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A ball of mass $2 \mathrm{~kg}$ and another of mass $4 \mathrm{~kg}$ 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:
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Verified Answer
The correct answer is:
$1: 2$
Let $E_1$ and $E_2$ be the K.E. of two bodies.
Then $\quad E_1=\frac{1}{2} m_1 v_1^2$
and $E_2=\frac{1}{2} m_2 v_2{ }^2$
$$
\therefore \quad \frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v^2}=\frac{m_1 v_1^2}{m_2 v_2^2}
$$
The initial velocity of both the bodies are zero.
$$
\begin{aligned}
& \therefore \quad v^2=2 g h \text { is same for both } \\
& \text { bodies } \\
& \therefore \quad \frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$$
Then $\quad E_1=\frac{1}{2} m_1 v_1^2$
and $E_2=\frac{1}{2} m_2 v_2{ }^2$
$$
\therefore \quad \frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v^2}=\frac{m_1 v_1^2}{m_2 v_2^2}
$$
The initial velocity of both the bodies are zero.
$$
\begin{aligned}
& \therefore \quad v^2=2 g h \text { is same for both } \\
& \text { bodies } \\
& \therefore \quad \frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$$
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