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(i)Find the acceleration of the centre of mass of two particle approaching towards each other under their own grabitational field.(ii)A boy of mass 30 kg is standing on a flat boat so that he is 20 meter from the shore. He walks 8 m on the boat towards the shore and then stops. The mass of the boat is 90 kg and friction between the boat and the water surface is negligible. How far is the boy from the shore now ? |
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Answer» Solution :(i)`a_(cm)=(|m_(1)a_(1)-m_(2)a_(2)|)/(m_(1)+m_(2))=(|F_(1)-F_(2)|)/(m_(1)+m_(2))` Since `F_(1)` & `F_(2)`are equal ln magnitude and opposite in direction, `a_(cm)=0` (ii)Let the boy move a distance `xx` towards left w.r.t ground, the DISPLACEMENT of plank `= 8-x` towards right. As shift in C.M. = 0 due to `F_(axt)=0`. `RARR 0 = (-30x + 90(8-x))/(30+90)` `rArr x = 6 m` `therefore` Required ANSWER is `(20-6)m=14 m` |
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