Collision: Practice Problem & Solution
A ball is thrown vertically downwards from a height of $20\text{ m}$ with an initial velocity $u_0$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $u_0$ is: (Take $g = 10\text{ ms}^{-2}$) (2015 Re)
Solution Explained:
To solve this problem, we apply the core principles of Collision. Understanding the underlying formula is key to arriving at the correct answer below:
The velocity just before impact is $v^2 = u_0^2 + 2gh$. Since it loses $50\%$ energy and reaches the same height $h$, the post-collision kinetic energy satisfies $mgh = \frac{1}{2}(\frac{1}{2}mv^2)$, leading to $v^2 = 4gh$. Solving gives $u_0 = \sqrt{2gh} = 20\text{ m/s}$.
Leave a Reply