Applying conservation of mechanical energy at the point A and B, we get21mu2+0=21mv2+mgL
⇒v=u2−2gL
Now in vector form. vi=ui^ and vf=vj^
Therefore, change in velocity will be,ΔV=vj^−ui^ and ∣ΔV∣=u2+v2
⇒ΔV=u2+(u2−2gL)
=2(u2−gL)
Hence, x=2
A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x(u2−gL). The value of x is
Held on 27 Jun 2022 · Verified 6 Jul 2026.
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