Question 8.7: The ballistic pendulum Now we will look at a classic physics...
The ballistic pendulum
Now we will look at a classic physics lab apparatus for measuring the speed of a projectile. Figure 8.13 shows a simple form of the apparatus, known as a ballistic pendulum, composed of a block and some string. A bullet with mass m_B is fired into a block of wood with mass m_W suspended like a pendulum. The bullet makes a completely inelastic collision with the block, becoming embedded in it. After the impact of the bullet, the block swings up to a maximum height h. Given values of h, m_B, and m_W, how can we find the initial speed υ of the bullet? What becomes of its initial kinetic energy?
SET UP We analyze this event in two stages: first, the embedding of the bullet in the block, and second, the subsequent swinging of the block on its strings. During the first stage, the bullet embeds itself in the block so quickly that the block doesn’t have time to swing appreciably away from its initial position. So, during the impact, the supporting strings remain very nearly vertical, there is negligible external horizontal force acting on the system, and the horizontal component of momentum is conserved. In the second stage, after the collision, the block and bullet move as a unit. The only forces are the weight (a conservative force) and the string tensions (which do no work). As the pendulum swings upward and to the right, mechanical energy is conserved.
To keep the notation simple, we’ll denote the x component of the bullet’s velocity just before impact as υ and the x component of the velocity of block plus bullet just after the impact as V.
SOLVE Figure 8.13 shows the situation just before impact, the situation just after impact, and the block (with the embedded bullet) at the highest point of its path. Conservation of momentum just before impact and just after gives us
m_B\upsilon = (m_B + m_W) V.
\upsilon= \frac{m_B + m_W}{m_B}V.
The kinetic energy of the system just after the collision is K=\frac{1}{2}(m_B +m_W)V^2. The pendulum comes to rest (for an instant) at a height h, where its kinetic energy \frac{1}{2}(m_B +m_W)V^2 has all become potential energy (m_B +m_W)gh ; then it swings back down. Energy conservation gives
\frac{1}{2}(m_B+m_W)V^2 (m_B+m_W)gh .
Assuming that h can be measured, we solve this equation for V:
V=\sqrt{2gh} h (velocity of block and bullet just after impact).
Now we substitute this result into the momentum equation to find υ:
\upsilon= \frac{m_B+m_W}{m_B}\sqrt{2gh} (velocity of bullet just before impact)
By measuring m_B, m_W, and h, we can compute the original velocity υ of the bullet. For example, if m_B = 5.00 g = 0.00500 kg, m_W = 2.00 kg, and h = 3.00 cm =0.0300 m, then
\upsilon =\frac{2.00 kg + 0.00500 kg}{0.00500 kg}\sqrt{2(9.80 m/s^2)(0.0300 m)}
= 307 m/s.
The x component V of velocity of the block just after impact is
V=\sqrt{2 gh}=\sqrt{2(9.80 m/s^2)(0.0300 m)}
= 0.767 m/s.
Once we have the needed velocities, we can compute the kinetic energies just before and just after impact. The total kinetic energy just before impact is K_i = \frac{1}{2}m_B\upsilon^2 = \frac{1}{2} (0.00500 kg)(307 m/s)^2 = 236 J. We find that just after impact, it is K_f = \frac{1}{2} (m_B + m_W)V^2 = \frac{1}{2} (2.005 kg) \times (0.767 m/s)^2 = 0.590 J. Only a small fraction of the initial kinetic energy remains.
REFLECT When an object collides inelastically with a stationary object that has a much larger mass, nearly all of the first object’s kinetic energy is lost. In this problem, the wood splinters, and the bullet and wood become hotter as mechanical energy is converted to internal energy.
Practice Problem: Suppose the mass of the bullet, the mass of the block, and the height of the block’s swing have the same values as above. If the bullet goes all the way through the block and emerges with half its initial velocity, what was its initial speed? Answer: \upsilon =\frac{2m_W}{m_B}\sqrt{2gh} = 615 m/s.