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A ball of mass `m` is released from A inside a smooth wedge of mass `m` as shown in figure. What is the speed of the wedge when the ball reaches point B? A. (a) `((gR)/(3sqrt2))^(1//2)`B. (b) `sqrt(2gR)`C. (c) `((5gR)/(2sqrt3))^(1//2)`D. (d) `sqrt(3/2gR)` |
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Answer» Correct Answer - A Let the velocity of wedge be v. Loss in PE=Gain in KE `mgR cos 45^@=1/2mv^2+1/2m(v_1cos45^@-v)^2+1/2m(v_1sin45^@)^2` From conservation of linear momentum `m(v_1cos45^@-v)=mv` Here `v_1` is the velocity of ball w.r.t. wedge, Solve to get `v=((gR)/(3sqrt2))^(1//2)` |
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