Falling balls from a building – Rankers Physics
Topic: Work Energy and Power
Subtopic: Kinetic Energy and Momentum

Falling balls from a building

A ball of mass \(2\text{ kg}\) and another ball of mass \(4\text{ kg}\) are dropped together from a \(60\text{ feet}\) tall building. After a fall of \(30\text{ feet}\) each towards earth, their respective kinetic energies will be in the ratio of:

(2004)

\(1 : 4\)
\(1 : 2\)
\(1 : \sqrt{2}\)
\(\sqrt{2} : 1\)

Solution:

When objects are dropped, potential energy is converted to kinetic energy. For a fall of height \(h\), the kinetic energy gained is \(KE = mgh\). For the two balls, \(KE_1 = m_1gh\) and \(KE_2 = m_2gh\). The ratio is \(\frac{KE_1}{KE_2} = \frac{m_1gh}{m_2gh} = \frac{m_1}{m_2} = \frac{2\text{ kg}}{4\text{ kg}} = \frac{1}{2}\). So, the ratio is \(1:2\).

Leave a Reply

Your email address will not be published. Required fields are marked *