1. **Problem Statement:** A 0.127 mm thick 304L flyer impacts HNS-1 at 2 km/s generating shocks in HNS at 6.182 km/s and in 304L at -5.141 km/s. We need to find the Energy Fluence in HNS and determine if it will detonate. 2. **Given Data:** - Thickness of flyer, $d = 0.127$ mm = $0.0127$ cm - Impact velocity, $U_g = 2$ km/s - Shock velocity in HNS, $U = 6.182$ km/s - Shock velocity in 304L, $U_{304L} = -5.141$ km/s - Density of HNS, $\rho = 1.55$ g/cm$^3$ - $C_0 = 1.0$ km/s, $s = 3.21$ - $E_c = 34$ cal/cm$^2$ 3. **Formulas:** - Particle velocity: $$U = C_0 + s u$$ - Pressure: $$P = \rho C_0 u + \rho s u^2$$ - Energy fluence: $$E = \frac{P^2 t}{\rho U}$$ - Time $t$ is found by $$t = \frac{d}{U_g}$$ 4. **Step 1: Solve for particle velocity $u$ in HNS:** $$U = C_0 + s u \Rightarrow u = \frac{U - C_0}{s} = \frac{6.182 - 1.0}{3.21} = \frac{5.182}{3.21} \approx 1.615 \, \text{km/s}$$ 5. **Step 2: Calculate pressure $P$ in HNS:** $$P = \rho C_0 u + \rho s u^2 = 1.55 \times 1.0 \times 1.615 + 1.55 \times 3.21 \times (1.615)^2$$ Calculate each term: $$1.55 \times 1.615 = 2.503$$ $$1.55 \times 3.21 \times 2.610 = 1.55 \times 3.21 \times 2.610 = 1.55 \times 8.375 = 12.981$$ Sum: $$P = 2.503 + 12.981 = 15.484 \, \text{GPa (approx)}$$ 6. **Step 3: Calculate time $t$ for shock to pass flyer:** $$t = \frac{d}{U_g} = \frac{0.0127}{2} = 0.00635 \, \text{seconds}$$ 7. **Step 4: Calculate Energy Fluence $E$:** $$E = \frac{P^2 t}{\rho U} = \frac{(15.484)^2 \times 0.00635}{1.55 \times 6.182}$$ Calculate numerator: $$15.484^2 = 239.77$$ $$239.77 \times 0.00635 = 1.522$$ Calculate denominator: $$1.55 \times 6.182 = 9.582$$ Energy fluence: $$E = \frac{1.522}{9.582} = 0.159 \, \text{cal/cm}^2$$ 8. **Step 5: Determine detonation:** Given $E_c = 34$ cal/cm$^2$, the calculated energy fluence $0.159$ cal/cm$^2$ is much less than $E_c$, so **the HNS will not detonate** under these conditions. **Final answer:** $$\boxed{E = 0.159 \, \text{cal/cm}^2, \quad \text{No detonation}}$$