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 $g = 10 m s^{- 2}$ )

8.5 m / s

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5 \sqrt{2} 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 A $8.5 m / s$
R.E.F. Image.
fall in height = 4 - 1.65
= 2.35 m
So gain in K.E
will be loss in PE
$\Rightarrow K E_{2} = K E_{1}$ + $Δ$ PE
$K . E_{2} = K, E_{1} + m g h$
$K, E_{2} = \frac{1}{2} m v_{1}^{2} + m g h$
$\frac{1}{2} m v_{2}^{2} = \frac{1}{2} m v_{1}^{2} + m g h$ $\Rightarrow v_{2}^{2} = v_{1}^{2} + 29 h$
$\Rightarrow v_{2} = \sqrt{v_{1}^{2} + 29 h} = \sqrt{25 + 2 \times 10 \times 2.35} = 8.48 m / s$
wire was frictionless so no loss against friction.

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