يحي الكهالي
22-06-2014, 18:11
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻧﻲ :
ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻤﻴﺔ ﺍﻟﺘﺤﺮﻙ ﻟﺼﺎﺭﻭﺥ 5 × 810 ﻛﺠﻢ. ﻡ/ ﺙ،
ﻭﻛﺎﻧﺖ ﺳﺮﻋﺘﻪ ×5 310ﻡ/ ﺙ . ﻓﻤﺎ ﻛﺘﻠﺘﻪ؟
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻧﻲ :
∵ ﻛﺖ = ﻙ × ﻉ
∴ 5 × 810 = ﻙ× 5 × 310
ﻙ = = 510 ﻛﺠﻢ.
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻟـﺚ :
ﺇﺫﺍ ﻛﺎﻧﺖ ﺳﺮﻋﺔ ﺍﻹِﻓﻼﺕ ﻟﺼﺎﺭﻭﺥ ﻣﻦ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ ﻫﻲ
11.2 ﻛﻢ/ ﺙ ﻭﻛﺎﻧﺖ ﻋﺠﻠﺔ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ 9.8 ﻡ/ ﺙ.2
ﻓﻤﺎ ﻣﻘﺪﺍﺭ ﻧﺼﻒ ﻗﻄﺮ ﺍﻷﺭﺽ.
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻟﺚ :
ﻉ = 11.2 ﻛﻢ/ ﺙ = 11.2 × 310 ﻡ/ ﺙ ﺩ = 9.8 ﻡ/ ﺙ2 ﻧﻖ
= ؟
∵ ﻉ = 2 ﺩ ﻧﻖ
11.2 × 310 = 2 × 9.8×ﻧﻖ
ﺑﺘﺮﺑﻴﻊ ﺍﻟﻄﺮﻓﻴﻦ 125.44 × 610 = 19.6 ﻧﻖ
∴ ﻧﻖ = = 6.4 × 610 ﻡ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺮﺍﺑـﻊ :
ﻳﺘﺤﺮﻙ ﺟﺰﻱﺀ ﻏﺎﺯ ﺑﺴﺮﻋﺔ 150ﻡ/ ﺙ ﻧﺤﻮ ﺟﺰﻱﺀ ﻏﺎﺯ ﺁﺧﺮ
ﺳﺎﻛﻦ " ﻓﺮﺿﺎً " ﻭﻣﺴﺎﻭ ﻟﻪ ﻓﻲ ﺃﻟﻜﺘﻠﻪ ، ﻭﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﺗﺤﺮﻙ
ﺍﻟﺠﺰﻱﺀ ﺍﻷﻭﻝ ﻓﻲ ﺍﺗﺠﺎﻩ ﻳﺼﻨﻊ ﺯﺍﻭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 530 ﻣﻊ ﺧﻂ
ﺣﺮﻛﺘﻪ ﺍﻻﺑﺘﺪﺍﺋﻴﺔ ﻭﻣﺘﻌﺎﻣﺪﺍً ﻣﻊ ﺍﺗﺠﺎﻩ ﺣﺮﻛﺔ ﺍﻟﺠﺰﻱﺀ
ﺍﻟﺜﺎﻧﻲ . ﺃﺣﺴﺐ -:
ﺃ ) ﺳﺮﻋﺘﻪ ﻭﺇﺗﺠﺎﻫﻪ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ .
