Two bodies of mass m_1 and m_2 are attached to threads of the same length with a common suspension point and are deflected-one to the left, the other to the right - at the same angle. The bodies are released simultaneously. When they hit each other, they stick together. Determine the ratio of the height to which the bodies rise after sticking together to the height from which they began their downward movement.

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    <h3 id="back-link"><a href="/#2.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.5.35.$ Two bodies of mass $m_1$ and $m_2$ are attached to threads of the same length with a common suspension point and are deflected-one to the left, the other to the right - at the same angle. The bodies are released simultaneously. When they hit each other, they stick together. Determine the ratio of the height to which the bodies rise after sticking together to the height from which they began their downward movement.
    </p>

    <h3>Solution</h3>
    <p>

Considering the fact that two bodies would stick together, the conservation of momentum right before and after the collision
$$|m_1v-m_2v| = (m_1+m_2)u$$
Where $v$ is a velocity of two bodies right before the collision. It could be found from the conservation of energy
$$\frac{m_1v^2}{2}=m_1gh_0 \Leftrightarrow v_1=v_2=\sqrt{2gh_0}$$
From where
$$u=\frac{|m_1-m_2|}{m_1+m_2}\cdot\sqrt{2gh_0}$$
After the sticked bodies with a total mass $m_1+m_2$ will rise to a height of $h$ found, alternatively, using energy conservation
$$h=\frac{u^2}{2g}=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\cdot h_0$$
Dividing by $h_0$, we obtain
$$\boxed{\frac{h}{h_0}=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2}$$

Lutfulloyev Shukurullo

    <h4>Answer</h4>
    <p>
        $$h/h_0=[(m_1-m_2)/(m_1+m_2)]^2$$
    </p>


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