Down she goes!
A thin light inextensible frictionless cord of length l wearing a small bead is tied between two nails that are in the same level at a distance 0.5l apart. Initially the bead is held close to a nail and released. Find the speed of the bead immediately after the cord becomes taut. If your answer is v, report 4v. Take g=10 m/s^2 and l=1/6 m
The answer is 2.
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1 solution
Initially the string was slack.So the bead fell straight,letting it fall by x ,the length becomes 2 a ,when x + a 2 − x 2 = 2 a ,yielding x = 4 3 a .Energy conservation yields v = 2 g ( 3 a / 4 ) = 1 . 5 a g .Hence its acted suddenly by two impulsive forces which are equal as its the same string attached on both sides of the bead.Thereby the speed along the resultant of these forces(which is the bisector of them)becomes zero.The speed of the bead just after is v ′ = v sin 2 6 . 5 = v / √ 5 = 2