪Ax ، Rx ، Ay ، Ry‬‬

‫‪By = (7.3 m) sin 135° = 5.16 m‬‬ ‫ﺍﺳﺘﻌﻤﻞ ﻣﺮﻛﺒﺎﺕ ﻛﻞ ﻣﻦ ‪ R‬ﻭ ‪ A‬ﻹﻳﺠﺎﺩ ﻣﺮﻛﺒﺎﺕ ﺍﻟﻤﺘﺠﻪ ‪.B‬‬

‫)‪Rx = Ax + Bx = 0 + (-5.16 m‬‬ ‫ﺍﻹﺷﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﺗﻌﻨﻲ ﺃﻥ ﺍﻟﻨﻘﺎﻁ ﺍﻹﺣﺪﺍﺛﻴﺔ ﻓﻲ ﺍﺗﺠﺎﻩ ﺍﻟﻐﺮﺏ‬
‫‪=-5.16 m‬‬
‫‪Bx = Rx - Ax = -11.5 km - 0.0 km = -11.5 km‬‬

‫‪By = Ry - Ay = 9.64 km - 5.0 km = 4.6 km‬‬ ‫ﺍﺳﺘﻌﻤﻞ ﻣﺮﻛﺒﺎﺕ ﺍﻟﻤﺘﺠﻪ ‪ B‬ﻹﻳﺠﺎﺩ ﻣﻘﺪﺍﺭ ﺍﻟﻤﺘﺠﻪ ‪.B‬‬

‫‪B = Bx2 + By2‬‬ ‫ﻋﻮﺽ ﻋﻦ ﻛﻞ ﻣﻦ ‪Bx ، By‬‬

‫‪= (-11.5km)2 + (4.6km)2‬‬ ‫‪By‬‬ ‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﻈﻞ ﻹﻳﺠﺎﺩ ﺍﺗﺠﺎﻩ ﺍﻟﻤﺘﺠﻪ ‪.B‬‬
‫ﻋﻮﺽ ﻋﻦ ﻛﻞ ﻣﻦ ‪Bx ، By‬‬
‫)‪Ry = Ay + By = (-4.0 m) + (5.16 m‬‬ ‫‪ = 12km‬‬
‫‪= 1.16 m‬‬ ‫‪θ = tan-1‬‬ ‫‪By‬‬ ‫‪θ‬‬
‫‪Bx‬‬ ‫‪Bx‬‬

‫‪R2 = Rx2 + Ry2‬‬ ‫)‪= tan-1 (-41.16.k5mkm‬‬
‫‪R2 = (-5.16 m)2 + (-1.16 m)2‬‬
‫)‪ (158°‬ﺃﻭ ‪= -22°‬‬

‫‪R = 5.3 m‬‬ ‫ﺿﻊ ﺫﻳﻞ ﺍﻟﻤﺘﺠﻪ ‪ B‬ﻋﻨﺪ ﻧﻘﻄﺔ ﺍﻷﺻﻞ ﻟﻨﻈﺎﻡ ﺇﺣﺪﺍﺛﻲ‪ ،‬ﻭﺍﺭﺳﻢ ﺍﻟﻤﺮﻛﺒﺘﻴﻦ ‪ BX‬ﻭ ‪ ،By‬ﻓﻴﻜﻮﻥ ﺍﻻﺗﺠﺎﻩ ﻓﻲ ﺍﻟﺮﺑﻊ ﺍﻟﺜﺎﻧﻲ ﻭﻓﻲ ﺍﺗﺠﺎﻩ‬
‫ﻳﺼﻨﻊ ﺯﺍﻭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ ‪ 22°‬ﺷﻤﺎﻝ ﺍﻟﻐﺮﺏ‪.‬‬
‫ﺍﻻﲡﺎﻩ‪:‬‬ ‫‪ 3‬ت‪‬و‪ ‬ا‪‬وا‪‬‬