ﺏ ) ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ ﺍﻟﻤﻔﻘﻮﺩﺓ ﻧﺘﻴﺠﺔ ﺍﻟﺘﺼﺎﺩﻡ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺮﺍﺑﻊ :
ﺃ ) ﻙ=1 ﻙ2 ، ﻉ 150=1 , ﻉ2 = ﺻﻔﺮ
ﻫـ =1 530 ﻫـ=2 90 ـــ 30 = 560
ﻉَ =1 ؟ ﻉَ2 = ؟
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ :
∵ ﻉ1 + ﻉ 2 = ﻉَ1 ﺟﺘﺎ ﻫـ1 + ﻉَ 2 ﺟﺘﺎ ﻫـ2
∴ 150 + ﺻﻔﺮ = ﻉَ1 × + ﻉَ2 ×
150 = ﻉَ1 3 + ﻉَ2
2
300 = ﻉَ1 3 + ﻉَ2 1 )
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻟﺮﺃﺳﻲ :
∵ ﻉَ1 ﺟﺎ ﻫـ1 = ﻉَ2 ﺟﺎ ﻫـ2
∴ ﻉَ1 × = ﻉَ2 ×
∴ ﻉَ1 = ﻉَ 2 3 2 )
ﺑﺎﻟﺘﻌﻮﻳﺾ ﻣﻦ 2 ﻓﻲ 1
∴ 300 = ﻉَ2 3 × 3 + ﻉَ2
300 = 4 ﻉَ2 ﻉَ 2 = = 75ﻡ/ ﺙ
ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ 2 ﻋﻦ ﻗﻴﻤﺔ ﻉَ2
∴ ﻉَ2 = 75 3 ﻡ/ ﺙ
ﺏ ) ﻃﺎﺡ = ﻃﺎﺡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ـــــ ﻃﺎﺡ ﻗﺒﻞ ﺍﻟﺘﺼﺎﺩﻡ
ﻃﺎﺡ = ﻙ ﻉَ ـــــ [ ( ) + ( )]
= × 9 × (10 )2 ــــ [ ( ×3× (18 )2 + ( × 6 × (12 )2 ]
= 450 ــــ ( 486 + 432 )
= 450 ــــ 918 = ـــ 468 ﺟﻮﻝ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺨﺎﻣﺲ :
ﻋﺮﺑﺔ ﻛﺘﻠﺘﻬﺎ 5 ﻃﻦ ﺗﺘﺤﺮﻙ ﺑﺴﺮﻋﺔ 36 ﻛﻢ/ ﺳﺎﻋﺔ ﻓﻲ
ﺍﺗﺠﺎﻩ ﺍﻟﺸﺮﻕ ﺗﺼﺎﺩﻣﺖ ﻣﻊ ﻋﺮﺑﻪ ﺃﺧﺮﻯ ﻛﺘﻠﺘﻬﺎ 4 ﻃﻦ ﺗﺘﺤﺮﻙ
ﺑﺴﺮﻋﺔ 72 ﻛﻢ/ﺳﺎﻋﺔ ﻓﻲ ﺍﺗﺠﺎﻩ ﺍﻟﺸﻤﺎﻝ ، ﺇﺫﺍ ﺍﻟﺘﺼﻘﺖ
ﺍﻟﻌﺮﺑﺘﺎﻥ ﻭﺗﺤﺮﻛﺘﺎ ﻣﻌﺎ ﻛﺤﻄﺎﻡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﻓﺄﺣﺴﺐ ﻣﺎ
ﻳﻠﻲ -:
ﺃ ) ﺍﻟﺴﺮﻋﺔ ﺍﻟﺘﻲ ﻳﺘﺤﺮﻙ ﺑﻬﺎ ﺍﻟﺤﻄﺎﻡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﻣﺒﺎﺷﺮﺓ .
ﺏ ) ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﺗﺼﻨﻌﻬﺎ ﻣﻊ ﺍﺗﺠﺎﻩ ﺍﻟﺸﺮﻕ .
