Chapter 7: Linear momentum

7.3 Ballistic pendulum

ballistic pendulum is a device that used to be used to determine the velocity of a bullet; it was invented in 1742 by English mathematician Benjamin Robins for this purpose. A simple version consists of a block of wood that hangs on a rod (the mass of the rod is small enough compared to the block to be ignored). You fire a bullet into the block, into which the bullet embeds itself. The block of wood, with the bullet lodged in it, then swings to some height that you can measure.

This device makes an excellent case study in conservation laws: there are two stages to this situation, each using different physical principles. The first stage is an inelastic collision between the bullet and block, in which momentum is conserved. After that, the bullet/block system has some velocity—so it has kinetic energy—that is converted into gravitational potential energy as the pendulum swings up.

Example

Consider a ballistic pendulum where the block of wood has a mass \(m_w = 1.3\ \textrm{kg}\) and is struck by a bullet with a mass of \(m_b = 0.01\ \textrm{kg}\). After the bullet lodges itself into the block, the two swing up to a height of 0.33 m. What was the speed of the bullet just before the collision?

To solve this, we will work backwards, starting with the conservation of energy portion of the problem. The final energy is the gravitational potential energy of the bullet/block combination. The initial energy is the kinetic energy of the bullet/block combination (note that this is after the bullet is in the block). We ignore air resistance, so there is no energy dissipated as “waste.”

\[
\begin{align*}
E_3 &= E_2 \\
(m_w + m_b)gy &= \frac{1}{2}(m_w + m_b)v_{w+b}^2 \\
\hookrightarrow v_{w+b} &= \sqrt{2gh}
\end{align*}
\]

Now, this is the velocity of the wood/bullet system after the collision. We know that in a collision, momentum is conserved. This allows us to determine the velocity of the bullet before it is embedded in the wood:

\[
\begin{align*}
p_2 &= p_1 \\
(m_w + m_b) v_{w+b} &= m_w (0) + m_b v_b \\
(m_w + m_b) v_{w+b} &= m_b v_b \\
\hookrightarrow v_b &= \frac{(m_w + m_b) v_{w+b}}{m_b} \\
&= \frac{(m_w + m_b)\sqrt{2gh}}{m_b} \\
&= 333.2\ \textrm{m/s}
\end{align*}
\]

This same process can apply to more than just pendulums. In the following practice problems, divide the scenario into a number of steps. Identify when momentum is conserved, and where energy is conserved.

“Ballistic pendulum-type” problems are those where you need to consider both conservation of energy and momentum in different phases of a multi-step problem.

Practice