Question

A block is released from a height $h$ on a track $AB$ as shown in the figure. From $C$ there is a circular track of radius $R=50\text{cm}$.

Assuming all the surfaces to be smooth, the minimum value of $h$ for which the block reaches the point $D$ is

• A. $1.25\text{m}$
• B. $2.5\text{m}$
• C. $3.5\text{m}$
• D. $4.25\text{m}$

Answer 1

The correct option is A $1.25\text{m}$

To reach the point $D$, the minimum velocity given to the block at point $C$ is $v_{min}=\sqrt{5gR}$

Since all the surfaces are smooth,
using law of conservation of energy
Potential energy at A = Kinetic energy at C
$\Rightarrow mgh=\frac{1}{2}m{v}^2$

$h=\frac{1}{2}\frac{(\sqrt{5gR}{)}^2}{g}=\frac{5gR}{2g}=\frac{5R}{2}=1.25\text{m}$ where $R=50\text{cm}=\frac{1}{2}\text{m}$