ﺝ ) ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ ﺍﻟﻤﻔﻘﻮﺩﺓ ﺃﺛﻨﺎﺀ ﺍﻟﺘﺼﺎﺩﻡ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺨﺎﻣﺲ :
ﻙ1 = 5ﻃﻦ = 5 × 1000 = 5000 ﻛﺠﻢ ﻉ 36=1ﻛﻢ/ ﺱ
36 × = 10 ﻡ/ ﺙ
ﻙ2 = 4ﻃﻦ = 4 × 1000 = 4000 ﻛﺠﻢ ﻉ =2 72 ﻛﻢ/ ﺱ
= 72 × = 20 ﻡ/ ﺙ
ﺃ ) ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ :
∵ ﻙ1 ﻉ1 + ﻙ2ﻉ2 = ( ﻙ1 + ﻙ2 ) ﻋَﺲ
∴ 5000 × 10 + ﺻﻔﺮ = 9000 ﻋَﺲ
ﻋَﺲ = = 5.6 ﻡ/ ﺙ
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻟﺮﺃﺳﻲ :
∵ ﻙ1 ﻉ1 + ﻙ2ﻉ2 = ( ﻙ1 + ﻙ2 ) ﻋَﺺ
∴ ﺻﻔﺮ + 4000 × 20 = 9000 ﻋَﺺ
ﻋَﺺ= = 8.9 ﻡ/ ﺙ
∵ ﻉَ = ﻉَ + ﻉَ
ﻉَ = ( 5.6 )2 + ( 8.9 )2 = 10.5 ﻡ/ ﺙ
ﺏ ) ∵ ﻋَﺲ = ﻉَ ﺟﺘﺎﻫـ
∴ 5.6 = 10.5 × ﺟﺘﺎﻫـ
∴ ﺟﺘﺎﻫـ = 5.6 = 0.53 ∴ ﻫـ = 585
10.5
ﺝ ) ﺟـ ﻃﺎﺡ = ﻃﺎﺡ ﺑﻌﺪ – ﻃﺎﺡ ﻗﺒﻞ
= ×9000× ( 10.5 )2 - [ ×5000× (10 )2 + ×4000× ( 20 )
2 ]
= 496.125 – 1050000
= - 1.049.503.875 ﺟﻮﻝ
ﺍﻹﺷﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﺗﻌﻨﻲ ﺃﻥ ﻫﻨﺎﻙ ﻓﻘﺪ ﻓﻲ ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ .
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺩﺱ -:
ﺃﺣﺴﺐ ﺍﻻﺭﺗﻔﺎﻉ ﻓﻮﻕ ﺳﻄﺢ ﺍﻷﺭﺽ ﻟﻘﻤﺮ ﺻﻨﺎﻋﻲ ﻳﺘﺤﺮﻙ
ﻓﻲ ﻣﺴﺎﺭ ﺩﺍﺋﺮﻱ ﺑﺴﺮﻋﺔ ﻣﺪﺍﺭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 4 ﻛﻢ/ ﺙ.
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺩﺱ -:
ﻉ = 4ﻛﻢ/ ﺙ = 4 × 310 ﻡ/ ﺙ = 6400ﻛﻢ = 510×64ﻡ
ﻝ = ؟؟
∵ ﻉ = ﺝ ﻙ ﻧﻖ
∴ 4 × 310 = 6.67 × × 6 ×
ﻧﻘﻤﺪﺍﺭ
∴ 16 × 610 = 40.02 ×
ﻧﻘﻤﺪﺍﺭ
ﻧﻘﻤﺪﺍﺭ = 40.02 × 1310 = 2.5 × 710 ﻡ
16 × 610
∵ ﻧﻘﻤﺪﺍﺭ = ﻝ + ﻧﻖ ﺭ
∴ 2.5 × 710 = ﻝ + 64 × 510
250 × 510 = ﻝ + 64 × 510
ﻝ = 250 × 510 – 64 × 510
∴ ﻝ = 186 × 510ﻡ = 18600 ﻛﻢ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺑﻊ -:
1 ) ﺃﻃﻠﻊ ﻣﺪﻓﻊ ﻗﺬﻳﻔﻪ ﺑﺴﺮﻋﺔ ﺇﺑﺘﺪﺍﺋﻴﺔ ﻣﻘﺪﺍﺭﻫﺎ 200 2
ﻡ/ ﺙ ﺑﺈﺗﺠﺎﻩ ﻳﺼﻨﻊ ﺯﺍﻭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 545 ﻣﻊ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ
ﺑﺈﻫﻤﺎﻝ ﻣﻘﺎﻭﻣﺔ ﺍﻟﻬﻮﺍﺀ ﻭﺍﻋﺘﺒﺎﺭ ﻋﺠﻠﺔ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ 10
ﻡ/ ﺙ2 ﺃﻭﺟﺪ :
ﺃ ) ﺫﺭﻭﺓ ﺍﻟﻘﺬﻑ ﺏ ) ﺍﻟﻤﺪﻯ ﺍﻷﻓﻘﻲ
ﺝ ) ﺳﺮﻋﺔ ﺍﻟﻘﺬﻳﻔﺔ ﺍﻟﻜﻠﻴﺔ ﺑﻌﺪ ﻣﺮﻭﺭ 35 ﺛﺎﻧﻴﺔ ﻣﻦ ﻟﺤﻈﺔ
ﺍﻟﻘﺬﻑ .
