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_B makes a completely inelastic collision with a block of wood of mass m_W, which is suspended like a pendulum. After the impact, the block swings up to a maximum m height h. In terms of h, m_B, and m_W, what is the initial speed v_1of 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. 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, 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 (mB + mW)gh; 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}Let’s plug in the realistic numbers m_B = 5.00 g = 0.00500 kg, m_W = 2.00 kg, and h = 3.00 cm = 0.0300m:
\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}= 0.590 J. Nearly all the kinetic energy disappears as the wood splinters and the bullet and block become warmer.