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A running man has half the KE that a body of half his mass has. The man speeds up by `1.0ms^(-1)` and then has the same energy as the boy. What were the original speeds of the man and the boy?A. `2.4, 4.8 ms^(-1)`B. `2.4, 3.4 ms^(-1)`C. `3.4, 4.8 ms^(-1)`D. `3.4, 6.8 ms^(-1)` |
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Answer» Let M = mass of man , m= mass of boy V = speed of man , v = speed of boy Given `(1)/(2) MV^(2) = (1)/(2) ((1)/(2)mv^(2))`, As `m = (M)/(2)` So `(1)/(2) MV^(2) = (1)/(2) ((1)/(2) xx (M)/(2) mv^(2))` Hence `v^(2) = 4V^(2)` or `v = 2V` When the man speeds up by `1 m//s`, then we get `(1)/(2) M (V + 1)^(2) = (1)/(2)mv^(2) = (1)/(2) (M)/(2) (4V^(2))` or `(V + 1)^(2) = 2V^(2)` or `V^(2) - 2V - 1 = 0` Solving we get `V = 2.4 ms^(-1)` and` v = 2V = 4.8 ms^(-1)` |
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