ﺩ ) ﺇﺭﺗﻔﺎﻉ ﺍﻟﻘﺬﻳﻔﺔ ﺑﻌﺪ ﻣﺮﻭﺭ 35 ﺛﺎﻧﻴﺔ ﻣﻦ ﻟﺤﻈﺔ ﺍﻟﻘﺬﻑ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺑﻊ -:
= 200 2 ﻫـ = 45
ﺃ ) ∵ = ( ﻉ ﺟﺎﻫـ ) 2 2- ﺩ ﻓﺺ
ﺻﻔﺮ = ( 200 2 × ) 2 2- × 10 ﻓﺺ
20 ﻓﺺ = 40000
ﻓﺺ = 40000 = 2000 ﻡ
20
ﺏ ) =
ﻉ ﺟﺎ ﻫـ
5
= 200 2 ×
5
∵ ﻑ ﺱ = ﻉ – ﺟﺘﺎ ﻫـ ﺯ
∴ ﻑ ﺱ = 200 2 × × 40
ﻑ ﺱ = 8000 ﻡ
ﺝ ) = ﺟﺘﺎﻫـ = 200 2 × = 200 ﻡ/ ﺙ
= ﺟﺎﻫـ - ﺩ ﺯ = 200 2 × - 10 × 35 = - 150 ﻡ/ ﺙ
ﻉ = + ﻉ = ( 200 ) 2 + ( 150- ) 2
ﻉ = 250 ﻡ/ ﺙ
ﺩ ) = ﺟﺎﻫـ ﺯ - ﺩ ﺯ 2
= 200 2 × × 35 - × 10 × ( 35 ) 2 = 875 ﻡ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻣﻦ -:
ﻣﻘﺬﻭﻑ ﺫﺭﻭﺓ ﻗﺬﻓﻪ 40 ﻣﺘﺮ ﻭﻣﺪﺍﻩ ﺍﻷﻓﻘﻲ 160 3 ﻣﺘﺮ
ﺃﺣﺴﺐ -:
ﺃ ) ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﻗﺬﻑ ﺑﻬﺎ .
ﺏ ) ﺳﺮﻋﺘﻪ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻣﻦ -:
= 40 = 160 3
ﺃ ) ∵ = ( ﺟﺎﻫـ ) 2 – 2 ﺩ
∴ ﺻﻔﺮ = ( ﺟﺎﻫـ ) 2 – 2 × 10 × 40
800 = ( ﺟﺎﻫـ ) 2 1 )
∵ = ﺟﺘﺎﻫـ
، ∵ = ﺟﺎﻫـ
5
∴ =
ﺟﺘﺎﻫـ × ﺟﺎﻫـ
5
160 3 =
800 3 = ﺟﺎﻫـ ﺟﺘﺎﻫـ 2 )
ﺑﻘﺴﻤﺔ 1 ) ﻋﻠﻰ 2 )
∴ =
∴ ﻫـ = 530
ﺏ ) ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ 1 ) 800 = ( × ) 2
= 800 × 4 = 3200
= 3200 = 56.6 ﻡ/ ﺙ
ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻤﻴﺔ ﺍﻟﺘﺤﺮﻙ ﻟﺼﺎﺭﻭﺥ 5 × 810 ﻛﺠﻢ. ﻡ/ ﺙ،
ﻭﻛﺎﻧﺖ ﺳﺮﻋﺘﻪ ×5 310ﻡ/ ﺙ . ﻓﻤﺎ ﻛﺘﻠﺘﻪ؟
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻧﻲ :
∵ ﻛﺖ = ﻙ × ﻉ
∴ 5 × 810 = ﻙ× 5 × 310
ﻙ = = 510 ﻛﺠﻢ.
