A bullet of mass m_1 having an initial velocity v penetrates a lead ball of mass m_2 suspended on a thread and flies out of it at half speed. How much of the bullet's kinetic energy was converted to heat?

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

    <h3> Statement </h3>
    <p>
        $2.5.36.$ A bullet of mass $m_1$ having an initial velocity $v$ penetrates a lead ball of mass $m_2$ suspended on a thread and flies out of it at half speed. How much of the bullet's kinetic energy was converted to heat?
    </p>

    <h3>Solution</h3>
    <p>

After the collision, the velocity of the lead ball $u$ could be found from the preservation of momentum
$$m_1 v = m_2 u + \frac{m_1v}{2}$$
From here the velocity of the lead ball $u$ is
$$u = v \frac{m_1}{2m_2}$$
Considering the energy $Q$ converted to heat, the conservation of energy could be written as
$$\frac{m_1v^2}{2} = \frac{m_1(v/2)^2}{2} + \frac{m_2u^2}{2} + Q \quad(1)$$
Where the initial kinetic energy
$$E_K = \frac{m_1v^2}{2}\quad(2)$$
From $(1)$, the amount of heat released
$$Q = \frac{m_1v^2}{8} + \frac{m_1^2v^2}{2m_2} - \frac{m_1v^2}{2}\quad(3)$$
Deviding $(2)$ and $(3)$ equations, we obtain
$$\boxed{\frac{E_K}{Q} = \frac{3- {m_1}/{m_2}}{4}}$$

Aliaksandr Melnichenka

    <h4>Answer</h4>
    <p>
        $$Q/K=(3-m_1/m_2)/4$$
    </p>


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