Worked example 7.5: Ballistic pendulum

Question: A bullet of mass $m=10 {\rm g}$ strikes a pendulum bob of mass $M=1.3 {\rm kg}$ horizontally with speed $v$, and then becomes embedded in the bob. The bob is initially at rest, and is suspended by a stiff rod of length $l=0.6 {\rm m}$ and negligible mass. The bob is free to rotate in the vertical direction. What is the minimum value of $v$ which causes the bob to execute a complete vertical circle? How does the answer change if the bob is suspended from a light flexible rod (of the same length), instead of a stiff rod?

Answer: When the bullet strikes the bob, and then sticks to it, the bullet and bob move off with a velocity $v'$ which is given by momentum conservation:

$$m v = (M+m) v'.$$

Hence,

$$v' = \frac{m v}{M+m}.$$

Consider the case where the bob is suspended by a rigid rod. If the bob and bullet only just manage to execute a vertical loop, then their initial kinetic energy $(1/2) (M+m) {v'}^2$ must only just be sufficient to lift them from the bottom to the top of the loop--a distance $2 l$. Hence, in this critical case, energy conservation yields

$$\frac{1}{2} (M+m) {v'}^2 = (M+m) 2 g l,$$

which implies

$${v'}^2 = 4 g l,$$

or

$$v = \frac{ (M+m) \sqrt{4 g l} }{ m } = \frac{1.31\times\sqrt{4\times 9.81\times 0.6}}{0.01} =635.6 {\rm m/s}.$$

Consider the case where the bob is suspended by a flexible rod. The velocity $v''$ of the bob and bullet at the top of the loop is obtained from energy conservation:

$$\frac{1}{2} (M+m) {v''}^2 = \frac{1}{2} (M+m) {v'}^2 - (M+m) 2 g l.$$

If the bob and bullet only just manage to execute a vertical loop, then the tension in the rod is zero at the top of the loop. Hence, the acceleration due to gravity $g$ must account exactly for the required acceleration ${v''}^2/l$ towards the centre of the loop:

$$\frac{{v''}^2}{l} = g.$$

It follows that, in this critical case,

$${v'}^2 = 5 g l,$$

or

$$v = \frac{(M+m) \sqrt{5 g l} }{ m } = \frac{1.31\times\sqrt{5\times 9.81\times 0.6}}{0.01} =710.7 {\rm m/s}.$$