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻟـﺚ :
ﺇﺫﺍ ﻛﺎﻧﺖ ﺳﺮﻋﺔ ﺍﻹِﻓﻼﺕ ﻟﺼﺎﺭﻭﺥ ﻣﻦ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ ﻫﻲ
11.2 ﻛﻢ/ ﺙ ﻭﻛﺎﻧﺖ ﻋﺠﻠﺔ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ 9.8 ﻡ/ ﺙ.2
ﻓﻤﺎ ﻣﻘﺪﺍﺭ ﻧﺼﻒ ﻗﻄﺮ ﺍﻷﺭﺽ.
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻟﺚ :
ﻉ = 11.2 ﻛﻢ/ ﺙ = 11.2 × 310 ﻡ/ ﺙ ﺩ = 9.8 ﻡ/ ﺙ2 ﻧﻖ
= ؟
∵ ﻉ = 2 ﺩ ﻧﻖ
11.2 × 310 = 2 × 9.8×ﻧﻖ
ﺑﺘﺮﺑﻴﻊ ﺍﻟﻄﺮﻓﻴﻦ 125.44 × 610 = 19.6 ﻧﻖ
∴ ﻧﻖ = = 6.4 × 610 ﻡ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺮﺍﺑـﻊ :
ﻳﺘﺤﺮﻙ ﺟﺰﻱﺀ ﻏﺎﺯ ﺑﺴﺮﻋﺔ 150ﻡ/ ﺙ ﻧﺤﻮ ﺟﺰﻱﺀ ﻏﺎﺯ ﺁﺧﺮ
ﺳﺎﻛﻦ " ﻓﺮﺿﺎً " ﻭﻣﺴﺎﻭ ﻟﻪ ﻓﻲ ﺃﻟﻜﺘﻠﻪ ، ﻭﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﺗﺤﺮﻙ
ﺍﻟﺠﺰﻱﺀ ﺍﻷﻭﻝ ﻓﻲ ﺍﺗﺠﺎﻩ ﻳﺼﻨﻊ ﺯﺍﻭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 530 ﻣﻊ ﺧﻂ
ﺣﺮﻛﺘﻪ ﺍﻻﺑﺘﺪﺍﺋﻴﺔ ﻭﻣﺘﻌﺎﻣﺪﺍً ﻣﻊ ﺍﺗﺠﺎﻩ ﺣﺮﻛﺔ ﺍﻟﺠﺰﻱﺀ
ﺍﻟﺜﺎﻧﻲ . ﺃﺣﺴﺐ -:
ﺃ ) ﺳﺮﻋﺘﻪ ﻭﺇﺗﺠﺎﻫﻪ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ .
