Question

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 radiallyfrom 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 speedahead 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. 16-48a. Part of the projectile’s energy goes into this motion and stretching. The transverse pulse, moving at a slower speed${\mathit{v}}_{\mathbf{t}}$, 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 16-48bis a graph of speed vversus time tfor a bullet of mass 10.2g fired from a .38 Special revolver directly into body armor. The scales of the vertical and horizontal axes are set by${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{300}\mathbf{m}\mathbf{/}\mathbf{s}$and${\mathbf{t}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\mu s}$. Take${\mathbf{v}}_{\mathbf{I}}\mathbf{=}\mathbf{2000}\mathbf{m}\mathbf{/}\mathbf{s}$, and assume that the half-angle $\mathit{\theta }$of the conical dent is$\mathbf{60}\mathbf{°}$. 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)?

Short answer

a) The radii of the thinned region at the end of the collision is$8.0\times {10}^{-2}m$

b) The radii of the dent at the end of the collision is$1.0\times {10}^{-2}\mathrm{m}$

Step-by-step solution

Identification of given data

i) Mass of bullet,$\mathrm{m}=10.2\mathrm{g}\mathrm{or}10.2\times {10}^{-3}\mathrm{kg}$

ii) Velocity,$v_I=2000\mathrm{m}/\mathrm{s}$

iii) Velocity,$v_s=300\mathrm{m}/\mathrm{s}$

iv) Time,${\mathrm{t}}_{\mathrm{s}}=40.0\mathrm{\mu s}\mathrm{or}40.0\times {10}^{-6}\mathrm{s}$

v) Half-angle of the conical dent,$\theta =60°$

Understanding the concept of kinematic equations

The longitudinal speed determines the distance from which we can find the radii of the thinned region at the end of the collision. Using the first and third kinematic equations, we can find the distance. From this distance, we can find the radii of the dent at the end of the collision.

Formula:

The first kinematic equation,$v_f=v_i+at$ …(i)

The third kinematic equation of motion,$v_f^2=v_i^2+2as$ …(ii)

The distance traveled by a body, d=vt …(iii)

(a) Determining the radii in the thinned region at the end of collision

Using equation (iii), the longitudinal distance is given as:

$$\begin{gathered} d=v_{i}t \\ =\left(2000\frac{m}{s}\right)\left(40.0\times {10}^{-6}s\right) \\ =8.0\times 0^{-2}\mathrm{m} \end{gathered}$$

$$\begin{gathered} radius=d \\ =8.0\times {10}^{-2}\mathrm{m} \end{gathered}$$

Therefore, the radii of the thinned region at the end of the collision is$8.0\times {10}^{-2}\mathrm{m}$

(b) Determining the radii of the dent

Using equation (i), we get the acceleration of the body in motion is given as:

$$\begin{gathered} a=\frac{v_f-v_i}{t} \\ =\frac{\left(300{\displaystyle \frac{m}{s}}\right)-\left(0{\displaystyle \frac{m}{s}}\right)}{40.0\times {10}^{-6}s} \\ =7.5\times {10}^6\frac{m}{s^2} \end{gathered}$$

Using equation (ii), we get the distance at which collision ended;

$$\begin{gathered} \mathrm{s}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2}{2\mathrm{a}} \\ =\frac{{\left(300{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)}^2-{\left(0{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)}^2}{2\times \left(7.5\times {10}^6{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}\right)} \\ =6.0\times {10}^{-3}\mathrm{m} \end{gathered}$$

Therefore, radius of the dent is given as:

$$\begin{gathered} r=s\tan \theta \\ =\left(6.0\times {10}^{-3}m\right)\tan 60° \\ =1.0\times {10}^2\mathrm{m} \end{gathered}$$

Therefore, the radii of the dent at the end of the collision is$1.0\times {10}^2\mathrm{m}$