‫‪_R_y‬‬ ‫__‪_1_._1_6__m‬‬ ‫• ﻫﻞ ﺍﻟﻮﺣﺪﺍﺕ ﺻﺤﻴﺤﺔ؟ ﺍﻟﻜﻴﻠﻮﻣﺘﺮﺍﺕ ﻭﺍﻟﺪﺭﺟﺎﺕ ﺻﺤﻴﺤﺔ‪.‬‬
‫‪Rx‬‬ ‫‪-5.16 m‬‬ ‫• ﻫﻞ ﺍﻹﺷﺎﺭﺍﺕ ﻣﻨﻄﻘﻴﺔ؟ ﺗﺘﻔﻖ ﺍﻹﺷﺎﺭﺍﺕ ﻣﻊ ﺍﻟﻤﺨﻄﻂ‪.‬‬
‫‪( ) ( )θ = tan-1‬‬
‫‪= tan-1‬‬ ‫• ﻫﻞ ﺍﻟﻤﻘﺪﺍﺭ ﻭﺍﻗﻌﻲ؟ ﺇﻥ ﻃﻮﻝ ﺍﻟﻤﺘﺠﻪ ‪ B‬ﺃﻛﺒﺮ ﻣﻦ ‪RX‬؛ ﻭﺫﻟﻚ ﻷﻥ ﺍﻟﺰﺍﻭﻳﺔ ﺑﻴﻦ ‪ A‬ﻭ ‪ B‬ﺃﻛﺒﺮ ﻣﻦ ‪.90°‬‬