ﺏ ) ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ ﺍﻟﻤﻔﻘﻮﺩﺓ ﻧﺘﻴﺠﺔ ﺍﻟﺘﺼﺎﺩﻡ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺮﺍﺑﻊ :
ﺃ ) ﻙ=1 ﻙ2 ، ﻉ 150=1 , ﻉ2 = ﺻﻔﺮ
ﻫـ =1 530 ﻫـ=2 90 ـــ 30 = 560
ﻉَ =1 ؟ ﻉَ2 = ؟
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ :
∵ ﻉ1 + ﻉ 2 = ﻉَ1 ﺟﺘﺎ ﻫـ1 + ﻉَ 2 ﺟﺘﺎ ﻫـ2
∴ 150 + ﺻﻔﺮ = ﻉَ1 × + ﻉَ2 ×
150 = ﻉَ1 3 + ﻉَ2
2
300 = ﻉَ1 3 + ﻉَ2 1 )
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻟﺮﺃﺳﻲ :
∵ ﻉَ1 ﺟﺎ ﻫـ1 = ﻉَ2 ﺟﺎ ﻫـ2
∴ ﻉَ1 × = ﻉَ2 ×
∴ ﻉَ1 = ﻉَ 2 3 2 )
ﺑﺎﻟﺘﻌﻮﻳﺾ ﻣﻦ 2 ﻓﻲ 1
∴ 300 = ﻉَ2 3 × 3 + ﻉَ2
300 = 4 ﻉَ2 ﻉَ 2 = = 75ﻡ/ ﺙ
ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ 2 ﻋﻦ ﻗﻴﻤﺔ ﻉَ2
∴ ﻉَ2 = 75 3 ﻡ/ ﺙ
ﺏ ) ﻃﺎﺡ = ﻃﺎﺡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ـــــ ﻃﺎﺡ ﻗﺒﻞ ﺍﻟﺘﺼﺎﺩﻡ
ﻃﺎﺡ = ﻙ ﻉَ ـــــ [ ( ) + ( )]
= × 9 × (10 )2 ــــ [ ( ×3× (18 )2 + ( × 6 × (12 )2 ]
= 450 ــــ ( 486 + 432 )
= 450 ــــ 918 = ـــ 468 ﺟﻮﻝ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺨﺎﻣﺲ :
ﻋﺮﺑﺔ ﻛﺘﻠﺘﻬﺎ 5 ﻃﻦ ﺗﺘﺤﺮﻙ ﺑﺴﺮﻋﺔ 36 ﻛﻢ/ ﺳﺎﻋﺔ ﻓﻲ
ﺍﺗﺠﺎﻩ ﺍﻟﺸﺮﻕ ﺗﺼﺎﺩﻣﺖ ﻣﻊ ﻋﺮﺑﻪ ﺃﺧﺮﻯ ﻛﺘﻠﺘﻬﺎ 4 ﻃﻦ ﺗﺘﺤﺮﻙ
ﺑﺴﺮﻋﺔ 72 ﻛﻢ/ﺳﺎﻋﺔ ﻓﻲ ﺍﺗﺠﺎﻩ ﺍﻟﺸﻤﺎﻝ ، ﺇﺫﺍ ﺍﻟﺘﺼﻘﺖ
ﺍﻟﻌﺮﺑﺘﺎﻥ ﻭﺗﺤﺮﻛﺘﺎ ﻣﻌﺎ ﻛﺤﻄﺎﻡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﻓﺄﺣﺴﺐ ﻣﺎ
ﻳﻠﻲ -:
ﺃ ) ﺍﻟﺴﺮﻋﺔ ﺍﻟﺘﻲ ﻳﺘﺤﺮﻙ ﺑﻬﺎ ﺍﻟﺤﻄﺎﻡ ﺑﻌﺪ ﺍﻟﺘﺼﺎﺩﻡ ﻣﺒﺎﺷﺮﺓ .
ﺏ ) ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﺗﺼﻨﻌﻬﺎ ﻣﻊ ﺍﺗﺠﺎﻩ ﺍﻟﺸﺮﻕ .
