A particle is placed at the point A of a frictionless track ABC as shown in figure. It is pushed slightly towards right. Its speed when it reaches the point B is ( $g = 10 m s^{- 2}$ )

10 m/s

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\sqrt{10} m / s

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20 m s^{- 1}

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\sqrt{20} m s^{- 1}

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Solution

The correct option is B $\sqrt{10} m / s$
Potential energy of A is $m g h = m \times 10 \times 1 = 10 m$
Kinetic energy at $A = 0$
Mechanical energy at $A = 10 m + 0 = 10 m$
Potential energy of $B = m g h = m \times 10 \times 5 = 5 m$
Kinetic energy at B is $10 m - 5 m = 5 m$
Hence Velocity (for speed)
$\frac{1}{2} m v^{2} = 5 m$
$\frac{1}{2} v^{2} = 5$
$v^{2} = 5 \times 2$
$v = \sqrt{10}$

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