λ™μ‹œ 좜발, λ™μ‹œ 도달과 같은 ν•œμ •λœ μƒν™©μ—μ„œλ§Œ μ“Έ 수 μžˆλ‹€λŠ” μ μ—μ„œ κ·Έλ ‡κ²Œ μœ μš©ν•œ 풀이법은 μ•„λ‹Œκ±° κ°™μ§€λ§Œ,

μ΅œκ·Όμ— λ™μ‹œ λ°œμ‚¬, λ™μ‹œ 도달을 많이 물어보기도 ν•˜κ³ 

200919와 같이 거의 λŒ€λ†“κ³  μƒλŒ€ 속도λ₯Ό λ¬Όμ–΄λ³Έ κ²½μš°λ„ μžˆμœΌλ‹ˆ μ–Έμ  κ°„ μ“Έ 수 μžˆμ„κ±° κ°™μ•„μ„œ 써봄

200919λŠ” λ„ˆλ¬΄ 유λͺ…ν•œ μ˜ˆμ‹œλ‹ˆκΉŒ μŠ€ν‚΅ν•¨


220619

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A에 λŒ€ν•œ B의 μƒλŒ€ μ†λ„λŠ” v + V

A의 xλ°©ν–₯ μš΄λ™μ„ μ‚΄νŽ΄λ³΄λ©΄, 평균 속도 2v/5둜 8d/5만큼 μš΄λ™ν•œλ‹€.

λ”°λΌμ„œ 5d만큼 μ΄λ™ν•˜λ €λ©΄ 5v/4 만큼의 속λ ₯이 ν•„μš”ν•˜λ―€λ‘œ V = v/4


210920

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λ°œμ‚¬ 직후 A와 B의 μƒλŒ€ 속도λ₯Ό λ²‘ν„°λ‘œ ν‘œν˜„ν•˜λ©΄

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λ™μΌν•œ μ‹œκ°„ λ™μ•ˆ λ™μΌν•œ κ°€μ†λ„λ‘œ μš΄λ™ν•˜λ―€λ‘œ A와 B의 μƒλŒ€μ†λ„λŠ” μΌμ •ν•˜λ‹€.

λ”°λΌμ„œ rμ—μ„œ A와 B의 μƒλŒ€μ†λ„λŠ”

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각을 ν‘œμ‹œν•΄λ³΄λ©΄

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그런데 κ°€μ†λ„λŠ” yλ°©ν–₯으둜만 μž‘μš©ν•˜λ―€λ‘œ A와 B의 xλ°©ν–₯ μ†λ„λŠ” μΌμ •ν•˜λ‹€.

λ”°λΌμ„œ v1κ³Ό v2λŠ” μ•„λž˜μ™€ 같이 ν‘œν˜„ν•  수 μžˆλ‹€.

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γ„±. o


γ„΄. v_2/v_1 = vsin75Β°/vcos75Β° = tan75Β°

μ €λ²ˆμ—λ„ μ„€λͺ…ν–ˆμ§€λ§Œ tan75Β°λŠ” μ•„λž˜μ™€ 같이 λΉ λ₯΄κ²Œ ꡬ할 수 μžˆλ‹€.

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λ”°λΌμ„œ o


γ„·. 0.5m(v_1Β² + v_2Β²)

= 0.5mvΒ²(sinΒ² 15Β° + cosΒ² 15Β°)/(sinΒ² 60Β°)

= 2mvΒ²/3

o