Question 8.8: The ballistic pendulum Figure 8.19 shows a ballistic pendulu......
The ballistic pendulum
Figure 8.19 shows a ballistic pendulum, a simple system for measuring the speed of a bullet. A bullet of mass m_{\mathrm{B}} makes a completely inelastic collision with a block of wood of mass m_{\mathrm{W}}, which is suspended like a pendulum. After the impact, the block swings up to a maximum height h. In terms of h, m_{\mathrm{B}}, and m_{\mathrm{W}}, what is the initial speed v_{1} of the bullet?
IDENTIFY We’ll analyze this event in two stages: (1) the bullet embeds itself in the block, and (2) the block swings upward. The first stage happens so quickly that the block does not move appreciably. The supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet-block system, and the horizontal component of momentum is conserved. Total mechanical energy is not conserved during this stage, however, because a nonconservative force does work (the force of friction between bullet and block).
In the second stage, the block and bullet move together. The only forces acting on this system are gravity (a conservative force) and the string tensions (which do no work). Thus, as the block swings, total mechanical energy is conserved. Momentum is not conserved during this stage, however, because there is a net external force (the forces of gravity and string tension don’t cancel when the strings are inclined).
SET UP We take the positive x-axis to the right and the positive y-axis upward. Our target variable is v_{1}. Another unknown quantity is the speed v_{2} of the system just after the collision. We’ll use momentum conservation in the first stage to relate v_{1} to v_{2}, and we’ll use energy conservation in the second stage to relate v_{2} to h.
EXECUTE In the first stage, all velocities are in the +x-direction. Momentum conservation gives
\begin{aligned} m_{\mathrm{B}} v_{1} & =\left(m_{\mathrm{B}}+m_{\mathrm{W}}\right) v_{2} \\ v_{1} & =\frac{m_{\mathrm{B}}+m_{\mathrm{W}}}{m_{\mathrm{B}}} v_{2} \end{aligned}
At the beginning of the second stage, the system has kinetic energy K=\frac{1}{2}\left(m_{\mathrm{B}}+m_{\mathrm{W}}\right) v_{2}^{2}. The system swings up and comes to rest for an instant at a height h, where its kinetic energy is zero and the potential energy is \left(m_{\mathrm{B}}+m_{\mathrm{W}}\right) g h; it then swings back down. Energy conservation gives
\begin{aligned} \frac{1}{2}\left(m_{\mathrm{B}}+m_{\mathrm{W}}\right) v_{2}^{2} & =\left(m_{\mathrm{B}}+m_{\mathrm{W}}\right) g h \\ v_{2} & =\sqrt{2 g h} \end{aligned}
We substitute this expression for v_{2} into the momentum equation:
v_{1}=\frac{m_{\mathrm{B}}+m_{\mathrm{W}}}{m_{\mathrm{B}}} \sqrt{2 g h}
EVALUATE Let’s plug in realistic numbers: m_{\mathrm{B}}=5.00 \mathrm{~g}=0.00500 \mathrm{~kg}, m_{\mathrm{W}}=2.00 \mathrm{~kg}, and h=3.00 \mathrm{~cm}=0.0300 \mathrm{~m} :
\begin{aligned} & v_{1}=\frac{0.00500 \mathrm{~kg}+2.00 \mathrm{~kg}}{0.00500 \mathrm{~kg}} \sqrt{2\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(0.0300 \mathrm{~m})}=307 \mathrm{~m} / \mathrm{s} \\ & v_{2}=\sqrt{2 g h}=\sqrt{2\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(0.0300 \mathrm{~m})}=0.767 \mathrm{~m} / \mathrm{s} \end{aligned}
The speed v_{2} of the block after impact is much lower than the initial speed v_{1} of the bullet. The kinetic energy of the bullet before impact is \frac{1}{2}(0.00500 \mathrm{~kg})(307 \mathrm{~m} / \mathrm{s})^{2}=236 \mathrm{~J}. Just after impact the kinetic energy of the system is \frac{1}{2}(2.005 \mathrm{~kg})(0.767 \mathrm{~m} / \mathrm{s})^{2}=0.589 \mathrm{~J}. Nearly all the kinetic energy disappears as the wood splinters and the bullet and block become warmer.
KEYCONCEPT Conservation of momentum holds true only when the net external force is zero. In some situations momentum is conserved during part of the motion (such as during a collision) but not during other parts.