InterviewSolution
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InterviewSolution
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Two equal masses of `6.40 kg` are separted by a distance of `0.16 m`. A small body is released from a point `P`, equidistant from the two masses and at a distance of `0.06 m` from the line joining them. Fig. (a) Calculate the velocity of this body when it passes through `Q`. (b) Calculate the acceleration of this body at `P` and `Q` if its mass is `0.1 kg`. Use `G = 6.67 xx 10^(-11) N m^(2) kg^(-2)` |
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Answer» In figure, `AP = BP = sqrt((0.08)^(2) + (0.06)^(2)) = 0.1 m` Here, `m_(1) = 6.4 kg = m_(2)` and `m_(3) = 0.1 kg` (a) When body falls from `P` to `Q`, then gain in `KE` of body is equal to loss in `PE` of the system. `:. (1)/(2) m_(3) upsilon^(2) = - [(Gm_(1)m_(3))/((0.1)) + (Gm_(2) m_(3))/(0.1)]` `+ [(Gm_(1)m_(3))/(0.08) + (Gm_(2) m_(3))/(0.08)]` or `(1)/(2) upsilon^(2) = - [(Gm_(1))/(0.1) + (Gm_(2) )/(0.1)] + [(Gm_(1))/(0.08) + (Gm_(2) )/(0.08)]` or `upsilon^(2) = - 2 xx G [((6.4)/(0.1) + (6.4)/(0.1)) - ((6.4)/(0.08) + (6.4)/(0.08))]` `= - 2 xx (6.67 xx 10^(-11)) xx (-32)` `= 64 xx 6.67 xx 10^(-11)` or `upsilon = (64 xx 6.67 xx 10^(-11))^(1//2) = 6.53 xx 10^(-5) ms^(-1)` |
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