Q100 / 200 · NEET 2004 · Physics

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:

  • A
    $\sqrt{2}+1$
  • B
    $1: 4$
  • Correct answer
    C
    $1: 2$
  • D
    $1: \sqrt{2}$

Solution

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. $$ \begin{aligned} & \therefore \quad v^2=2 g h \text { is same for both } \\ & \text { bodies } \\ & \therefore \quad \frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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