ﺝ ) ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ ﺍﻟﻤﻔﻘﻮﺩﺓ ﺃﺛﻨﺎﺀ ﺍﻟﺘﺼﺎﺩﻡ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺨﺎﻣﺲ :
ﻙ1 = 5ﻃﻦ = 5 × 1000 = 5000 ﻛﺠﻢ ﻉ 36=1ﻛﻢ/ ﺱ
36 × = 10 ﻡ/ ﺙ
ﻙ2 = 4ﻃﻦ = 4 × 1000 = 4000 ﻛﺠﻢ ﻉ =2 72 ﻛﻢ/ ﺱ
= 72 × = 20 ﻡ/ ﺙ
ﺃ ) ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ :
∵ ﻙ1 ﻉ1 + ﻙ2ﻉ2 = ( ﻙ1 + ﻙ2 ) ﻋَﺲ
∴ 5000 × 10 + ﺻﻔﺮ = 9000 ﻋَﺲ
ﻋَﺲ = = 5.6 ﻡ/ ﺙ
ﺍﻟﺤﺮﻛﺔ ﻓﻲ ﺍﻻﺗﺠﺎﻩ ﺍﻟﺮﺃﺳﻲ :
∵ ﻙ1 ﻉ1 + ﻙ2ﻉ2 = ( ﻙ1 + ﻙ2 ) ﻋَﺺ
∴ ﺻﻔﺮ + 4000 × 20 = 9000 ﻋَﺺ
ﻋَﺺ= = 8.9 ﻡ/ ﺙ
∵ ﻉَ = ﻉَ + ﻉَ
ﻉَ = ( 5.6 )2 + ( 8.9 )2 = 10.5 ﻡ/ ﺙ
ﺏ ) ∵ ﻋَﺲ = ﻉَ ﺟﺘﺎﻫـ
∴ 5.6 = 10.5 × ﺟﺘﺎﻫـ
∴ ﺟﺘﺎﻫـ = 5.6 = 0.53 ∴ ﻫـ = 585
10.5
ﺝ ) ﺟـ ﻃﺎﺡ = ﻃﺎﺡ ﺑﻌﺪ – ﻃﺎﺡ ﻗﺒﻞ
= ×9000× ( 10.5 )2 - [ ×5000× (10 )2 + ×4000× ( 20 )
2 ]
= 496.125 – 1050000
= - 1.049.503.875 ﺟﻮﻝ
ﺍﻹﺷﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﺗﻌﻨﻲ ﺃﻥ ﻫﻨﺎﻙ ﻓﻘﺪ ﻓﻲ ﺍﻟﻄﺎﻗﺔ ﺍﻟﺤﺮﻛﻴﺔ .
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺩﺱ -:
ﺃﺣﺴﺐ ﺍﻻﺭﺗﻔﺎﻉ ﻓﻮﻕ ﺳﻄﺢ ﺍﻷﺭﺽ ﻟﻘﻤﺮ ﺻﻨﺎﻋﻲ ﻳﺘﺤﺮﻙ
ﻓﻲ ﻣﺴﺎﺭ ﺩﺍﺋﺮﻱ ﺑﺴﺮﻋﺔ ﻣﺪﺍﺭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 4 ﻛﻢ/ ﺙ.
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺩﺱ -:
ﻉ = 4ﻛﻢ/ ﺙ = 4 × 310 ﻡ/ ﺙ = 6400ﻛﻢ = 510×64ﻡ
ﻝ = ؟؟
∵ ﻉ = ﺝ ﻙ ﻧﻖ
∴ 4 × 310 = 6.67 × × 6 ×
ﻧﻘﻤﺪﺍﺭ
∴ 16 × 610 = 40.02 ×
ﻧﻘﻤﺪﺍﺭ
ﻧﻘﻤﺪﺍﺭ = 40.02 × 1310 = 2.5 × 710 ﻡ
16 × 610
∵ ﻧﻘﻤﺪﺍﺭ = ﻝ + ﻧﻖ ﺭ
∴ 2.5 × 710 = ﻝ + 64 × 510
250 × 510 = ﻝ + 64 × 510
ﻝ = 250 × 510 – 64 × 510
∴ ﻝ = 186 × 510ﻡ = 18600 ﻛﻢ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺑﻊ -:
1 ) ﺃﻃﻠﻊ ﻣﺪﻓﻊ ﻗﺬﻳﻔﻪ ﺑﺴﺮﻋﺔ ﺇﺑﺘﺪﺍﺋﻴﺔ ﻣﻘﺪﺍﺭﻫﺎ 200 2
ﻡ/ ﺙ ﺑﺈﺗﺠﺎﻩ ﻳﺼﻨﻊ ﺯﺍﻭﻳﺔ ﻣﻘﺪﺍﺭﻫﺎ 545 ﻣﻊ ﺍﻻﺗﺠﺎﻩ ﺍﻷﻓﻘﻲ
ﺑﺈﻫﻤﺎﻝ ﻣﻘﺎﻭﻣﺔ ﺍﻟﻬﻮﺍﺀ ﻭﺍﻋﺘﺒﺎﺭ ﻋﺠﻠﺔ ﺍﻟﺠﺎﺫﺑﻴﺔ ﺍﻷﺭﺿﻴﺔ 10
ﻡ/ ﺙ2 ﺃﻭﺟﺪ :
ﺃ ) ﺫﺭﻭﺓ ﺍﻟﻘﺬﻑ ﺏ ) ﺍﻟﻤﺪﻯ ﺍﻷﻓﻘﻲ
ﺝ ) ﺳﺮﻋﺔ ﺍﻟﻘﺬﻳﻔﺔ ﺍﻟﻜﻠﻴﺔ ﺑﻌﺪ ﻣﺮﻭﺭ 35 ﺛﺎﻧﻴﺔ ﻣﻦ ﻟﺤﻈﺔ
ﺍﻟﻘﺬﻑ .
