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Two bodies having masses in the ratio $2: 3$ fall freely under gravity from heights which are in the ratio $9: 16$. The ratio of their linear momenta on touching the ground is
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Verified Answer
The correct answer is:
$1: 2$
The velocity acquired by a mass $m$ falling freely under the influence of gravity from the heigh $h$ is
$$
v=\sqrt{2 g h}
$$
Momentum of first body, $p_1=m_1 v_1$
$$
\Rightarrow \quad=m_1 \sqrt{2 g h_1}
$$
Momentum of second body $p_2=m_2 v_2$
$$
\Rightarrow \quad=m_2 \sqrt{2 g h_2}
$$
The ratio of their linear momenta on touching the ground is
$$
\begin{aligned}
\frac{p_1}{p_2} & =\frac{m_1 \sqrt{2 g h_1}}{m_2 \sqrt{2 g h_2}} \\
& =\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2} \text { or } 1: 2
\end{aligned}
$$
$$
v=\sqrt{2 g h}
$$
Momentum of first body, $p_1=m_1 v_1$
$$
\Rightarrow \quad=m_1 \sqrt{2 g h_1}
$$
Momentum of second body $p_2=m_2 v_2$
$$
\Rightarrow \quad=m_2 \sqrt{2 g h_2}
$$
The ratio of their linear momenta on touching the ground is
$$
\begin{aligned}
\frac{p_1}{p_2} & =\frac{m_1 \sqrt{2 g h_1}}{m_2 \sqrt{2 g h_2}} \\
& =\frac{2}{3} \times \sqrt{\frac{9}{16}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2} \text { or } 1: 2
\end{aligned}
$$
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