Question

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

• A. $\sqrt 2 :1$
• B. $1 : 4$
• C. $1 : 2$
• D. $1 : \sqrt 2$

Answer 1

The Correct Option is C

To determine the ratio of the kinetic energies of the two balls after falling 30 feet, we first need to understand the relationship between gravitational potential energy and kinetic energy. When an object falls under gravity, its potential energy is converted into kinetic energy.

The potential energy lost by an object of mass \(m\) falling from height \(h\) is given by: 

\(PE = mgh\)

Where:

• \(m\) is the mass of the object,
• \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity), and
• \(h\) is the height of the fall.

Assuming no energy is lost to air resistance, the potential energy lost is equal to the kinetic energy gained. Therefore, the kinetic energy (\(KE\)) is:

\(\text{KE} = \text{PE} = mgh\)

Given:

• Mass of first ball, \(m_1 = 2 \, \text{kg}\)
• Mass of second ball, \(m_2 = 4 \, \text{kg}\)
• Height fallen, \(h = 30 \, \text{ft} = 30 \times 0.3048 \, \text{m} = 9.14 \, \text{m}\) (since 1 ft = 0.3048 m)

Now, calculate the kinetic energy for each ball after falling 30 feet:

1. First Ball (\(m_1\)):
   • \(\text{KE}_1 = m_1gh = 2 \times 9.8 \times 9.14 = 179.064 \, \text{Joules}\)
2. Second Ball (\(m_2\)):
   • \(\text{KE}_2 = m_2gh = 4 \times 9.8 \times 9.14 = 358.128 \, \text{Joules}\)

Now, find the ratio of kinetic energies:

\(\frac{\text{KE}_1}{\text{KE}_2} = \frac{179.064}{358.128} = \frac{1}{2}\)

Thus, the ratio of the kinetic energies of the two balls is \(1:2\).

The correct answer is therefore: \(1:2\)