Question

A stationary ball of mass $m$ and potential energy $U$ is dropped from a height. If the total mechanical energy is constant throughout the fall, then the velocity with which it reaches the ground is:

• A. $\sqrt{\frac{U}{m}}$
• B. $\sqrt{\frac{U}{2m}}$
• C. $\sqrt{\frac{2U}{m}}$
• D. $\sqrt{\frac{U}{m}}$

Answer 1

The correct option is C $\sqrt{\frac{2U}{m}}$

Given: mass of ball $=m$, and potential energy at the dropped height $=U$.
Let the velocity with which the ball reaches the ground be $v$.

When the stationary ball is at some height, all of its mechanical energy exists as potential energy.
But when it reaches the ground, all that potential energy gets converted to kinetic energy.
Thus, the kinetic energy of ball as it hits the ground, $\frac{1}{2}m{v}^2=U$
$\Rightarrow v^2=\frac{2U}{m}\Rightarrow v=\sqrt{\frac{2U}{m}}$
Therefore, the velocity with which the ball reaches the ground is $\sqrt{\frac{2U}{m}}$.