Body armor. When a high-speed projectile such as a bullet or bomb fragment strikes modern body armor, the fabric of
the armor stops the projectile and prevents penetration by quickly
spreading the projectile's energy over a large area. This spreading
is done by longitudinal and transverse pulses that move radially from the impact point, where the projectile pushes a cone-shaped
dent into the fabric. The longitudinal pulse, racing along the fibers
of the fabric at speed $v_{l}$ ahead of the denting, causes the fibers to thin and stretch, with material flowing radially inward into the
dent. One such radial fiber is shown in Fig. $16-48 a$ . Part of the projectile's energy goes into this motion and stretching. The transverse pulse, moving at a slower speed $v_{r},$ is due to the denting. As the
projectile increases the dent's depth, the dent increases in radius,
causing the material in the fibers to move in the same direction as the projectile (perpendicular to the transverse pulse's direction of
travel). The rest of the projectile's energy goes into this motion. All
the energy that does not eventually go into permanently deforming
the fibers ends up as thermal energy. Figure $16-48 b$ is a graph of speed $v$ versus time $t$ for a bullet of mass 10.2 g fired from a .38 Special revolver directly into body armor. The scales of the vertical and horizontal axes are set by $v_{s}=$
300 $\mathrm{m} / \mathrm{s}$ and $t_{s}=40.0$ \mus. Take $v_{l}=2000 \mathrm{m} / \mathrm{s},$ and assume that the half-angle $\theta$ of the conical dent is $60^{\circ} .$ At the end of the collision,
what are the radii of (a) the thinned region and (b) the dent (assuming that the person wearing the armor remains stationary)?