ﺩ ) ﺇﺭﺗﻔﺎﻉ ﺍﻟﻘﺬﻳﻔﺔ ﺑﻌﺪ ﻣﺮﻭﺭ 35 ﺛﺎﻧﻴﺔ ﻣﻦ ﻟﺤﻈﺔ ﺍﻟﻘﺬﻑ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺑﻊ -:
= 200 2 ﻫـ = 45
ﺃ ) ∵ = ( ﻉ ﺟﺎﻫـ ) 2 2- ﺩ ﻓﺺ
ﺻﻔﺮ = ( 200 2 × ) 2 2- × 10 ﻓﺺ
20 ﻓﺺ = 40000
ﻓﺺ = 40000 = 2000 ﻡ
20
ﺏ ) =
ﻉ ﺟﺎ ﻫـ
5
= 200 2 ×
5
∵ ﻑ ﺱ = ﻉ – ﺟﺘﺎ ﻫـ ﺯ
∴ ﻑ ﺱ = 200 2 × × 40
ﻑ ﺱ = 8000 ﻡ
ﺝ ) = ﺟﺘﺎﻫـ = 200 2 × = 200 ﻡ/ ﺙ
= ﺟﺎﻫـ - ﺩ ﺯ = 200 2 × - 10 × 35 = - 150 ﻡ/ ﺙ
ﻉ = + ﻉ = ( 200 ) 2 + ( 150- ) 2
ﻉ = 250 ﻡ/ ﺙ
ﺩ ) = ﺟﺎﻫـ ﺯ - ﺩ ﺯ 2
= 200 2 × × 35 - × 10 × ( 35 ) 2 = 875 ﻡ
ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻣﻦ -:
ﻣﻘﺬﻭﻑ ﺫﺭﻭﺓ ﻗﺬﻓﻪ 40 ﻣﺘﺮ ﻭﻣﺪﺍﻩ ﺍﻷﻓﻘﻲ 160 3 ﻣﺘﺮ
ﺃﺣﺴﺐ -:
ﺃ ) ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﻗﺬﻑ ﺑﻬﺎ .
ﺏ ) ﺳﺮﻋﺘﻪ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ .
ﺇﺟﺎﺑﺔ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺜﺎﻣﻦ -:
= 40 = 160 3
ﺃ ) ∵ = ( ﺟﺎﻫـ ) 2 – 2 ﺩ
∴ ﺻﻔﺮ = ( ﺟﺎﻫـ ) 2 – 2 × 10 × 40
800 = ( ﺟﺎﻫـ ) 2 1 )
∵ = ﺟﺘﺎﻫـ
، ∵ = ﺟﺎﻫـ
5
∴ =
ﺟﺘﺎﻫـ × ﺟﺎﻫـ
5
160 3 =
800 3 = ﺟﺎﻫـ ﺟﺘﺎﻫـ 2 )
ﺑﻘﺴﻤﺔ 1 ) ﻋﻠﻰ 2 )
∴ =
∴ ﻫـ = 530
ﺏ ) ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ 1 ) 800 = ( × ) 2
= 800 × 4 = 3200
= 3200 = 56.6 ﻡ/ ﺙ