If block A is moving horizontally with velocity vA, then find the velocity of block B at the instant shown in figure.
A
hvA2√x2+h2
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B
xvA√x2+h2
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C
xvA2√x2+h2
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D
hvA√x2+h2
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Solution
The correct option is CxvA2√x2+h2
Let the vertical distance of block B from the ground be y.
Hence, total length of string L is: L=2h−2y+√x2+h2
Differentiating the equation w.r.t time, we get: 0=0−2dydt+12√x2+h2×2x×dxdt
(since h is constant) ⇒dydt=x2√x2+h2dxdt....(i)
But dydt=vB&dxdt=vA....(ii)
because vB= time rate of change of vertical distance y between ground and block B vA= time rate of change of horizontal distance x between block A and block B
From Eq. (i) and Eq. (ii): vB=xvA2√x2+h2
Hence option (c) is correct.