‫‪=167°‬‬ ‫‪13‬‬

‫‪=12.7°‬‬ ‫ﺃﻭ ﰲ ﺍﲡﺎﻩ ﺷﲈﻝ ﺍﻟﻐﺮﺏ‬

‫ﻣﺴﺎﻋﺪة اﻟﻄﻼب ذوي ﺻﻌﻮﺑﺎت اﻟﺘﻌ ّﻠﻢ‬ ‫ﻧﺸﺎط‬

‫اﻟﺘﻔﻜﻴﺮ اﻟﻨﺎﻗﺪ‬ ‫‪‬م‪‬ع ا‪ ‬تجهات ﺍﳌﺠﻤـﻮﻉ ﺍﻻﲡﺎﻫﻲ ﻟﻀﻠﻌﻲ ﺯﺍﻭﻳﺔ ﻻ ﻳﺴـﺎﻭﻱ ﳎﻤﻮﻋﻬﲈ ﺍﳉـﱪﻱ‪ .‬ﲢﺘﺎﺝ ﺇﱃ‬
‫‪‬مع ‪‬لا‪‬ة م ‪‬تجهات ﺫ ﹼﻛﺮ ﺍﻟﻄﻼﺏ ﺑﺄﳖﻢ ﺩﺭﺳﻮﺍ ﲨﻊ‬
‫ﻣﻜﺎﻥﻭﺍﺳﻊﻟﺘﻨﻔﻴﺬﻫﺬﺍﺍﻟﻨﺸﺎﻁ‪،‬ﻭﴍﻳﻂﻻﺻﻖﻭﻣﺴﻄﺮﺓﻣﱰ ﹼﻳﺔﻭﺁﻟﺔﺣﺎﺳﺒﺔ‪.‬ﺍﺧﱰﻣﺴﺎﺣﺔﻛﺒﲑﺓ‬
‫ﻣ ﹼﺘﺠﻬﲔ‪ ،‬ﺛﻢ ﺍﺳـﺄﳍﻢ‪ :‬ﻛﻴﻒ ﻳﻤﻜﻦ ﺍﺳـﺘﻌﲈﻝ ﺫﻟﻚ ﳊ ﹼﻞ‬ ‫ﻭﺍﺿﺤﺔ ﻟﻠﻄﻼﺏ ﺛﻢ ﺍﺭﺳـﻢ ﻣﺜﻠ ﹰﺜﺎ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻛﺒ ﹰﲑﺍ ﻧﺴـﺒ ﹼﹰﻴﺎ ﻋﲆ ﺃﺭﺿ ﹼﻴﺔ ﺍﻟﻘﺎﻋﺔ ﺑﺎﺳﺘﻌﲈﻝ ﺍﻟﴩﻳﻂ‬
‫ﺍﻟﻼﺻـﻖ‪ .‬ﺍﻃﻠـﺐ ﺇﱃ ﻃﺎﻟﺒﲔ ﺍﻟﺘـﺪ ﹼﺭﺏ ﻋﲆ ﺍﳌﴚ ﺑﺨﻄﻮﺍﺕ ﻣﺘﺴـﺎﻭﻳﺔ ﺍﻟﻄﻮﻝ‪ ،‬ﺛـ ﹼﻢ ﺍﻃﻠﺐ ﺇﻟﻴﻬﲈ‬
‫ﻣﺴـﺄﻟﺔ ﺗﺸـﺘﻤﻞ ﻋﲆ ﺛﻼﺛﺔ ﻣ ﹼﺘﺠﻬـﺎﺕ؟ ﺍﲨـﻊ ﺍﳌ ﹼﺘﺠﻬﲔ‬ ‫ﺍﻟﻮﻗـﻮﻑ ﻋﲆ ﺭﺃﺱ ﺍﳌﺜﻠﺚ ﻋﻨـﺪ ﺇﺣﺪ￯ ﺍﻟﺰﻭﺍﻳﺎ ﺍﳊﺎ ﹼﺩﺓ‪ ،‬ﺛ ﹼﻢ ﺍﻃﻠـﺐ ﺇﻟﻴﻬﲈ ﺍﻟﺒﺪﺀ ﺑﺎﳌﴚ ﰲ ﺍﻟﻠﺤﻈﺔ‬
‫ﻧﻔﺴـﻬﺎ ﺑﺤﻴـﺚ ﻳ ﹼﺘﺠﻪ ﺃﺣﺪﳘﺎ ﺇﱃ ﺍﻟﺰﺍﻭﻳـﺔ ﺍﳊﺎﺩﺓ ﺍﻷﺧﺮ￯ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﻮﺗـﺮ‪ ،‬ﺑﻴﻨﲈ ﻳﺘﺠﻪ ﺍﻵﺧﺮ ﺇﱃ‬
‫ﺛـ ﹼﻢ ﺍﲨﻊ ﺍﳌ ﹼﺘﺠﻪ ﺍﻟﺜﺎﻟﺚ ﻣـﻊ ﳏﺼﻠﺔ ﺍﳌﺘﺠﻬﲔ‪ .‬ﺛﻢ ﻭ ﱢﺿﺢ‬
‫ﻟﻠﻄـﻼﺏ ﺃﻧﻪ ﻳﻤﻜﻨﻬﻢ ﺇﳚﺎﺩ ﻣﻘﺪﺍﺭ ﳏﺼﻠﺔ ﺃﻭﻝ ﻣ ﹼﺘﺠﻬﲔ‬ ‫ﺍﻟﺰﺍﻭﻳﺔ ﻧﻔﺴﻬﺎ ﻭﻟﻜﻦ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﻀﻠﻌﲔ ﺍﻵﺧﺮﻳﻦ ﻟﻠﻤﺜﻠﺚ‪ .‬ﺗﺄﻛﺪ ﻣﻦ ﺃﻥ ﺧﻄﻮﺍﲥﲈ ﻣﺘﺴﺎﻭﻳﺔ ﺃﺛﻨﺎﺀ‬

‫ﺑﺎﺳـﺘﻌﲈﻝ ﻧﻈﺮ ﹼﻳـﺔ ﻓﻴﺜﺎﻏـﻮﺭﺱ ﺃﻭ ﻗﺎﻧﻮﻥ ﺟﻴـﻮﺏ ﺍﻟﺘﲈﻡ‬ ‫ﺍﳌﴚ‪ .‬ﺳﻴﺼﻞ ﺃﻭﻻﹰ ﺍﻟﻄﺎﻟﺐ ﺍﻟﺬﻱ ﻳﺴﲑ ﻋﲆ ﺍﻟﻮﺗﺮ‪ 1 .‬ر‪‬‬
‫ﻭﻟﻜﻦ ﻻ ﻳﻤﻜﻨﻬﻢ ﺇﳚﺎﺩ ﺍﲡﺎﻫﻬﺎ‪2 .‬‬

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