Ballistic pendulum

Question:
A 0.010-kg bullet is fired toward a 2.0-kg pendulum. The bullet is sucked in the pendulum block and both are raised up to 0.20-m height together. What is the initial velocity of the bullet?

Answer:
The masses of the bullet and the block can be denoted as $m$ and $M$, respectively. The initial velocity of the bullet and the final velocity of both the bullet and block are $v$ and $V$. Consider the states (a) through (b) in the figure. According to the conservation of linear momentum, the total initial momentum must be equal to the total final momentum.
$$\begin{equation} mv = (m+M)V \end{equation}$$
Now, consider the states (b) to (c). For those states, we use the conservation of mechanical energy.
$$\begin{equation} \frac{1}{2}(m+M)V^2 = (m+M)gh \end{equation}$$
$v$ and $V$ are unknown. In order to obtain the initial velocity, solve for $V$ first from the equation (1).
$$\begin{equation} V = \frac{mv}{m+M} \end{equation}$$
From equation (2), we have
$$\begin{equation} V = \sqrt{2gh} \end{equation}$$
Therefore,
$$\begin{equation} \frac{mv}{m+M} = \sqrt{2gh} \end{equation}$$
The initial velocity is
$$\begin{equation} v = \frac{m+M}{m}\sqrt{2gh} = \frac{0.010+2.00}{0.010}\sqrt{2\cdot 9.8\cdot 0.20} = 398\quad\mathrm{m/s} \end{equation}$$

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