Two bodies having masses in the ratio 2:3 fal | vedclass

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(C) The velocity $v$ acquired by a mass $m$ falling freely under gravity from a height $h$ is given by $v = \sqrt{2gh}$.The linear momentum $p$ is def

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$1:2$

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(C) The velocity $v$ acquired by a mass $m$ falling freely under gravity from a height $h$ is given by $v = \sqrt{2gh}$.The linear momentum $p$ is defined as $p = mv$.For the first body:$p_1 = m_1 v_1 = m_1 \sqrt{2gh_1}$For the second body:$p_2 = m_2 v_2 = m_2 \sqrt{2gh_2}$The ratio of their linear momenta is:$\frac{p_1}{p_2} = \frac{m_1 \sqrt{2gh_1}}{m_2 \sqrt{2gh_2}} = \frac{m_1}{m_2} \sqrt{\frac{h_1}{h_2}}$Given $\frac{m_1}{m_2} = \frac{2}{3}$ and $\frac{h_1}{h_2} = \frac{9}{16}$,we substitute these values:$\frac{p_1}{p_2} = \frac{2}{3} 	imes \sqrt{\frac{9}{16}} = \frac{2}{3} 	imes \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$Thus,the ratio is $1:2$.

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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