Question 59 of 183

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

A$20m{s}^{-1}$B$\sqrt{10}m{s}^{-1}$C$2\sqrt{10}m{s}^{-1}$D$10m{s}^{-1}$

Show answer and solution▾

Answer: (B) $\sqrt{10}m{s}^{-1}$

Applying conservation of mechanical energy between point $A$ & $B$, we get
$K{E}_{A}+{U}_{A}=K{E}_{B}+{U}_{B}$
$\Rightarrow 0+mg(1)=\dfrac{1}{2}m{v}^{2}+mg\times 0.5$
$\Rightarrow v=\sqrt{g}=\sqrt{10}m{s}^{-1}$