A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 $m s^{- 1}$ . Find the velocity of bead at point R (take g = 10 $m s^{- 2}$ )

7 m/s

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\sqrt{105}

m/s

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6 \sqrt{2}

m/s

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6 m/s

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Solution

The correct option is C $6 \sqrt{2}$ m/s
$\begin{matrix} Given-* Initial velocity of the Bead v_{0} = 5 m / s By conservation of energy- Total Energy at point p = Total Energy at point R \Rightarrow p E_{p} + k E_{p} = p E_{R} + k E_{R} \Rightarrow m g h_{p} + \frac{1}{2} m v_{0}^{2} = m g h_{R} + \frac{1}{2} m v_{R}^{2} \end{matrix}$
$\begin{matrix} \Rightarrow g \cdot (4) + \frac{v_{0}^{2}}{2} = g \cdot (1 \cdot 65) + \frac{v_{R}^{2}}{2} \Rightarrow v_{R}^{2} = 72 \Rightarrow v_{R} = \sqrt{72} m / 8 = 6 \sqrt{2} m / s Hence, option (c) is correct. \end{matrix}$

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A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 ms−1. Find the velocity of bead at point R (take g = 10 ms−2)

The correct option is C 6√2 m/s Given-* Initial velocity of the Bead v0=5 m/s By conservation of energy- Total Energy at point p= Total Energy at point R⇒pEp+kEp=pER+kER⇒mghp+12mv20=mghR+12mv2R⇒g⋅(4)+v202=g⋅(1⋅65)+v2R2⇒v2R=72⇒vR=√72m/8=6√2 m/s Hence, option (c) is correct.

A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 $m s^{- 1}$ . Find the velocity of bead at point R (take g = 10 $m s^{- 2}$ )