Worked Exercise — the Bulletproof Vest

Problem

A bullet of mass $m_1 = 20\;\text{g}$ travels at $v_0 = 400\;\text{m/s}$ when it strikes a soldier’s bulletproof vest (mass $m_2 = 80\;\text{kg}$, initially at rest). The bullet comes to a complete stop after penetrating the vest by $\Delta s = 5\;\text{cm}$. Calculate: (a) the acceleration of the bullet; (b) the stopping time; (c) the force exerted by the vest on the bullet; (d) the force (equal and opposite, by the third law) exerted by the bullet on the soldier; (e) how far the soldier recoils in the same time interval.

Solution

(a) Acceleration of the bullet. We use the kinematic relation without time, $v^2 - v_0^2 = 2a\,\Delta s$, with $v = 0$ and $\Delta s = 0{,}05\;\text{m}$:

$a = \frac{0 - v_0^2}{2\,\Delta s} = \frac{-400^2}{2 \cdot 0{,}05}$

$\ev{a = -1{,}6 \cdot 10^6\;\text{m/s}^2}$

A deceleration of about $160\,000\,g$, roughly one hundred thousand times the acceleration of gravity on Earth.

(b) Stopping time. From the definition of acceleration, $v = v_0 + a\,\Delta t$ with $v = 0$:

$\Delta t = \frac{0 - v_0}{a} = \frac{-400}{-1{,}6 \cdot 10^6}$

$\ev{\Delta t = 2{,}5 \cdot 10^{-4}\;\text{s}}$

A quarter of a millisecond.

(c) Force on the bullet. By Newton’s second law, $F = m_1\,|a|$ with $m_1 = 0{,}020\;\text{kg}$:

$|F| = m_1\,|a| = 0{,}020 \cdot 1{,}6 \cdot 10^6$

$\ev{|F| = 32\,000\;\text{N}}$

(d) Force on the soldier (third law). By the action–reaction principle, the bullet exerts on the soldier an equal and opposite force, $|F| = 32\,000\;\text{N} \approx 3{,}3$ tonnes-force. This produces an acceleration of the soldier:

$a_2 = \frac{F}{m_2} = \frac{32\,000}{80}$

$\ev{a_2 = 400\;\text{m/s}^2}$

(e) Recoil of the soldier. Over the same interval $\Delta t = 2{,}5 \cdot 10^{-4}\;\text{s}$, starting from rest:

$\Delta s_2 = \tfrac{1}{2}\,a_2\,(\Delta t)^2 = \tfrac{1}{2} \cdot 400 \cdot (2{,}5 \cdot 10^{-4})^2$

$\ev{\Delta s_2 \approx 1{,}25 \cdot 10^{-5}\;\text{m} = 12{,}5\;\mu\text{m}}$

The soldier recoils by just a few tens of micrometres during the impact, but the force of $32\,000\;\text{N}$ leaves its mark for that instant: this is precisely why people hit on a bulletproof vest are thrown violently backwards, even though the bullet does not penetrate them. Newton’s third law gives no quarter.

Links

Topics: Dynamics Concepts: Newton’s third law · Newton’s second law · Uniformly accelerated motion Objects: Projectile